Samacheer Kalvi Class 11 Chemistry Solutions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Textual Questions:

 

Question 1. 40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature
(a) 40 ml CO2
(b) 40 ml CO2 gas and 80 ml H2O gas
(c) 60 ml CO2 gas and 60 ml H2O gas
(d) 120 ml CO2 gas
Answer: (a) 40 ml CO2
Solution:
\( \text{CH}_4(\text{g}) + 2\text{O}_2 \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O} (1) \)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.
In simple words: When methane burns with oxygen, it makes carbon dioxide and water. Because the mixture is cooled, the water turns into a liquid, so only the carbon dioxide gas is left.

🎯 Exam Tip: Remember that water becomes liquid at room temperature, which affects the final volume of gas collected.

 

Question 2. The following isotopic composition \(^{200}\text{X} = 90\%\), \(^{199}\text{X} = 8\%\) and \(^{202}\text{X} = 2\%\). The weighted average atomic mass of the element X is closest to
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
Answer: (d) 200 u
Solution:
\( \text{X} = \frac{(200 \times 90) + (199 \times 8) + (202 \times 2)}{100} = \frac{18000 + 1592 + 404}{100} = \frac{19996}{100} = 199.96 = 200 \text{ u} \)
In simple words: To find the average atomic mass, multiply the mass of each isotope by its percentage, add them up, and then divide by 100. This gives a weighted average that is closest to 200 u.

🎯 Exam Tip: Always use the exact isotopic mass and abundance in calculations, then round the final answer to the nearest whole number if specified.

 

Question. Assertion: Two mole of glucose contains \(12.044 \times 10^{23}\) molecules of glucose
Reason: Total number of entities present in one mole of any substance is equal to \(6.02 \times 10^{22}\)
(a) both assertion and reason are true and the reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer: (c) assertion is true but reason is false
Solution:
Correct reason:
Total number of entities present in one mole of any substance is equal to \(6.022 \times 10^{23}\).
In simple words: The first statement (assertion) is true because two moles have double the number of molecules in one mole. However, the second statement (reason) is false because Avogadro's number, which tells us how many particles are in one mole, is \(6.022 \times 10^{23}\), not \(6.02 \times 10^{22}\).

🎯 Exam Tip: Remember Avogadro's number exactly: it's \(6.022 \times 10^{23}\) for one mole of any substance. This value is crucial for many calculations.

 

Question 4. Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) both carbon and oxygen
(d) neither carbon nor oxygen
Answer: (b) Oxygen
Solution:
Reaction 1:
\( 2\text{C} + \text{O}_2 \rightarrow 2\text{CO} \)
\( 2 \times 12\text{g} \) carbon combines with \( 32\text{g} \) of oxygen. Hence, Equivalent mass of carbon \( = \frac{2 \times 12}{32} \times 8 = 6 \)
Reaction 2:
\( \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \)
\( 12\text{ g} \) carbon combines with \( 32\text{ g} \) of oxygen. Hence, Equivalent mass of carbon \( = \frac{12}{32} \times 8 = 3 \)
In simple words: Oxygen's equivalent mass stays the same no matter which carbon oxide is formed, because it always combines with other elements in a consistent ratio. Carbon's equivalent mass changes depending on whether it forms carbon monoxide or carbon dioxide.

🎯 Exam Tip: The equivalent mass of an element depends on its valency or combining capacity in a specific reaction. For elements like oxygen, which typically form stable oxides, the equivalent mass often remains constant.

 

Question 5. The equivalent mass of a trivalent metal element is \(9\text{ g eq}^{-1}\) the molar mass of its anhydrous oxide is
(a) 102 g
(b) 27 g
(c) 270 g
(d) 78 g
Answer: (a) 102 g
Solution:
Let the trivalent metal be \( \text{M}^3 \)
Equivalent mass \( = \text{mass of the metal / 3 eq} \)
\( 9\text{ g eq}^{-1} = \text{mass of the metal / 3 eq} \)
Mass of the metal \( = 27\text{ g} \)
Oxide formed \( \text{M}_2\text{O}_3 \)
Mass of the oxide \( = (2 \times 27) + (3 \times 16) = 54 + 48 = 102\text{ g} \)
In simple words: Since the metal has an equivalent mass of \(9\text{ g}\) and is trivalent, its atomic mass is \(27\text{ g}\). When it forms an oxide (\(\text{M}_2\text{O}_3\)), we add the masses of two metal atoms and three oxygen atoms to get the molar mass of the oxide, which is \(102\text{ g}\).

🎯 Exam Tip: Remember that equivalent mass is atomic mass divided by valency. Use this to find the atomic mass of the metal, then combine it with oxygen's mass to find the oxide's molar mass.

 

Question 6. The number of water molecules in a drop of water weighing 0.018 g is
(a) \(6.022 \times 10^{26}\)
(b) \(6.022 \times 10^{23}\)
(c) \(6.022 \times 10^{20}\)
(d) \(9.9 \times 10^{22}\)
Answer: (c) \(6.022 \times 10^{20}\)
Solution:
Weight of the water drop \( = 0.018\text{ g} \)
No. of moles of water in the drop \( = \text{Mass of water / molar mass} = 0.018/18 = 10^{-3}\text{ mole} \)
No of water molecules present in \( 1\text{ mole} \) of water \( = 6.022 \times 10^{23} \)
No. water molecules in one drop of water (\(10^{-3}\text{ mole}\)) \( = 6.022 \times 10^{23} \times 10^{-3} \)
\( = 6.022 \times 10^{20} \)
In simple words: First, find how many moles are in the water drop by dividing its weight by the molar mass of water. Then, multiply this number of moles by Avogadro's number (which is \(6.022 \times 10^{23}\) molecules per mole) to find the total number of water molecules.

🎯 Exam Tip: Always convert mass to moles first, then use Avogadro's number to find the number of molecules or atoms. Molar mass of water is \(18\text{ g/mol}\).

 

Question 7. 1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is
(b) 4.4%
(c) 16%
(d) 8.4%
Answer: (c) 16%
Solution:
\( \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2\uparrow \)
Molar mass of \( \text{MgCO}_3 \):
\( (1 \times 24) + (1 \times 12) + (3 \times 16) = 84\text{ g} \)
Molar mass of \( \text{CO}_2 \): \( (1 \times 12) + (2 \times 16) = 44\text{g} \)
\( 100\% \) pure \( 84\text{ g MgCO}_3 \) on heating gives \( 44\text{ g CO}_2 \)
Given that \( 1\text{ g} \) of \( \text{MgCO}_3 \) on heating gives \( 0.44\text{ g} \) of \( \text{CO}_2 \)
Therefore, \( 84\text{ g MgCO}_3 \) sample on heating gives \( \frac{44\text{ g CO}_2}{84\text{ g MgCO}_3} \times 1\text{ g MgCO}_3 = 0.44\text{ g CO}_2 \). For a pure sample, \(1\text{ g}\) of \( \text{MgCO}_3 \) would yield \( \frac{44}{84} \times 1 \approx 0.5238\text{ g} \) of \( \text{CO}_2 \). Since only \(0.44\text{ g}\) is produced, the sample is impure.
From the given information, a \(1\text{ g}\) sample yields \(0.44\text{ g}\) of \( \text{CO}_2 \). To find the mass of pure \( \text{MgCO}_3 \) that produced \(0.44\text{ g}\) of \( \text{CO}_2 \):
Mass of pure \( \text{MgCO}_3 = \frac{84\text{ g MgCO}_3}{44\text{ g CO}_2} \times 0.44\text{ g CO}_2 = 0.84\text{ g} \)
Percentage of purity of the sample \( = \frac{\text{Mass of pure MgCO}_3}{\text{Total mass of sample}} \times 100\% = \frac{0.84\text{ g}}{1\text{ g}} \times 100\% = 84\% \)
Percentage of impurity \( = 100\% - 84\% = 16\% \)
In simple words: When magnesium carbonate breaks down, it makes carbon dioxide. We find out how much pure magnesium carbonate would have made the \(0.44\text{ g}\) of carbon dioxide. Since the total sample was \(1\text{ g}\), the difference between the pure amount and \(1\text{ g}\) tells us the impurities. This works out to \(16\%\) impurity.

🎯 Exam Tip: For percentage purity problems, first use stoichiometry to determine the theoretical yield from a pure sample, then compare it to the actual yield to find the pure substance amount, and finally calculate the impurity percentage.

 

Question 8. When \(6.3\text{ g}\) of sodium bicarbonate is added to \(30\text{ g}\) of acetic acid solution, the residual solution is found to weigh \(33\text{ g}\). The number of moles of carbon dioxide released in the reaction is
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3
Answer: (c) 0.075
Solution:
\( \text{NaHCO}_3 + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \)
According to the law of conservation of mass:
Total mass of reactants \( = \text{Mass of NaHCO}_3 + \text{Mass of CH}_3\text{COOH solution} \)
\( = 6.3\text{ g} + 30\text{ g} = 36.3\text{ g} \)
Total mass of products \( = \text{Mass of residual solution} + \text{Mass of CO}_2 \)
Let \( x \) be the mass of \( \text{CO}_2 \) released.
\( 36.3\text{ g} = 33\text{ g} + x \)
The amount of \( \text{CO}_2 \) released, \( x = 36.3\text{ g} - 33\text{ g} = 3.3\text{ g} \)
Now, calculate the moles of \( \text{CO}_2 \). Molar mass of \( \text{CO}_2 = 12 + (2 \times 16) = 44\text{ g/mol} \)
No. of moles of \( \text{CO}_2 \) released \( = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} = \frac{3.3\text{ g}}{44\text{ g/mol}} = 0.075\text{ mol} \)
In simple words: First, use the law of conservation of mass to find the weight of carbon dioxide gas that escaped during the reaction. Then, divide the weight of the carbon dioxide by its molar mass to get the number of moles.

🎯 Exam Tip: Always apply the law of conservation of mass carefully, ensuring that the mass of any gas released is accounted for when comparing reactant and product masses.

 

Question 9. When \(22.4\text{ litres}\) of \( \text{H}_2 (\text{g}) \) is mixed with \(11.2\text{ litres}\) of \( \text{Cl}_2 (\text{g}) \), each at \(273\text{ K}\) at \(1\text{ atm}\) the moles of \( \text{HCl} (\text{g}) \), formed is equal to
(a) 2 moles of \( \text{HCl} (\text{g}) \)
(b) 0.5 moles of \( \text{HCl} (\text{g}) \)
(c) 1.5 moles of \( \text{HCl} (\text{g}) \)
(d) 1 moles of \( \text{HCl} (\text{g}) \)
Answer: (d) 1 moles of HCl (g)
Solution:
\( \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{HCl}(\text{g}) \)

Amount of \( \text{HCl} \) formed \( = 1\text{ mol} \)
In simple words: Since \(22.4\text{ L}\) of gas at standard conditions is 1 mole, \(22.4\text{ L}\) of \( \text{H}_2 \) is 1 mole and \(11.2\text{ L}\) of \( \text{Cl}_2 \) is 0.5 moles. The reaction shows that 1 mole of \( \text{H}_2 \) reacts with 1 mole of \( \text{Cl}_2 \) to make 2 moles of \( \text{HCl} \). Because \( \text{Cl}_2 \) is the limiting reagent (we have less of it), only 0.5 moles of \( \text{Cl}_2 \) will react with 0.5 moles of \( \text{H}_2 \) to produce 1 mole of \( \text{HCl} \).

🎯 Exam Tip: Always identify the limiting reagent first in stoichiometry problems involving multiple reactants; the amount of product formed is determined by the reactant that runs out first.

 

Question 10. Hot concentrated sulphuric acid is a moderately strong oxidising agent. Which of the following reactions does not show oxidising behaviour?
(a) \( \text{Cu} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{SO}_2 +2\text{H}_2\text{O} \)
(b) \( \text{C} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CO}_2 + 2\text{SO}_2 +2\text{H}_2\text{O} \)
(c) \( \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl} \)
(d) none of the options
Answer: (c) \( \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl} \)
Solution:
In reaction (c), \( \text{H}_2\text{SO}_4 \) acts as an acid in a double displacement reaction, not as an oxidizing agent. The oxidation states of the elements remain unchanged.
\( \text{Ba}^{\text{+2}}\text{Cl}_{2}^{\text{-1}} + \text{H}_{2}^{\text{+1}}\text{S}^{\text{+6}}\text{O}_{4}^{\text{-2}} \rightarrow \text{Ba}^{\text{+2}}\text{S}^{\text{+6}}\text{O}_{4}^{\text{-2}} + 2\text{H}^{\text{+1}}\text{Cl}^{\text{-1}} \)
In simple words: Sulphuric acid can sometimes oxidize other substances, meaning it takes electrons from them. In the given options, only reaction (c) is a simple swap of ions (double displacement) where no element changes its oxidation state, so sulphuric acid is not acting as an oxidizer here.

🎯 Exam Tip: To identify if a reaction shows oxidizing behavior, look for changes in the oxidation states of the elements. If a reactant's oxidation state increases, it has been oxidized, and the other reactant is the oxidizing agent.

 

Question 11. Choose the disproportionation reaction among the following redox reactions.
(a) \( 3\text{Mg}(\text{s}) + \text{N}_2(\text{g}) \rightarrow \text{Mg}_3\text{N}_2(\text{s}) \)
(b) \( \text{P}_4(\text{s}) + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{PH}_3(\text{g}) + 3\text{NaH}_2\text{PO}_2(\text{aq}) \)
(c) \( \text{Cl}_2(\text{g}) + 2\text{KI} (\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{I}_2(\text{s}) \)
(d) \( \text{Cr}_2\text{O}_3(\text{s}) + 2\text{Al}(\text{s}) \rightarrow \text{Al}_2\text{O}_3(\text{s}) + 2\text{Cr}(\text{s}) \)
Answer: (b) \( \text{P}_4(\text{s}) + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{PH}_3(\text{g}) + 3\text{NaH}_2\text{PO}_2(\text{aq}) \)
Solution:
A disproportionation reaction is a type of redox reaction where an element in one oxidation state is simultaneously oxidized and reduced.
In option (b), phosphorus (P) has an oxidation state of 0 in \( \text{P}_4 \). In \( \text{PH}_3 \), the oxidation state of P is -3. In \( \text{NaH}_2\text{PO}_2 \), the oxidation state of P is +1.
So, phosphorus is reduced from 0 to -3 and oxidized from 0 to +1.
\( \text{P}_{4}^{\text{0}} + 3\text{NaOH} + 3\text{H}_2\text{O} \rightarrow \text{P}^{\text{-3}}\text{H}_3(\text{g}) + 3\text{NaH}_2\text{P}^{\text{+1}}\text{O}_2(\text{aq}) \)
In simple words: A disproportionation reaction is special because one element changes its oxidation number in two ways: it gets both oxidized (loses electrons) and reduced (gains electrons) in the same reaction. In option (b), phosphorus starts at 0 and ends up as -3 and +1, showing both processes.

🎯 Exam Tip: To spot a disproportionation reaction, look for a single element on the reactant side that appears in two different oxidation states on the product side (one higher, one lower than its initial state).

 

Question 12. The equivalent mass of potassium permanganate in an alkaline medium is \( \text{MnO}_4 + 2\text{H}_2\text{O} + 3\text{e} \rightarrow \text{MnO}_2 + 4\text{OH}^{-} \)
(a) 31.6
(b) 52.7
Answer: (b) 52.7
Solution:
The reduction reaction of the oxidising agent (\( \text{MnO}_4^{-} \)) involves the gain of 3 electrons.
The oxidation state of Mn in \( \text{MnO}_4^{-} \) is \(+7\). In \( \text{MnO}_2 \), the oxidation state of Mn is \(+4\). The change in oxidation state is \(7-4 = 3\). So, it gains 3 electrons.
Hence the equivalent mass \( = \frac{\text{Molar mass of KMnO}_4}{\text{Number of electrons gained}} = \frac{158.1}{3} = 52.7 \)
In simple words: To find the equivalent mass of potassium permanganate in an alkaline solution, we look at how many electrons it gains during the reaction. In this case, it gains 3 electrons. So, we divide its molar mass by 3 to get the equivalent mass.

🎯 Exam Tip: The equivalent mass of an oxidizing agent is calculated by dividing its molar mass by the number of electrons gained in the reaction. Pay attention to the oxidation state change of the central metal ion.

 

Question 13. Which one of the following represents 180g of water?
(a) 5 Moles of water
(b) 90 moles of water
(c) \( \frac{6.022 \times 10^{23}}{180} \)
(d) \( \frac{6.022 \times 10^{23}}{1.7} \)
Answer: (c) 10 moles of water / \(6.022 \times 10^{24}\) water molecules
Solution:
Molar mass of water (\( \text{H}_2\text{O} \)) \( = (2 \times 1) + 16 = 18\text{ g/mol} \)
No. of moles of water present in \( 180\text{ g} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{180\text{ g}}{18\text{ g mol}^{-1}} = 10\text{ moles} \)
One mole of water contains \( = 6.022 \times 10^{23} \) water molecules (Avogadro's number)
10 moles of water contains \( = 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24} \) water molecules
Since option (c) and (d) are fractions, and the result is 10 moles, option (c) is likely a typo in question text/options. The correct numerical value for 10 moles would be \(6.022 \times 10^{24}\) molecules.
Comparing with the options, (a) 5 moles is incorrect, (b) 90 moles is incorrect. If the choices were meant to be number of molecules, \(6.022 \times 10^{24}\) would be the answer. If the choices are moles, then 10 moles is the correct answer. The options as given in the source are problematic. Assuming the question intended to ask for the number of moles or total molecules, and given the structure, 10 moles or \(6.022 \times 10^{24}\) molecules are the correct derived values. Given the provided answer (d) \( \frac{6.022 \times 10^{23}}{1.7} \), which is numerically \(3.54 \times 10^{23}\) (incorrect), there seems to be a mismatch in the source's question/answer.
Based on standard calculations for 180g of water, it corresponds to 10 moles, which is \(10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24}\) molecules. Neither (c) nor (d) directly match 10 moles or \(6.022 \times 10^{24}\) molecules.
However, if we strictly adhere to the given *solution* implying the result, and if there's an OCR error in the options, none of them directly present 10 moles or \(6.022 \times 10^{24}\) molecules. Given the answer is (d) and the provided solution text, there's an inconsistency. I will re-derive based on the calculation.
Let's re-evaluate the source answer for Q13, which is stated as (d) \( \frac{6.022 \times 10^{23}}{1.7} \). This is mathematically incorrect based on 180g of water. The solution explicitly calculates 10 moles and \(6.022 \times 10^{24}\) molecules. I will provide the answer based on the correct calculation and note the discrepancy if it arises from the choice selection.
The calculation clearly shows 10 moles, which means \(6.022 \times 10^{24}\) molecules. None of the options correctly represent this. I will assume there's a typo in the provided answer or options and will choose the option that is numerically closest to the correct calculation if a simple typo correction makes it match, or provide the correct value if no options match reasonably. Here, 10 moles or \(6.022 \times 10^{24}\) molecules are the correct calculations. None of the choices (a), (b), (c), or (d) match this. The provided "Answer: (d) \frac{6.022 \times 10^{23}}{1.7}" from the source is clearly wrong based on the solution provided. I will output the *correct* calculation and a simple explanation, acknowledging the correct number of moles and molecules.
Since the provided answer choice `(d) \frac{6.022 \times 10^{23}}{1.7}` is numerically incorrect for 180g of water (it should be 10 moles or \(6.022 \times 10^{24}\) molecules), I will state the correct result of the calculation as the answer, rather than trying to match a clearly erroneous option from the source. This falls under the rule to produce a clean, confident solution.
Answer: 180g of water represents 10 moles of water, which contains \(6.022 \times 10^{24}\) water molecules.
In simple words: To find out how many moles are in 180 grams of water, divide the mass (180 g) by the molar mass of water (18 g/mol). This gives 10 moles. Each mole has Avogadro's number of molecules, so 10 moles means \(6.022 \times 10^{24}\) molecules.

🎯 Exam Tip: Always remember that one mole of any substance contains Avogadro's number (\(6.022 \times 10^{23}\)) of particles, and mass-to-mole conversions are fundamental.

 

Question 14. 7.5 g of a gas occupies a volume of 5.6 litres at \(0^{\circ}\text{ C}\) and \(1\text{ atm}\) pressure. The gas is
(a) NO
(b) \( \text{N}_2\text{O} \)
(d) \( \text{CO}_2 \)
Answer: (a) NO
Solution:
At Standard Temperature and Pressure (STP), \(0^{\circ}\text{ C}\) (\(273\text{ K}\)) and \(1\text{ atm}\) pressure, 1 mole of any ideal gas occupies \(22.4\text{ litres}\).
Given: 7.5 g of gas occupies 5.6 litres at \(273\text{ K}\) and \(1\text{ atm}\) pressure.
Therefore, the mass of gas that occupies a volume of \(22.4\text{ litres}\) (which is 1 mole) is:
Molar mass \( = \frac{7.5\text{ g}}{5.6\text{ L}} \times 22.4\text{ L} \)
\( = 7.5 \times 4 = 30\text{ g/mol} \)
Now, check the molar mass of the options:
(a) NO (\text{Nitrogen monoxide}): Molar mass \( = 14 + 16 = 30\text{ g/mol} \)
(b) \( \text{N}_2\text{O} \) (\text{Dinitrogen monoxide}): Molar mass \( = (2 \times 14) + 16 = 28 + 16 = 44\text{ g/mol} \)
(d) \( \text{CO}_2 \) (\text{Carbon dioxide}): Molar mass \( = 12 + (2 \times 16) = 12 + 32 = 44\text{ g/mol} \)
The calculated molar mass (30 g/mol) matches that of NO.
In simple words: At standard conditions, one mole of any gas takes up \(22.4\text{ litres}\) of space. Since \(7.5\text{ g}\) of the gas fills \(5.6\text{ litres}\), we can figure out how much \(22.4\text{ litres}\) of this gas would weigh, which gives us its molar mass. The molar mass turns out to be \(30\text{ g/mol}\), which is the same as for nitrogen monoxide (NO).

🎯 Exam Tip: Remember the molar volume of a gas at STP (\(22.4\text{ L/mol}\)). This conversion factor is key for identifying unknown gases from their mass and volume at standard conditions.

 

Question 15. Total number of electrons present in 1.7 g of ammonia is
(a) \(6.022 \times 10^{23}\)
(b) \( \frac{6.022 \times 10^{22}}{1.7} \)
(c) \( \frac{6.022 \times 10^{24}}{1.7} \)
(d) \( \frac{6.022 \times 10^{23}}{1.7} \)
Answer: (a) \(6.022 \times 10^{23}\)
Solution:
First, find the number of electrons in one ammonia (\( \text{NH}_3 \)) molecule:
Nitrogen (N) has 7 electrons.
Each Hydrogen (H) has 1 electron, and there are 3 H atoms.
Total electrons in one \( \text{NH}_3 \) molecule \( = 7 + (3 \times 1) = 10 \text{ electrons} \)
Next, find the number of moles of ammonia in 1.7 g:
Molar mass of \( \text{NH}_3 = 14 + (3 \times 1) = 17\text{ g/mol} \)
No. of moles of ammonia \( = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.7\text{ g}}{17\text{ g/mol}} = 0.1\text{ mol} \)
Now, calculate the total number of \( \text{NH}_3 \) molecules in 0.1 mol:
Number of molecules \( = \text{Moles} \times \text{Avogadro's number} = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \) molecules
Finally, calculate the total number of electrons:
Total electrons \( = \text{Number of molecules} \times \text{Electrons per molecule} \)
\( = 6.022 \times 10^{22} \text{ molecules} \times 10 \text{ electrons/molecule} \)
\( = 6.022 \times 10^{23} \text{ electrons} \)
In simple words: First, count the electrons in one ammonia molecule (Nitrogen has 7, plus 3 from three Hydrogens, so 10 electrons total). Then, find how many moles are in \(1.7\text{ g}\) of ammonia by dividing its mass by its molar mass. This gives \(0.1\) moles. Multiply this by Avogadro's number to get the total molecules, then multiply by 10 (electrons per molecule) to find all the electrons.

🎯 Exam Tip: For problems involving total electrons, always determine the number of electrons per molecule first, then convert the given mass to moles and finally to the total number of molecules before multiplying by the electrons per molecule.

 

Question 16. The correct increasing order of the oxidation state of sulphur in the anions \( \text{SO}_4^{2-} \), \( \text{SO}_3^{2-} \), \( \text{S}_2\text{O}_4^{2-} \), \( \text{S}_2\text{O}_6^{2-} \) is
(a) \( \text{SO}_3^{2-} < \text{SO}_4^{2-} < \text{S}_2\text{O}_4^{2-} < \text{S}_2\text{O}_6^{2-} \)
(c) \( \text{S}_2\text{O}_4^{2-} < \text{SO}_3^{2-} < \text{S}_2\text{O}_6^{2-} < \text{SO}_4^{2-} \)
(d) \( \text{S}_2\text{O}_6^{2-} < \text{S}_2\text{O}_4^{2-} < \text{SO}_4^{2-} < \text{SO}_3^{2-} \)
Answer: (c) \( \text{S}_2\text{O}_4^{2-} < \text{SO}_3^{2-} < \text{S}_2\text{O}_6^{2-} < \text{SO}_4^{2-} \)
Solution:
Calculate the oxidation state of sulfur in each anion:
1. In \( \text{SO}_4^{2-} \): Let the oxidation state of S be \( x \).
\( x + 4(-2) = -2 \)
\( x - 8 = -2 \)
\( x = +6 \)
2. In \( \text{SO}_3^{2-} \): Let the oxidation state of S be \( x \).
\( x + 3(-2) = -2 \)
\( x - 6 = -2 \)
\( x = +4 \)
3. In \( \text{S}_2\text{O}_4^{2-} \): Let the oxidation state of S be \( x \).
\( 2x + 4(-2) = -2 \)
\( 2x - 8 = -2 \)
\( 2x = +6 \)
\( x = +3 \)
4. In \( \text{S}_2\text{O}_6^{2-} \): Let the oxidation state of S be \( x \).
\( 2x + 6(-2) = -2 \)
\( 2x - 12 = -2 \)
\( 2x = +10 \)
\( x = +5 \)
The oxidation states are: \( \text{S}_2\text{O}_4^{2-} (+3) \), \( \text{SO}_3^{2-} (+4) \), \( \text{S}_2\text{O}_6^{2-} (+5) \), \( \text{SO}_4^{2-} (+6) \).
Therefore, the increasing order is \( \text{S}_2\text{O}_4^{2-} < \text{SO}_3^{2-} < \text{S}_2\text{O}_6^{2-} < \text{SO}_4^{2-} \).
In simple words: To compare the oxidation states of sulfur in these compounds, calculate the oxidation number for sulfur in each one. Assuming oxygen is \(-2\), we find the oxidation states are \(+3\) for \( \text{S}_2\text{O}_4^{2-} \), \(+4\) for \( \text{SO}_3^{2-} \), \(+5\) for \( \text{S}_2\text{O}_6^{2-} \), and \(+6\) for \( \text{SO}_4^{2-} \). Arranging these from smallest to largest gives the correct order.

🎯 Exam Tip: When calculating oxidation states in polyatomic ions, remember that the sum of the oxidation states of all atoms must equal the overall charge of the ion.

 

Question 17. The equivalent mass of ferrous oxalate is
(a) \( \frac{\text{molar mass of ferrous oxalate}}{1} \)
(b) \( \frac{\text{molar mass of ferrous oxalate}}{2} \)
(c) \( \frac{\text{molar mass of ferrous oxalate}}{3} \)
(d) none of these
Answer: (c) \( \frac{\text{molar mass of ferrous oxalate}}{3} \)
Solution:
Ferrous oxalate is \( \text{FeC}_2\text{O}_4 \). In this compound, iron is in the \(+2\) oxidation state (\( \text{Fe}^{2+} \)) and oxalate is \( \text{C}_2\text{O}_4^{2-} \).
When ferrous oxalate acts as a reducing agent in an acidic medium, both \( \text{Fe}^{2+} \) and \( \text{C}_2\text{O}_4^{2-} \) get oxidized.
\( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + 1\text{e}^{-} \)
\( \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^{-} \)
Total electrons lost per mole of \( \text{FeC}_2\text{O}_4 \) is \( 1 \text{ (from Fe)} + 2 \text{ (from C}_2\text{O}_4^{2-}) = 3 \)
Thus, the n-factor (number of electrons exchanged) is 3.
Equivalent mass \( = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{\text{molar mass of ferrous oxalate}}{3} \)
In simple words: To find the equivalent mass of ferrous oxalate, we need to know how many electrons it loses when it reacts (its n-factor). Ferrous oxalate has iron (II) and an oxalate ion. Iron loses one electron to become iron (III), and the carbon in oxalate loses two electrons (each carbon goes from \(+3\) to \(+4\)). So, in total, three electrons are lost per molecule, making the n-factor 3. Therefore, the equivalent mass is the molar mass divided by 3.

🎯 Exam Tip: For compounds like ferrous oxalate that have both an oxidizable metal ion and an oxidizable anion, remember to sum the electron changes from both parts to find the total n-factor for equivalent mass calculation.

 

Question 18. If Avogadro number were changed from \(6.022 \times 10^{23}\) to \(6.022 \times 10^{20}\), this would change
(a) the ratio of chemical species to each other in a balanced equation
(b) the ratio of elements to each other in a compound
(c) the definition or mass in units of grams
(d) the mass of one mole of carbon
Answer: (d) the mass of one mole of carbon
Solution:
Avogadro's number defines the number of particles in one mole. If Avogadro's number changes, the definition of a mole changes. However, the ratio of atoms in a compound or chemical species in a balanced equation (which are fundamental stoichiometric ratios) would not change.
The mass of one mole of a substance is defined as the mass in grams that contains Avogadro's number of particles. If Avogadro's number becomes smaller, then one mole would contain fewer particles, and thus the mass of one mole would also decrease proportionately. For example, the atomic mass unit (u) is related to the mass of carbon-12. If Avogadro's number changes, the mass of 1 mole of carbon (which is 12 g if Avogadro's number is \(6.022 \times 10^{23}\)) would change because 1 mole would represent a different number of atoms.
Mass of one atom of carbon-12 \( = \frac{12\text{ g}}{6.022 \times 10^{23} \text{ atoms}} \)
If Avogadro number \( \text{N}_A' = 6.022 \times 10^{20} \), then mass of one mole of carbon \( = 12 \times \frac{6.022 \times 10^{20}}{6.022 \times 10^{23}} = 12 \times 10^{-3}\text{ g} = 0.012\text{ g} \).
In simple words: Avogadro's number tells us how many particles are in one mole. If this number becomes smaller, then a "mole" would mean fewer particles. This would directly change how much one mole of carbon weighs, because the weight of one mole is based on having a specific number of atoms. The ratios in chemical reactions or within a compound, however, are based on the individual atoms, not on how many are in a mole.

🎯 Exam Tip: Understand that while the relative atomic masses of elements remain constant, the absolute mass of a mole of any element is directly dependent on Avogadro's number.

 

Question 19. Two \(22.4\text{ liter}\) containers A and B contains \(8\text{ g}\) of \( \text{O}_2 \) and \(8\text{ g}\) of \( \text{SO}_2 \) respectively at \(273\text{ K}\) and \(1\text{ atm}\) pressure, then
(a) Number of molecules in A and B are the same
(b) Number of molecules in B is more than that in A.
(c) The ratio between the number of molecules in A to number of molecules in B is 2:1
(d) Number of molecules in B is three times greater than the number of molecules in A.
Answer: (c) The ratio between the number of molecules in A to number of molecules in B is 2:1
Solution:
At \(273\text{ K}\) and \(1\text{ atm}\) pressure (STP), 1 mole of any gas occupies \(22.4\text{ litres}\).
Container A: \(22.4\text{ litres}\) of \( \text{O}_2 \)
Molar mass of \( \text{O}_2 = 2 \times 16 = 32\text{ g/mol} \)
No. of moles of oxygen in container A \( = \frac{\text{Mass}}{\text{Molar mass}} = \frac{8\text{ g}}{32\text{ g/mol}} = 0.25\text{ moles} \)
Container B: \(22.4\text{ litres}\) of \( \text{SO}_2 \)
Molar mass of \( \text{SO}_2 = 32 + (2 \times 16) = 32 + 32 = 64\text{ g/mol} \)
No. of moles of sulfur dioxide in container B \( = \frac{\text{Mass}}{\text{Molar mass}} = \frac{8\text{ g}}{64\text{ g/mol}} = 0.125\text{ moles} \)
The number of molecules is directly proportional to the number of moles.
Ratio between the no. of molecules in A to no. of molecules in B \( = \text{Moles of O}_2 : \text{Moles of SO}_2 \)
\( = 0.25 : 0.125 \)
To simplify the ratio, divide both sides by the smaller number (0.125):
\( = \frac{0.25}{0.125} : \frac{0.125}{0.125} \)
\( = 2 : 1 \)
In simple words: Both containers have the same volume at the same temperature and pressure. We first find how many moles of gas are in each container by dividing the given mass by the gas's molar mass. Container A has \(0.25\) moles of oxygen, and container B has \(0.125\) moles of sulfur dioxide. Since the number of molecules is directly linked to the number of moles, the ratio of molecules in A to B is \(0.25\) to \(0.125\), which simplifies to 2:1.

🎯 Exam Tip: For gases at identical conditions of temperature, pressure, and volume, the number of moles (and thus molecules) is proportional to the mass divided by molar mass. Always calculate moles first to compare quantities.

 

Question 20. What is the mass of precipitate formed when \(50\text{ ml}\) of \(8.5\%\) solution of \( \text{AgNO}_3 \) is mixed with \(100\text{ ml}\) of \(1.865\%\) potassium chloride solution?
(a) 3.59 g
(b) 7 g
(c) 14 g
(d) 28 g
Answer: (a) 3.59 g
Solution:
The reaction is:
\( \text{AgNO}_3 + \text{KCl} \rightarrow \text{KNO}_3 + \text{AgCl} \downarrow \)
(\( \text{AgCl} \) is the precipitate)
First, calculate the mass of \( \text{AgNO}_3 \) in \(50\text{ mL}\) of \(8.5\%\) solution:
\(8.5\%\) solution means \(8.5\text{ g}\) of \( \text{AgNO}_3 \) in \(100\text{ mL}\) of solution.
So, in \(50\text{ mL}\), mass of \( \text{AgNO}_3 = \frac{8.5\text{ g}}{100\text{ mL}} \times 50\text{ mL} = 4.25\text{ g} \)
Molar mass of \( \text{AgNO}_3 = 108 (\text{Ag}) + 14 (\text{N}) + (3 \times 16) (\text{O}) = 108 + 14 + 48 = 170\text{ g/mol} \)
No. of moles of \( \text{AgNO}_3 = \frac{4.25\text{ g}}{170\text{ g/mol}} = 0.025\text{ moles} \)
Next, calculate the mass of \( \text{KCl} \) in \(100\text{ mL}\) of \(1.865\%\) solution:
\(1.865\%\) solution means \(1.865\text{ g}\) of \( \text{KCl} \) in \(100\text{ mL}\) of solution.
So, in \(100\text{ mL}\), mass of \( \text{KCl} = 1.865\text{ g} \)
Molar mass of \( \text{KCl} = 39 (\text{K}) + 35.5 (\text{Cl}) = 74.5\text{ g/mol} \)
No. of moles of \( \text{KCl} = \frac{1.865\text{ g}}{74.5\text{ g/mol}} = 0.025\text{ moles} \)
From the balanced equation, 1 mole of \( \text{AgNO}_3 \) reacts with 1 mole of \( \text{KCl} \) to produce 1 mole of \( \text{AgCl} \).
Since we have \(0.025\text{ moles}\) of \( \text{AgNO}_3 \) and \(0.025\text{ moles}\) of \( \text{KCl} \), they will react completely to form \(0.025\text{ moles}\) of \( \text{AgCl} \).
Molar mass of \( \text{AgCl} = 108 (\text{Ag}) + 35.5 (\text{Cl}) = 143.5\text{ g/mol} \)
Mass of \( \text{AgCl} \) formed \( = \text{No. of moles} \times \text{Molar mass} = 0.025\text{ moles} \times 143.5\text{ g/mol} = 3.5875\text{ g} \)
Rounding to two decimal places, the mass of \( \text{AgCl} \) formed is \(3.59\text{ g}\).
In simple words: First, figure out how many moles of silver nitrate and potassium chloride we have. Since both solutions provide \(0.025\) moles, they will completely react to form silver chloride. Then, calculate the mass of this \(0.025\) moles of silver chloride using its molar mass, which gives us \(3.59\text{ g}\) of precipitate.

🎯 Exam Tip: Always calculate the moles of all reactants first to identify any limiting reagents. In this case, both reactants were in stoichiometric amounts, simplifying the calculation of the product.

 

Question 21. The mass of a gas that occupies a volume of \(612.5\text{ ml}\) at room temperature and pressure (\(25^{\circ}\text{C}\) and \(1\text{ atm}\) pressure) is \(1.1\text{ g}\). The molar mass of the gas is
(a) \(66.25\text{ g mol}^{-1}\)
(b) \(44\text{ g mol}^{-1}\)
(c) \(24.5\text{ g mol}^{-1}\)
(d) \(662.5\text{ g mol}^{-1}\)
Answer: (b) \(44\text{ g mol}^{-1}\)
Solution:
At room temperature (\(25^{\circ}\text{C}\) or \(298\text{ K}\)) and \(1\text{ atm}\) pressure, 1 mole of any ideal gas occupies \(24.5\text{ litres}\) (molar volume at RTP).
Given: Volume of gas \( = 612.5\text{ mL} = 0.6125\text{ L} \)
Mass of gas \( = 1.1\text{ g} \)
First, find the number of moles of the gas:
No. of moles \( = \frac{\text{Volume of gas}}{\text{Molar volume at RTP}} = \frac{0.6125\text{ L}}{24.5\text{ L/mol}} = 0.025\text{ moles} \)
Now, calculate the molar mass of the gas:
Molar mass \( = \frac{\text{Mass of gas}}{\text{No. of moles}} = \frac{1.1\text{ g}}{0.025\text{ mol}} = 44\text{ g mol}^{-1} \)
In simple words: We know that at room temperature and pressure, one mole of any gas takes up \(24.5\text{ litres}\). Since \(1.1\text{ g}\) of our gas fills \(0.6125\text{ litres}\), we can calculate how many moles of gas that is. Then, by dividing the mass of the gas by the number of moles, we find its molar mass, which is \(44\text{ g mol}^{-1}\).

🎯 Exam Tip: Be careful to use the correct molar volume for the given conditions (STP is \(22.4\text{ L/mol}\), RTP is \(24.5\text{ L/mol}\)). This is a common source of error.

 

Question 22. Which of the following contain the same number of carbon atoms as in \(6\text{ g}\) of carbon -12.
(a) 7.5 g ethane
(b) 8 g methane
(c) both (a) and (b)
(d) none of these
Answer: (c) both (a) and (b)
Solution:
First, calculate the number of carbon atoms in \(6\text{ g}\) of Carbon-12:
Molar mass of Carbon-12 \( = 12\text{ g/mol} \)
No. of moles of carbon \( = \frac{6\text{ g}}{12\text{ g/mol}} = 0.5\text{ moles} \)
Number of carbon atoms \( = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \text{ carbon atoms} \)
Now, check options:
(a) \(7.5\text{ g}\) ethane (\( \text{C}_2\text{H}_6 \))
Molar mass of ethane \( = (2 \times 12) + (6 \times 1) = 24 + 6 = 30\text{ g/mol} \)
No. of moles of ethane \( = \frac{7.5\text{ g}}{30\text{ g/mol}} = 0.25\text{ moles} \)
Each molecule of ethane (\( \text{C}_2\text{H}_6 \)) contains 2 carbon atoms.
Number of carbon atoms \( = 0.25\text{ moles} \times 2 \text{ carbon atoms/molecule} \times 6.022 \times 10^{23} \text{ molecules/mole} \)
\( = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \text{ carbon atoms} \)
(b) \(8\text{ g}\) methane (\( \text{CH}_4 \))
Molar mass of methane \( = 12 + (4 \times 1) = 16\text{ g/mol} \)
No. of moles of methane \( = \frac{8\text{ g}}{16\text{ g/mol}} = 0.5\text{ moles} \)
Each molecule of methane (\( \text{CH}_4 \)) contains 1 carbon atom.
Number of carbon atoms \( = 0.5\text{ moles} \times 1 \text{ carbon atom/molecule} \times 6.022 \times 10^{23} \text{ molecules/mole} \)
\( = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \text{ carbon atoms} \)
Both (a) and (b) contain the same number of carbon atoms as in \(6\text{ g}\) of Carbon-12.
In simple words: First, find the number of carbon atoms in \(6\text{ g}\) of carbon-12, which is half a mole. Then, for each option, calculate the moles of the compound and multiply by the number of carbon atoms in each molecule. Both \(7.5\text{ g}\) of ethane and \(8\text{ g}\) of methane result in the same number of carbon atoms as in \(6\text{ g}\) of carbon-12.

🎯 Exam Tip: When comparing the number of specific atoms in different compounds, always convert the given mass to moles of the compound, then use the chemical formula to find the moles of the specific atom, and finally convert to the number of atoms using Avogadro's number.

 

Question 23. Which of the following compound(s) has/have a percentage of carbon same as that in ethylene (\( \text{C}_2\text{H}_4 \))
(a) Propene
(b) ethyne
(c) benzene
(d) ethane
Answer: (a) Propene
Solution:
Calculate the percentage of carbon in ethylene (\( \text{C}_2\text{H}_4 \)):
Molar mass of \( \text{C}_2\text{H}_4 = (2 \times 12) + (4 \times 1) = 24 + 4 = 28\text{ g/mol} \)
Mass of carbon in \( \text{C}_2\text{H}_4 = 2 \times 12 = 24\text{ g} \)
Percentage of carbon in \( \text{C}_2\text{H}_4 = \frac{\text{Mass of carbon}}{\text{Molar mass of C}_2\text{H}_4} \times 100\% = \frac{24}{28} \times 100\% \approx 85.71\% \)
Now, calculate the percentage of carbon for each option:
(a) Propene (\( \text{C}_3\text{H}_6 \))
Molar mass of \( \text{C}_3\text{H}_6 = (3 \times 12) + (6 \times 1) = 36 + 6 = 42\text{ g/mol} \)
Mass of carbon in \( \text{C}_3\text{H}_6 = 3 \times 12 = 36\text{ g} \)
Percentage of carbon in \( \text{C}_3\text{H}_6 = \frac{36}{42} \times 100\% \approx 85.71\% \)
(b) Ethyne (\( \text{C}_2\text{H}_2 \))
Molar mass of \( \text{C}_2\text{H}_2 = (2 \times 12) + (2 \times 1) = 24 + 2 = 26\text{ g/mol} \)
Mass of carbon in \( \text{C}_2\text{H}_2 = 2 \times 12 = 24\text{ g} \)
Percentage of carbon in \( \text{C}_2\text{H}_2 = \frac{24}{26} \times 100\% \approx 92.31\% \)
(c) Benzene (\( \text{C}_6\text{H}_6 \))
Molar mass of \( \text{C}_6\text{H}_6 = (6 \times 12) + (6 \times 1) = 72 + 6 = 78\text{ g/mol} \)
Mass of carbon in \( \text{C}_6\text{H}_6 = 6 \times 12 = 72\text{ g} \)
Percentage of carbon in \( \text{C}_6\text{H}_6 = \frac{72}{78} \times 100\% \approx 92.31\% \)
(d) Ethane (\( \text{C}_2\text{H}_6 \))
Molar mass of \( \text{C}_2\text{H}_6 = (2 \times 12) + (6 \times 1) = 24 + 6 = 30\text{ g/mol} \)
Mass of carbon in \( \text{C}_2\text{H}_6 = 2 \times 12 = 24\text{ g} \)
Percentage of carbon in \( \text{C}_2\text{H}_6 = \frac{24}{30} \times 100\% = 80\% \)
Propene has the same percentage of carbon as ethylene.
In simple words: To find the answer, calculate the percentage of carbon in ethylene first. Then, do the same calculation for each of the options. Propene turns out to have the exact same carbon percentage as ethylene. This is because both ethylene (\( \text{C}_2\text{H}_4 \)) and propene (\( \text{C}_3\text{H}_6 \)) have the same general formula ratio of carbon to hydrogen as \( \text{CH}_2 \).

🎯 Exam Tip: Compounds with the same empirical formula (simplest whole-number ratio of atoms) will have the same percentage composition of elements.

 

II. Write Brief Answer to the Following Questions:

 

Question 26. Define relative atomic mass.
Answer: Relative atomic mass is a way to compare the average mass of an atom to a standard unit, which is the unified atomic mass unit. This unit helps scientists define atomic weights across different elements. It's like saying how much heavier or lighter an atom is compared to this tiny standard weight.
In simple words: It tells us how heavy an atom is compared to a special tiny unit of mass.

🎯 Exam Tip: Remember that relative atomic mass is a ratio, so it has no units, but it is often expressed in "u" (unified atomic mass unit) for clarity.

 

Question 27. What do you understand by the term mole?
Answer: A mole is a specific amount of a substance. It contains \( 6.023 \times 10^{23} \) particles, which can be atoms, molecules, or ions. This huge number helps us count very tiny particles in chemistry, like using a "dozen" for eggs. It's represented by a symbol like 'mol'.
In simple words: A mole is a way to measure how much of a substance you have, counting a very large number of tiny particles.

🎯 Exam Tip: Always associate the term 'mole' with Avogadro's number \( 6.023 \times 10^{23} \) to easily recall its definition.

 

Question 28. Define equivalent mass.
Answer: Gram equivalent mass of an element, compound, or ion is the amount that either combines with or displaces \( 1.008 \text{ g} \) of hydrogen, \( 8 \text{ g} \) of oxygen, or \( 35.5 \text{ g} \) of chlorine. While equivalent mass itself has no unit, gram equivalent mass uses the unit 'g eq\(^{-1}\)'. This concept is crucial for understanding stoichiometry in reactions where substances react in certain fixed ratios.
In simple words: Equivalent mass is how much of a substance reacts with a fixed amount of hydrogen, oxygen, or chlorine.

🎯 Exam Tip: Remember the fixed values (\( 1.008 \text{ g} \) H, \( 8 \text{ g} \) O, \( 35.5 \text{ g} \) Cl) as they are key to defining and calculating equivalent mass.

 

Question 29. Distinguish between oxidation and reduction.
Answer:

Oxidation and reduction are opposite processes that always happen together in redox reactions. For instance, in a combustion reaction, one substance is oxidized while another is reduced.
In simple words: Oxidation is losing electrons, removing hydrogen, or gaining oxygen. Reduction is gaining electrons, adding hydrogen, or losing oxygen.

🎯 Exam Tip: Use the mnemonic "OIL RIG" (Oxidation Is Loss, Reduction Is Gain of electrons) to easily recall the electron concept of oxidation and reduction.

 

Question 30. What do you understand by the term oxidation number.
Answer:
Oxidation: Traditionally, oxidation means either adding oxygen or removing hydrogen. For example, in the reaction \( 2\text{H}_2\text{S} + \text{O}_2 \rightarrow \text{H}_2\text{O} + 2\text{S} \), hydrogen is removed from \( \text{H}_2\text{S} \). In terms of electrons, oxidation is the loss of electrons. During oxidation, the oxidation number of an element increases. When a substance undergoes oxidation, it acts as a reducing agent.
Reduction: Reduction is the opposite, meaning removing oxygen or adding hydrogen. For instance, in \( \text{Ca} + \text{H}_2 \rightarrow \text{CaH}_2 \), hydrogen is added to calcium. In terms of electrons, reduction is the gain of electrons. During reduction, the oxidation number of an element decreases. A common example is zinc gaining electrons to form a neutral zinc atom. An important example is the rusting of iron, where iron metal is oxidized to iron oxide.
In simple words: Oxidation is when an atom loses electrons or gains oxygen, making its charge number go up. Reduction is when an atom gains electrons or loses oxygen, making its charge number go down.

🎯 Exam Tip: For oxidation numbers, remember the general rules: hydrogen is usually \( +1 \) (except in metal hydrides), oxygen is usually \( -2 \) (except in peroxides or with fluorine), and the sum of oxidation numbers in a neutral compound is zero.

 

Question 31. Calculate the molar mass of the following compounds.
(i) urea [CO(NH2)2]:
(ii) acetone [CH3COCH3]
(iii) boric acid [H3BO3]:
(iv) sulphuric acid [H2SO4]
Answer:
(i) Urea [CO(NH2)2]:
To find the molar mass, we add up the atomic masses of all atoms present in the molecule.
Molar mass of urea = \( (4 \times \text{Atomic mass of hydrogen}) + (1 \times \text{Atomic mass of carbon}) + (2 \times \text{Atomic mass of nitrogen}) + (1 \times \text{Atomic mass of oxygen}) \)
\( = (4 \times 1.008) + (1 \times 12) + (2 \times 14) + (1 \times 16) \)
\( = 4.032 + 12 + 28 + 16 = 60.032 \text{ g mol}^{-1} \)

(ii) Acetone [CH3COCH3]:
Molar mass of Acetone = \( (6 \times \text{Atomic mass of hydrogen}) + (3 \times \text{Atomic mass of carbon}) + (1 \times \text{Atomic mass of oxygen}) \)
\( = (6 \times 1.008) + (3 \times 12) + (1 \times 16) \)
\( = 6.048 + 36 + 16 = 52.048 \text{ g mol}^{-1} \)

(iii) Boric acid [H3BO3]:
Molar mass of Boric acid = \( (3 \times \text{Atomic mass of hydrogen}) + (1 \times \text{Atomic mass of boron}) + (3 \times \text{Atomic mass of oxygen}) \)
\( = (3 \times 1.008) + (3 \times 11) + (1 \times 16) \)
\( = 3.024 + 33 + 16 = 52.024 \text{ g mol}^{-1} \)

(iv) Sulphuric acid [H2SO4]:
Molar mass of Sulphuric acid = \( (2 \times \text{Atomic mass of hydrogen}) + (1 \times \text{Atomic mass of sulphur}) + (4 \times \text{Atomic mass of oxygen}) \)
\( = (2 \times 1.008) + (1 \times 32) + (4 \times 16) \)
\( = 2.016 + 32 + 64 = 98.016 \text{ g mol}^{-1} \)
In simple words: To find the molar mass of each compound, you just add up the atomic weights of all the atoms in its chemical formula. It's like finding the total weight of ingredients in a recipe.

🎯 Exam Tip: Always use the precise atomic masses of elements from the periodic table to ensure accurate molar mass calculations, especially for multiple-choice questions where small differences matter.

 

Question 32. The density of carbon dioxide is equal to \( 1.965 \text{ kgm}^{-3} \) at 273 K and 1 atm pressure. Calculate the molar mass of CO2.
Answer:
Given:
Density of \( \text{CO}_2 \) at 273 K and 1 atm pressure \( = 1.965 \text{ kgm}^{-3} \)
At 273 K and 1 atm pressure, 1 mole of any gas occupies a volume of 22.4 L. This is known as the molar volume for ideal gases under standard conditions.
1 mole of \( \text{CO}_2 \) occupies a volume of \( 22.4 \text{ L} \)
Since \( 1 \text{ L} = 10^{-3} \text{ m}^3 \), then \( 22.4 \text{ L} = 22.4 \times 10^{-3} \text{ m}^3 \).
Now, we calculate the mass of 1 mole of \( \text{CO}_2 \):
Mass of 1 mole of \( \text{CO}_2 \) \( = \text{Density} \times \text{Volume of 1 mole} \)
\( = 1.965 \text{ kgm}^{-3} \times 22.4 \times 10^{-3} \text{ m}^3 \)
\( = 0.043996 \text{ kg} \)
To convert to grams:
\( = 0.043996 \times 1000 \text{ g} = 43.996 \text{ g} \)
This is approximately \( 44.00 \text{ g} \).
So, the molar mass of \( \text{CO}_2 \) \( = 44.00 \text{ g mol}^{-1} \).
In simple words: We used the given density of carbon dioxide and the known volume that one mole of gas takes up at that temperature and pressure. Multiplying these two numbers gives us the total mass of one mole of carbon dioxide.

🎯 Exam Tip: Remember that at standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters (or \( 22.4 \times 10^{-3} \text{ m}^3 \)).

 

Question 33. Which contains the greatest number of moles of oxygen atoms.
1 mol of ethanol
1 mol of formic acid
1 mol of \( \text{H}_2\text{O} \)
Answer:

Based on the table, 1 mol of formic acid contains two oxygen atoms per molecule, so it will have the greatest number of oxygen atoms compared to ethanol and water, which each have one oxygen atom per molecule. Therefore, Formic acid has the greatest number of moles of oxygen atoms.
In simple words: Formic acid has more oxygen atoms in each molecule than ethanol or water. So, if you have one mole of each, formic acid will have the most oxygen atoms.

🎯 Exam Tip: Always check the chemical formula for the number of atoms of a specific element before calculating total atoms or moles, as it directly impacts the result.

 

Question 34. Calculate the average atomic mass of naturally occurring magnesium using the following data:
Answer:

Average atomic mass \( = \frac{\sum (\text{Isotopic atomic mass} \times \text{Abundance})}{100} \)
\( = \frac{(23.99 \times 78.99) + (24.99 \times 10.00) + (25.98 \times 11.01)}{100} \)
\( = \frac{1895.07 + 249.9 + 286.06}{100} \)
\( = \frac{2431.03}{100} \)
\( = 24.31 \text{ u} \)
The average atomic mass of magnesium is approximately \( 24.31 \text{ u} \). This calculation considers the natural proportions of each isotope.
In simple words: To find the average atomic mass, you multiply each isotope's mass by how common it is, then add those values up and divide by 100. This gives you a weighted average, which is the atomic mass you see on the periodic table.

🎯 Exam Tip: Ensure that the sum of the percentage abundances for all isotopes is 100% before performing calculations, as this is a common point of error.

 

Question 35. In a reaction \( x + y + z_2 \rightarrow xyz_2 \), identify the Limiting reagent if any, in the following reaction mixtures.
(a) 200 atoms of x + 200 atoms of y + 50 molecules of \( z_2 \).
(b) 1 mol of x + 1 mol of y + 3 mol of \( z_2 \).
(c) 50 atoms of x + 25 atoms of y + 50 molecules of \( z_2 \).
(d) 2.5 mol of x + 5 mol of y + 5 mol of \( z_2 \).
Answer:
The given reaction is: \( x + y + z_2 \rightarrow xyz_2 \). This means 1 atom of x reacts with 1 atom of y and 1 molecule of \( z_2 \).

(a) For 200 atoms of x + 200 atoms of y + 50 molecules of \( z_2 \):
According to the reaction, 1 atom of x reacts with 1 atom of y and 1 molecule of \( z_2 \).
Here, 50 molecules of \( z_2 \) would need 50 atoms of x and 50 atoms of y to react completely. Since there are 200 atoms of x and 200 atoms of y, both x and y are in excess. \( z_2 \) is the limiting reagent because it will be used up first.

(b) For 1 mol of x + 1 mol of y + 3 mol of \( z_2 \):
1 mole of x reacts with 1 mole of y and 1 mole of \( z_2 \). Here, we have 3 moles of \( z_2 \), which is more than enough for 1 mole of x and 1 mole of y. So, x and y will be completely used up before \( z_2 \). In this case, either x or y could be considered the limiting reagent if only one were present, but since both are equally limited compared to \( z_2 \), both x and y are limiting reagents in this scenario (or it could be seen that if one runs out, the other cannot react further with \( z_2 \)).

(c) For 50 atoms of x + 25 atoms of y + 50 molecules of \( z_2 \):
Here, 25 atoms of y would need 25 atoms of x and 25 molecules of \( z_2 \). We have 50 atoms of x and 50 molecules of \( z_2 \), which are both more than 25. So, y is the limiting reagent because it will be fully consumed first.

(d) For 2.5 mol of x + 5 mol of y + 5 mol of \( z_2 \):
2.5 moles of x would need 2.5 moles of y and 2.5 moles of \( z_2 \). We have 5 moles of y and 5 moles of \( z_2 \), which are more than 2.5 moles. Therefore, x is the limiting reagent as it will be used up first.
In simple words: The limiting reagent is the one that gets used up first in a chemical reaction. It stops the reaction from making more product, even if other ingredients are left over.

🎯 Exam Tip: To identify the limiting reagent, calculate how much product each reactant could form if it were completely consumed. The reactant that produces the smallest amount of product is the limiting reagent.

 

Question 36. Mass of one atom of an element is \( 6.645 \times 10^{-23} \text{ g} \). How many moles of element are there in 0.320 kg.
Answer:
Given:
Mass of one atom \( = 6.645 \times 10^{-23} \text{ g} \)
First, calculate the mass of 1 mole of atoms (molar mass):
Molar mass \( = \text{Mass of one atom} \times \text{Avogadro's number} \)
Avogadro's number \( = 6.022 \times 10^{23} \text{ atoms/mol} \)
Molar mass \( = 6.645 \times 10^{-23} \text{ g/atom} \times 6.022 \times 10^{23} \text{ atoms/mol} \)
\( = 39.99 \text{ g/mol} \)
Next, convert the given mass of the element from kg to g:
Mass of element \( = 0.320 \text{ kg} = 0.320 \times 1000 \text{ g} = 320 \text{ g} \)
Finally, calculate the number of moles:
Number of moles \( = \frac{\text{Total mass of element}}{\text{Molar mass of element}} \)
\( = \frac{320 \text{ g}}{39.99 \text{ g/mol}} \)
\( \approx 8 \text{ moles} \)
There are approximately 8 moles of the element in 0.320 kg. This calculation shows how a very small mass of a single atom can build up to a macroscopic amount like moles.
In simple words: We first found the weight of one whole mole of the element, then divided the total given weight by the weight of one mole to find out how many moles are present.

🎯 Exam Tip: Remember to convert all mass units to be consistent (e.g., all to grams or all to kilograms) before performing calculations to avoid errors.

 

Question 37. What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
Answer:
Molecular mass:
• Relative molecular mass is the ratio of the mass of a molecule to the unified atomic mass unit. It tells us how heavy one molecule is compared to a standard unit.
• It can be calculated by adding the relative atomic masses of all the atoms that make up the molecule. For example, a water molecule is made of two hydrogen atoms and one oxygen atom.
• For carbon monoxide (CO), molecular mass \( = \text{Atomic mass of carbon} + \text{Atomic mass of oxygen} = 12 + 16 = 28 \text{ u} \).

Molar mass:
• It is the mass of one mole of a substance. A mole is a specific large number of particles. This is a practical way to measure quantities in the lab.
• The molar mass of a compound is the sum of the relative atomic masses of its components, but expressed in grams per mole (\( \text{g mol}^{-1} \)).
• For carbon monoxide (CO), the molar mass is \( 12 + 16 = 28 \text{ g mol}^{-1} \).
Both molecular mass and molar mass have the same numerical value for a substance, but their units are different: 'u' for molecular mass (for one molecule) and 'g mol\(^{-1}\)' for molar mass (for one mole of molecules).
In simple words: Molecular mass is the weight of just one molecule (in 'u' units), while molar mass is the weight of a huge group of molecules (one mole) and is given in grams per mole. For the same substance, these numbers are the same, but the units tell you if it's one tiny particle or a large collection.

🎯 Exam Tip: Clearly distinguish between the units 'u' (atomic mass unit) for individual particles and 'g mol\(^{-1}\)' (grams per mole) for a mole of particles; this is a fundamental concept in chemistry.

 

Question 38. What is the empirical formula of the following?
(i) Fructose (C6H12O6) found in honey
(ii) Caffeine (C8H10N4O2 )a substance found in tea and coffee.
Answer:

The empirical formula gives the simplest whole-number ratio of atoms in a compound. This formula is useful for initial analysis of unknown substances.
In simple words: The empirical formula is like simplifying a fraction; it shows the smallest possible whole-number ratio of atoms in a compound, not the exact number in the molecule.

🎯 Exam Tip: To find the empirical formula, divide the subscripts in the molecular formula by their greatest common divisor. Always simplify to the lowest whole number ratio.

 

Question 39. The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals. (Atomic mass of Al = 27 u Atomic mass of O = 16 u)
\( 2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe} \); If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide
(i) Calculate the mass of \( \text{Al}_2\text{O}_3 \) formed.
(ii) How much of the excess reagent is left at the end of the reaction?
Answer:
Given reaction: \( 2\text{Al} + \text{Fe}_2\text{O}_3 \rightarrow \text{Al}_2\text{O}_3 + 2\text{Fe} \)
Atomic masses: Al = 27 u, Fe = 56 u, O = 16 u
Molar mass of Al \( = 27 \text{ g/mol} \)
Molar mass of \( \text{Fe}_2\text{O}_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \text{ g/mol} \)
Molar mass of \( \text{Al}_2\text{O}_3 = (2 \times 27) + (3 \times 16) = 54 + 48 = 102 \text{ g/mol} \)

Step 1: Calculate the initial moles of each reactant.
Moles of Al \( = \frac{324 \text{ g}}{27 \text{ g/mol}} = 12 \text{ mol} \)
Mass of \( \text{Fe}_2\text{O}_3 = 1.12 \text{ kg} = 1120 \text{ g} \)
Moles of \( \text{Fe}_2\text{O}_3 = \frac{1120 \text{ g}}{160 \text{ g/mol}} = 7 \text{ mol} \)

Step 2: Determine the limiting reagent.
From the balanced equation, 2 moles of Al react with 1 mole of \( \text{Fe}_2\text{O}_3 \).
For 12 moles of Al, we would need \( \frac{12}{2} = 6 \) moles of \( \text{Fe}_2\text{O}_3 \). We have 7 moles of \( \text{Fe}_2\text{O}_3 \), so \( \text{Fe}_2\text{O}_3 \) is in excess.
For 7 moles of \( \text{Fe}_2\text{O}_3 \), we would need \( 7 \times 2 = 14 \) moles of Al. We only have 12 moles of Al. Therefore, Al is the limiting reagent.

Step 3: Calculate the mass of \( \text{Al}_2\text{O}_3 \) formed (i).
Since Al is the limiting reagent, the amount of product formed depends on Al.
From the equation, 2 moles of Al produce 1 mole of \( \text{Al}_2\text{O}_3 \).
So, 12 moles of Al will produce \( \frac{12}{2} = 6 \) moles of \( \text{Al}_2\text{O}_3 \).
Mass of \( \text{Al}_2\text{O}_3 \) formed \( = 6 \text{ mol} \times 102 \text{ g/mol} = 612 \text{ g} \).

(i) The mass of \( \text{Al}_2\text{O}_3 \) formed is \( 612 \text{ g} \).

Step 4: Calculate the amount of excess reagent left (ii).
Moles of \( \text{Fe}_2\text{O}_3 \) reacted with 12 moles of Al \( = \frac{12 \text{ mol Al}}{2 \text{ mol Al}} \times 1 \text{ mol Fe}_2\text{O}_3 = 6 \text{ mol} \).
Initial moles of \( \text{Fe}_2\text{O}_3 = 7 \text{ mol} \).
Moles of \( \text{Fe}_2\text{O}_3 \) left \( = 7 \text{ mol} - 6 \text{ mol} = 1 \text{ mol} \).
Mass of excess \( \text{Fe}_2\text{O}_3 \) left \( = 1 \text{ mol} \times 160 \text{ g/mol} = 160 \text{ g} \).

(ii) The mass of the excess reagent (\( \text{Fe}_2\text{O}_3 \)) left at the end of the reaction is \( 160 \text{ g} \). This type of reaction is highly exothermic, meaning it releases a lot of heat, which is why it's used for welding.

In simple words: First, we figure out which reactant runs out first. Then, we use that reactant's amount to calculate how much product is made and how much of the other, leftover reactant remains.

🎯 Exam Tip: Always identify the limiting reagent first, as it dictates the maximum amount of product that can be formed and the amount of excess reagent remaining.

 

Question 40. How many moles of ethane is required to produce 44 g of CO2 (g) after combustion. Balanced equation for the combustion of ethane.
Answer:
The balanced equation for the combustion of ethane is:
\( \text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \)
To get rid of the fraction, multiply the entire equation by 2:
\( 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \)
From the balanced equation, we see that 2 moles of ethane (\( \text{C}_2\text{H}_6 \)) produce 4 moles of carbon dioxide (\( \text{CO}_2 \)).
We want to produce 44 g of \( \text{CO}_2 \). First, convert this mass to moles.
Molar mass of \( \text{CO}_2 = 12 + (2 \times 16) = 12 + 32 = 44 \text{ g/mol} \).
Moles of \( \text{CO}_2 \) to be produced \( = \frac{44 \text{ g}}{44 \text{ g/mol}} = 1 \text{ mol} \).
Now, use the stoichiometry from the balanced equation:
If 4 moles of \( \text{CO}_2 \) are produced from 2 moles of \( \text{C}_2\text{H}_6 \), then 1 mole of \( \text{CO}_2 \) will be produced from \( \frac{2}{4} \) moles of \( \text{C}_2\text{H}_6 \).
Moles of \( \text{C}_2\text{H}_6 \) required \( = \frac{1}{2} = 0.5 \text{ mol} \).
Therefore, 0.5 moles of ethane are required to produce 44 g of \( \text{CO}_2 \). This shows how chemical equations are used to predict reactant amounts.
In simple words: First, we write the balanced chemical reaction. Then, we find out how many moles of carbon dioxide we need. Finally, using the reaction, we calculate how many moles of ethane are needed to make that amount of carbon dioxide.

🎯 Exam Tip: Always make sure your chemical equation is balanced before performing any stoichiometric calculations; an unbalanced equation will lead to incorrect mole ratios.

 

Question 41. Hydrogen peroxide is an oxidising agent. It oxidises ferrous ion to ferric ion and reduced itself to water. Write a balanced equation.
Answer:
Here, hydrogen peroxide (\( \text{H}_2\text{O}_2 \)) acts as an oxidizing agent, which means it itself gets reduced. It oxidizes ferrous ion (\( \text{Fe}^{2+} \)) to ferric ion (\( \text{Fe}^{3+} \)).

1. Oxidation half-reaction:
Ferrous ion loses an electron to become ferric ion.
\( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^{-} \)

2. Reduction half-reaction:
Hydrogen peroxide gets reduced to water. In \( \text{H}_2\text{O}_2 \), oxygen has an oxidation state of \( -1 \), and in \( \text{H}_2\text{O} \), oxygen has an oxidation state of \( -2 \). So, each oxygen atom gains one electron.
\( \text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} \)
Balance oxygen atoms (already balanced in this case by forming \( 2\text{H}_2\text{O} \)).
Balance hydrogen atoms by adding \( \text{H}^{+} \):
\( \text{H}_2\text{O}_2 + 2\text{H}^{+} \rightarrow 2\text{H}_2\text{O} \)
Balance charge by adding electrons. The left side has a total charge of \( +2 \), and the right side is neutral. So, add 2 electrons to the left:
\( \text{H}_2\text{O}_2 + 2\text{H}^{+} + 2\text{e}^{-} \rightarrow 2\text{H}_2\text{O} \)

3. Combine the half-reactions:
To combine, the number of electrons lost must equal the number of electrons gained. Multiply the oxidation half-reaction by 2:
\( 2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2\text{e}^{-} \)
Now, add the two balanced half-reactions:
\( 2\text{Fe}^{2+} + \text{H}_2\text{O}_2 + 2\text{H}^{+} + 2\text{e}^{-} \rightarrow 2\text{Fe}^{3+} + 2\text{e}^{-} + 2\text{H}_2\text{O} \)
Cancel the electrons:
\( 2\text{Fe}^{2+} + \text{H}_2\text{O}_2 + 2\text{H}^{+} \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O} \)
This is the balanced equation. Hydrogen peroxide is a strong oxidizing agent often used in bleaching and disinfection.
In simple words: Hydrogen peroxide makes iron change from one form to another by taking away its electrons. At the same time, the hydrogen peroxide itself changes into water by gaining electrons.

🎯 Exam Tip: When balancing redox reactions in acidic medium, add \( \text{H}^{+} \) to balance hydrogen atoms and \( \text{H}_2\text{O} \) to balance oxygen atoms.

 

Question 42. Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38% hydrogen and rest oxygen its vapour density is 47.
Answer:
Given: Carbon = 76.6%, Hydrogen = 6.38%
Remaining percentage is oxygen: Oxygen \( = 100\% - (76.6\% + 6.38\%) = 100\% - 82.98\% = 17.02\% \).

Step 1: Convert percentages to grams (assume 100 g sample).
Carbon: 76.6 g
Hydrogen: 6.38 g
Oxygen: 17.02 g

Step 2: Convert grams to moles.
Atomic mass of C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol.
Moles of C \( = \frac{76.6}{12.01} \approx 6.38 \text{ mol} \)
Moles of H \( = \frac{6.38}{1.008} \approx 6.33 \text{ mol} \)
Moles of O \( = \frac{17.02}{16.00} \approx 1.06 \text{ mol} \)

Step 3: Find the simplest mole ratio by dividing by the smallest number of moles (1.06).
Ratio for C \( = \frac{6.38}{1.06} \approx 6 \)
Ratio for H \( = \frac{6.33}{1.06} \approx 6 \)
Ratio for O \( = \frac{1.06}{1.06} = 1 \)
The empirical formula is \( \text{C}_6\text{H}_6\text{O} \). This calculation reveals the basic building block ratio of atoms.

Step 4: Calculate the empirical formula mass.
Empirical formula mass of \( \text{C}_6\text{H}_6\text{O} = (6 \times 12) + (6 \times 1) + (1 \times 16) = 72 + 6 + 16 = 94 \text{ g/mol} \).

Step 5: Calculate the molecular mass from vapour density.
Molecular mass \( = 2 \times \text{Vapour density} \)
Molecular mass \( = 2 \times 47 = 94 \text{ g/mol} \).

Step 6: Determine the integer 'n'.
\( \text{n} = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{94}{94} = 1 \)

Step 7: Calculate the molecular formula.
Molecular formula \( = \text{n} \times (\text{Empirical formula}) \)
Molecular formula \( = 1 \times (\text{C}_6\text{H}_6\text{O}) = \text{C}_6\text{H}_6\text{O} \).
In simple words: We first find the simplest ratio of atoms in the compound using the percentages given. This gives us the empirical formula. Then, we use the vapor density to find the actual molecular weight and compare it to the empirical formula weight to get the exact molecular formula.

🎯 Exam Tip: Always remember the relationship between molecular mass and vapour density: Molecular mass \( = 2 \times \text{Vapour density} \). This is a quick way to find molecular mass from experimental data.

 

Question 43. A Compound on analysis gave Na = 14.31% S = 9.97% H= 6.22% and O= 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization, (molecular mass of the compound is 322).
Answer:
Given percentages: Na = 14.31%, S = 9.97%, H = 6.22%, O = 69.5%
Atomic masses: Na = 23, S = 32, H = 1, O = 16.

Step 1: Calculate the relative number of atoms (moles) for each element.
For Na: \( \frac{14.31}{23} = 0.622 \)
For S: \( \frac{9.97}{32} = 0.312 \)
For H: \( \frac{6.22}{1} = 6.22 \)
For O: \( \frac{69.5}{16} = 4.344 \)

Step 2: Find the simple ratio by dividing by the smallest value (0.312).
For Na: \( \frac{0.622}{0.312} \approx 2 \)
For S: \( \frac{0.312}{0.312} = 1 \)
For H: \( \frac{6.22}{0.312} \approx 20 \)
For O: \( \frac{4.344}{0.312} \approx 14 \)
The empirical formula is \( \text{Na}_2\text{SH}_{20}\text{O}_{14} \). This step identifies the basic elemental composition. The problem states that all hydrogen is present as water of crystallization.

Step 3: Calculate the empirical formula mass for \( \text{Na}_2\text{SH}_{20}\text{O}_{14} \).
Empirical formula mass \( = (2 \times 23) + (1 \times 32) + (20 \times 1) + (14 \times 16) \)
\( = 46 + 32 + 20 + 224 = 322 \text{ g/mol} \).

Step 4: Calculate 'n' (the ratio of molecular mass to empirical formula mass).
Given molecular mass of the compound is 322.
\( \text{n} = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{322}{322} = 1 \)

Step 5: Determine the molecular formula.
Molecular formula \( = \text{n} \times (\text{Empirical formula}) = 1 \times (\text{Na}_2\text{SH}_{20}\text{O}_{14}) = \text{Na}_2\text{SH}_{20}\text{O}_{14} \).

The problem states that all hydrogen is present as water of crystallization (\( \text{H}_2\text{O} \)).
We have 20 hydrogen atoms, which means there are 10 molecules of water (\( 10 \times \text{H}_2\text{O} \)). This accounts for 10 oxygen atoms.
The remaining oxygen atoms are \( 14 - 10 = 4 \).
So, the compound can be written as \( \text{Na}_2\text{S}\text{O}_4 \cdot 10\text{H}_2\text{O} \). This is sodium sulfate decahydrate, commonly known as Glauber's salt.
In simple words: We find the simplest ratio of elements, then calculate the total weight of this simple formula. Since this weight matches the given molecular weight, the molecular formula is the same as the simple formula, with hydrogen being part of water molecules.

🎯 Exam Tip: When a problem specifies water of crystallization, make sure to account for both the hydrogen and oxygen atoms within the water molecules to derive the correct main compound formula.

 

Question 44. Balance the following equations by oxidation number method
(i) \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{I}_2 + \text{H}_2\text{O} \)
Answer:
(i) Balancing \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{I}_2 + \text{H}_2\text{O} \)

Step 1: Assign oxidation numbers to all elements.
\( \text{K}_2\text{Cr}_2\text{O}_7 \): K is \( +1 \), O is \( -2 \). For Cr: \( 2(+1) + 2x + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6 \).
\( \text{KI} \): K is \( +1 \), I is \( -1 \).
\( \text{H}_2\text{SO}_4 \): H is \( +1 \), O is \( -2 \). For S: \( 2(+1) + y + 4(-2) = 0 \implies 2 + y - 8 = 0 \implies y = +6 \).
\( \text{K}_2\text{SO}_4 \): K is \( +1 \), O is \( -2 \). For S: \( 2(+1) + z + 4(-2) = 0 \implies 2 + z - 8 = 0 \implies z = +6 \).
\( \text{Cr}_2(\text{SO}_4)_3 \): \( \text{SO}_4 \) is \( -2 \), so \( (\text{SO}_4)_3 \) is \( -6 \). For Cr: \( 2x + (-6) = 0 \implies 2x = 6 \implies x = +3 \).
\( \text{I}_2 \): I is \( 0 \) (elemental state).
\( \text{H}_2\text{O} \): H is \( +1 \), O is \( -2 \).

Changes in oxidation number:
Cr: \( +6 \rightarrow +3 \) (gain of 3 electrons per Cr atom. Since there are two Cr atoms, total gain = \( 2 \times 3 = 6 \) electrons).
I: \( -1 \rightarrow 0 \) (loss of 1 electron per I atom. Since \( \text{I}_2 \) is formed, \( 2 \times -1 \rightarrow 2 \times 0 \), total loss = \( 2 \times 1 = 2 \) electrons).

Step 2: Balance the electron changes.
To balance, the total electrons gained must equal total electrons lost.
Gain: 6 electrons (Cr)
Loss: 2 electrons (I)
Multiply the iodine change by 3 to balance electrons: \( (2\text{I}^{-} \rightarrow \text{I}_2 + 2\text{e}^{-}) \times 3 = 6\text{I}^{-} \rightarrow 3\text{I}_2 + 6\text{e}^{-} \)

Step 3: Write the partial balanced equation with coefficients for changing species.
\( \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{KI} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{I}_2 + \text{H}_2\text{O} \)

Step 4: Balance other atoms by inspection (K, S, H, O).
Balance K: On the left, \( \text{K}_2\text{Cr}_2\text{O}_7 \) has 2 K, and \( 6\text{KI} \) has 6 K. Total 8 K. On the right, \( \text{K}_2\text{SO}_4 \) has 2 K. Need \( 4\text{K}_2\text{SO}_4 \).
\( \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{KI} + \text{H}_2\text{SO}_4 \rightarrow 4\text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{I}_2 + \text{H}_2\text{O} \)

Balance S: On the right, \( 4\text{K}_2\text{SO}_4 \) has 4 S, and \( \text{Cr}_2(\text{SO}_4)_3 \) has 3 S. Total 7 S. On the left, \( \text{H}_2\text{SO}_4 \) has 1 S. Need \( 7\text{H}_2\text{SO}_4 \).
\( \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{KI} + 7\text{H}_2\text{SO}_4 \rightarrow 4\text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{I}_2 + \text{H}_2\text{O} \)

Balance H: On the left, \( 7\text{H}_2\text{SO}_4 \) has \( 7 \times 2 = 14 \) H. On the right, \( \text{H}_2\text{O} \) has 2 H. Need \( 7\text{H}_2\text{O} \).
\( \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{KI} + 7\text{H}_2\text{SO}_4 \rightarrow 4\text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{I}_2 + 7\text{H}_2\text{O} \)

Balance O: On the left, \( \text{K}_2\text{Cr}_2\text{O}_7 \) has 7 O, \( 7\text{H}_2\text{SO}_4 \) has \( 7 \times 4 = 28 \) O. Total \( 7 + 28 = 35 \) O. On the right, \( 4\text{K}_2\text{SO}_4 \) has \( 4 \times 4 = 16 \) O, \( \text{Cr}_2(\text{SO}_4)_3 \) has \( 3 \times 4 = 12 \) O, \( 7\text{H}_2\text{O} \) has 7 O. Total \( 16 + 12 + 7 = 35 \) O. Oxygen is balanced.
The final balanced equation is: \( \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{KI} + 7\text{H}_2\text{SO}_4 \rightarrow 4\text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{I}_2 + 7\text{H}_2\text{O} \).

(ii) Balancing \( \text{KMnO}_4 + \text{Na}_2\text{SO}_3 \rightarrow \text{MnO}_2 + \text{Na}_2\text{SO}_4 + \text{KOH} \)
This reaction occurs in a basic medium, but it can be balanced using oxidation numbers first.
Oxidation numbers:
\( \text{KMnO}_4 \): Mn is \( +7 \)
\( \text{Na}_2\text{SO}_3 \): S is \( +4 \)
\( \text{MnO}_2 \): Mn is \( +4 \)
\( \text{Na}_2\text{SO}_4 \): S is \( +6 \)

Changes:
Mn: \( +7 \rightarrow +4 \) (Gain of 3 electrons)
S: \( +4 \rightarrow +6 \) (Loss of 2 electrons)

Multiply Mn change by 2 and S change by 3 to balance electrons (LCM of 3 and 2 is 6).
\( 2\text{KMnO}_4 + 3\text{Na}_2\text{SO}_3 \rightarrow 2\text{MnO}_2 + 3\text{Na}_2\text{SO}_4 \)

Balance K: 2K on left, need 2K on right, so add \( 2\text{KOH} \).
\( 2\text{KMnO}_4 + 3\text{Na}_2\text{SO}_3 \rightarrow 2\text{MnO}_2 + 3\text{Na}_2\text{SO}_4 + 2\text{KOH} \)

Balance Na: 6 Na on left, 6 Na on right (from \( 3\text{Na}_2\text{SO}_4 \)).
Balance H: 2 H on right from \( 2\text{KOH} \), no H on left. Add \( \text{H}_2\text{O} \) to the left.
\( 2\text{KMnO}_4 + 3\text{Na}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{Na}_2\text{SO}_4 + 2\text{KOH} \)

Balance O: Left: \( (2 \times 4) + (3 \times 3) + 1 = 8 + 9 + 1 = 18 \) O. Right: \( (2 \times 2) + (3 \times 4) + 2 = 4 + 12 + 2 = 18 \) O. Oxygen is balanced.
The final balanced equation is: \( 2\text{KMnO}_4 + 3\text{Na}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 3\text{Na}_2\text{SO}_4 + 2\text{KOH} \).

(iii) Balancing \( \text{Cu} + \text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + \text{NO}_2 + \text{H}_2\text{O} \)
This reaction occurs in an acidic medium.
Oxidation numbers:
Cu: \( 0 \rightarrow +2 \)
N in \( \text{HNO}_3 \): \( +5 \rightarrow +4 \) (in \( \text{NO}_2 \))

Changes:
Cu: \( 0 \rightarrow +2 \) (Loss of 2 electrons)
N: \( +5 \rightarrow +4 \) (Gain of 1 electron)

Multiply N change by 2 to balance electrons.
\( \text{Cu} + 2\text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + \text{H}_2\text{O} \)
Wait, there are more nitrogen atoms on the product side (2 in \( \text{Cu}(\text{NO}_3)_2 \) and 2 in \( \text{NO}_2 \)). This means we need 4 \( \text{HNO}_3 \) on the reactant side, as some \( \text{HNO}_3 \) acts as the oxidizing agent and some provides the nitrate ions.
\( \text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + \text{H}_2\text{O} \)

Balance H: 4 H on left, 2 H on right in \( \text{H}_2\text{O} \). Need \( 2\text{H}_2\text{O} \).
\( \text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O} \)

Balance O: Left: \( 4 \times 3 = 12 \) O. Right: \( (2 \times 3) + (2 \times 2) + (2 \times 1) = 6 + 4 + 2 = 12 \) O. Oxygen is balanced.
The final balanced equation is: \( \text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu}(\text{NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O} \).

(iv) Balancing \( \text{KMnO}_4 + \text{H}_2\text{C}_2\text{O}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{MnSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \)
This reaction occurs in an acidic medium.
Oxidation numbers:
Mn in \( \text{KMnO}_4 \): \( +7 \rightarrow +2 \) (in \( \text{MnSO}_4 \)). Gain of 5 electrons.
C in \( \text{H}_2\text{C}_2\text{O}_4 \): \( 2(+1) + 2x + 4(-2) = 0 \implies 2 + 2x - 8 = 0 \implies 2x = 6 \implies x = +3 \).
C in \( \text{CO}_2 \): \( x + 2(-2) = 0 \implies x = +4 \). Loss of 1 electron per C atom. Since \( \text{H}_2\text{C}_2\text{O}_4 \) has 2 C atoms, total loss \( = 2 \times 1 = 2 \) electrons.

Multiply Mn change by 2 and C change by 5 to balance electrons (LCM of 5 and 2 is 10).
\( 2\text{KMnO}_4 \) (gain 10e) and \( 5\text{H}_2\text{C}_2\text{O}_4 \) (loss 10e, forms \( 10\text{CO}_2 \)).
Partial equation: \( 2\text{KMnO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 10\text{CO}_2 + \text{H}_2\text{O} \)

Balance K: 2K on left, 2K on right (from \( \text{K}_2\text{SO}_4 \)). Balanced.
Balance Mn: 2Mn on left, 2Mn on right. Balanced.
Balance C: 10C on left, 10C on right. Balanced.

Balance S: On the right, \( \text{K}_2\text{SO}_4 \) has 1 S, \( 2\text{MnSO}_4 \) has 2 S. Total 3 S. On the left, \( \text{H}_2\text{SO}_4 \) has 1 S. Need \( 3\text{H}_2\text{SO}_4 \).
\( 2\text{KMnO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 10\text{CO}_2 + \text{H}_2\text{O} \)

Balance H: On the left, \( 5\text{H}_2\text{C}_2\text{O}_4 \) has 10 H, \( 3\text{H}_2\text{SO}_4 \) has 6 H. Total 16 H. On the right, \( \text{H}_2\text{O} \) has 2 H. Need \( 8\text{H}_2\text{O} \).
\( 2\text{KMnO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 10\text{CO}_2 + 8\text{H}_2\text{O} \)

Balance O: Left: \( (2 \times 4) + (5 \times 4) + (3 \times 4) = 8 + 20 + 12 = 40 \) O. Right: \( 4 + (2 \times 4) + (10 \times 2) + 8 = 4 + 8 + 20 + 8 = 40 \) O. Oxygen is balanced.
The final balanced equation is: \( 2\text{KMnO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 10\text{CO}_2 + 8\text{H}_2\text{O} \).
In simple words: To balance these reactions, we look at how the 'charge numbers' of key atoms change. We make sure the total loss of charge equals the total gain of charge by adjusting the numbers of molecules. Then, we balance all other atoms like hydrogen and oxygen using water and hydrogen ions.

🎯 Exam Tip: When balancing by the oxidation number method, always identify the atoms that change oxidation states first, then balance the electron transfer before balancing the remaining atoms by inspection.

 

Question 45. Balance the following equations by ion electron method.
(i) \( \text{KMnO}_4 + \text{SnCl}_2 + \text{HCl} \rightarrow \text{MnCl}_2 \text{ SnCl}_4 + \text{H}_2\text{O} + \text{KCl} \)
Answer:
(i) Balancing \( \text{KMnO}_4 + \text{SnCl}_2 + \text{HCl} \rightarrow \text{MnCl}_2 \text{ SnCl}_4 + \text{H}_2\text{O} + \text{KCl} \)
This is an acidic medium reaction.
Write the ionic equation by removing spectator ions (K\(^{+}\), Cl\(^{-}\)).
\( \text{MnO}_4^{-} + \text{Sn}^{2+} + \text{H}^{+} \rightarrow \text{Mn}^{2+} + \text{Sn}^{4+} + \text{H}_2\text{O} \)

Step 1: Write the two half-reactions.
Oxidation: \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} \)
Reduction: \( \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} \)

Step 2: Balance atoms other than O and H.
Sn is balanced. Mn is balanced.

Step 3: Balance O atoms by adding \( \text{H}_2\text{O} \).
Oxidation: \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} \) (No O atoms)
Reduction: \( \text{MnO}_4^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

Step 4: Balance H atoms by adding \( \text{H}^{+} \).
Oxidation: \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} \)
Reduction: \( \text{MnO}_4^{-} + 8\text{H}^{+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

Step 5: Balance charge by adding electrons.
Oxidation: \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2\text{e}^{-} \) (Loss of 2 electrons)
Reduction: \( \text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \) (Gain of 5 electrons)

Step 6: Equalize the number of electrons.
Multiply oxidation half-reaction by 5: \( 5\text{Sn}^{2+} \rightarrow 5\text{Sn}^{4+} + 10\text{e}^{-} \)
Multiply reduction half-reaction by 2: \( 2\text{MnO}_4^{-} + 16\text{H}^{+} + 10\text{e}^{-} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \)

Step 7: Add the two half-reactions and simplify.
\( 2\text{MnO}_4^{-} + 16\text{H}^{+} + 5\text{Sn}^{2+} + 10\text{e}^{-} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 5\text{Sn}^{4+} + 10\text{e}^{-} \)
Cancel electrons:
\( 2\text{MnO}_4^{-} + 5\text{Sn}^{2+} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 5\text{Sn}^{4+} + 8\text{H}_2\text{O} \)

(ii) Balancing \( \text{C}_2\text{O}_4^{2-} + \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} + \text{CO}_2 \) (in acid medium)

Step 1: Write the two half-reactions.
Oxidation: \( \text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_2 \)
Reduction: \( \text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+} \)

Step 2: Balance atoms other than O and H.
Oxidation: \( \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 \) (Balance C)
Reduction: \( \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} \) (Balance Cr)

Step 3: Balance O atoms by adding \( \text{H}_2\text{O} \).
Oxidation: \( \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 \) (O is already balanced)
Reduction: \( \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)

Step 4: Balance H atoms by adding \( \text{H}^{+} \).
Oxidation: \( \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 \)
Reduction: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)

Step 5: Balance charge by adding electrons.
Oxidation: \( \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^{-} \) (Left: \( -2 \), Right: \( -2 \). Loss of 2 electrons)
Reduction: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6\text{e}^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \) (Left: \( -2 + 14 = +12 \). Right: \( +6 \). Gain of 6 electrons)

Step 6: Equalize the number of electrons.
Multiply oxidation half-reaction by 3: \( 3\text{C}_2\text{O}_4^{2-} \rightarrow 6\text{CO}_2 + 6\text{e}^{-} \)
Reduction half-reaction remains the same: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6\text{e}^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)

Step 7: Add the two half-reactions and simplify.
\( 3\text{C}_2\text{O}_4^{2-} + \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6\text{e}^{-} \rightarrow 6\text{CO}_2 + 6\text{e}^{-} + 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)
Cancel electrons:
\( 3\text{C}_2\text{O}_4^{2-} + \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} \rightarrow 6\text{CO}_2 + 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)

(iii) Balancing \( \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + \text{NaI} \) (in acid medium)
Write the ionic equation:
\( \text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + \text{I}^{-} \)

Step 1: Write the two half-reactions.
Oxidation: \( \text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} \)
Reduction: \( \text{I}_2 \rightarrow \text{I}^{-} \)

Step 2: Balance atoms other than O and H.
Oxidation: \( 2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} \) (Balance S)
Reduction: \( \text{I}_2 \rightarrow 2\text{I}^{-} \) (Balance I)

Step 3 & 4: Balance O and H atoms.
Oxidation: Oxygen is balanced (6 on left, 6 on right). No H atoms.
Reduction: No O or H atoms.

Step 5: Balance charge by adding electrons.
Oxidation: \( 2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{e}^{-} \) (Left: \( -4 \), Right: \( -2 + (-2) = -4 \). Loss of 2 electrons)
Reduction: \( \text{I}_2 + 2\text{e}^{-} \rightarrow 2\text{I}^{-} \) (Left: \( -2 \), Right: \( -2 \). Gain of 2 electrons)

Step 6: Equalize electrons (already equal).

Step 7: Add the two half-reactions.
\( 2\text{S}_2\text{O}_3^{2-} + \text{I}_2 + 2\text{e}^{-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{e}^{-} + 2\text{I}^{-} \)
Cancel electrons:
\( 2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^{-} \)

Converting back to molecular form:
\( 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI} \).

(iv) Balancing \( \text{Zn} + \text{NO}_3^{-} \rightarrow \text{Zn}^{+2} + \text{NO} \) (in acid medium)

Step 1: Write the two half-reactions.
Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} \)
Reduction: \( \text{NO}_3^{-} \rightarrow \text{NO} \)

Step 2: Balance atoms other than O and H.
Zn is balanced. N is balanced.

Step 3: Balance O atoms by adding \( \text{H}_2\text{O} \).
Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} \)
Reduction: \( \text{NO}_3^{-} \rightarrow \text{NO} + 2\text{H}_2\text{O} \)

Step 4: Balance H atoms by adding \( \text{H}^{+} \).
Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} \)
Reduction: \( \text{NO}_3^{-} + 4\text{H}^{+} \rightarrow \text{NO} + 2\text{H}_2\text{O} \)

Step 5: Balance charge by adding electrons.
Oxidation: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^{-} \) (Loss of 2 electrons)
Reduction: \( \text{NO}_3^{-} + 4\text{H}^{+} + 3\text{e}^{-} \rightarrow \text{NO} + 2\text{H}_2\text{O} \) (Left: \( -1+4 = +3 \). Right: \( 0 \). Gain of 3 electrons)

Step 6: Equalize the number of electrons.
Multiply oxidation half-reaction by 3: \( 3\text{Zn} \rightarrow 3\text{Zn}^{2+} + 6\text{e}^{-} \)
Multiply reduction half-reaction by 2: \( 2\text{NO}_3^{-} + 8\text{H}^{+} + 6\text{e}^{-} \rightarrow 2\text{NO} + 4\text{H}_2\text{O} \)

Step 7: Add the two half-reactions and simplify.
\( 3\text{Zn} + 2\text{NO}_3^{-} + 8\text{H}^{+} + 6\text{e}^{-} \rightarrow 3\text{Zn}^{2+} + 6\text{e}^{-} + 2\text{NO} + 4\text{H}_2\text{O} \)
Cancel electrons:
\( 3\text{Zn} + 2\text{NO}_3^{-} + 8\text{H}^{+} \rightarrow 3\text{Zn}^{2+} + 2\text{NO} + 4\text{H}_2\text{O} \)
In simple words: For each reaction, we split it into two parts: one where electrons are lost (oxidation) and one where electrons are gained (reduction). We balance these parts separately for all atoms and charges, then combine them so that the number of electrons lost matches the number gained.

🎯 Exam Tip: The ion-electron method is robust: always follow the steps (half-reactions, non-O/H atoms, O with \( \text{H}_2\text{O} \), H with \( \text{H}^{+} \) or \( \text{OH}^{-} \), then charge with \( \text{e}^{-} \)) systematically to avoid errors, especially in acidic or basic media.

I. Choose the best answer:

 

Question 1. ______ consists of more than one chemical entity present without any chemical interactions.
(a) Mixtures
(b) Pure substances
(c) Compounds
(d) Elements
Answer: (a) Mixtures
In simple words: Mixtures are formed when different substances are put together but do not chemically combine with each other. You can still see or separate the original parts.

🎯 Exam Tip: Remember that in a mixture, the substances keep their own properties and can be easily separated, unlike in a compound where they bond chemically.

 

Question 2. ______ are made up of molecules which contain two or more atoms of different elements.
(a) Mixtures
(b) Compounds
(c) Pure substances
(d) Elements
Answer: (b) Compounds
In simple words: Compounds are substances formed when atoms of different elements join together chemically to make new molecules. Water (H2O) is a common example of a compound.

🎯 Exam Tip: The key difference for compounds is that atoms are chemically bonded, forming new substances with different properties than their original elements.

 

Question 3. Match the correct pair:

(a) A – ii, B – i, C – iii
(b) A – i, B – ii, C – iii
(c) A – ii, B – iii, C – i
(d) A – iii, B – ii, C – i
Answer: (c) A – ii, B – iii, C – i
In simple words: Glucose is a compound, air is a mixture, and sulfur (S8) is an element. Each type of matter has its own way of being put together.

🎯 Exam Tip: When matching, remember that compounds are chemically combined substances, mixtures are physically combined, and elements are the simplest pure forms of matter.

 

Question 4. ______ atom is considered as standard by the IUPAC for calculating atomic masses.
(a) C – 13
(b) C – 15
(c) C – 14
(d) C – 12
Answer: (d) C – 12
In simple words: Carbon-12 is used as a standard to measure the weight of all other atoms. It helps us compare and understand how heavy different atoms are.

🎯 Exam Tip: Carbon-12 is the international standard for atomic mass units, providing a consistent reference for all elements.

 

Question 5. The value of unified mass is equal to
(a) \( 1.6605 \times 10^{-27} \) kg
(b) \( 1.6736 \times 10^{-27} \) kg
(c) \( 1.6605 \times 10^{27} \) kg
(d) \( 1.6736 \times 10^{-29} \) kg
Answer: (a) \( 1.6605 \times 10^{-27} \) kg
In simple words: One unified atomic mass unit is a very tiny amount of mass. It's almost the same as the mass of a single proton or neutron.

🎯 Exam Tip: Remember this specific value for the unified atomic mass unit (amu or u) as it's a fundamental constant in chemistry calculations.

 

Question 6. Chlorine consists of two naturally occurring isotopes \( _{17}Cl^{35} \) and \( _{17}Cl^{37} \) in the ratio 77 : 23. The average relative atomic mass of chlorine is
(a) 36.45 u
(b) 35.56 u
(c) 35.46 u
(d) 35.65 u
Answer: (c) 35.46 u
In simple words: Chlorine atoms come in two slightly different weights. By taking how much of each weight exists naturally, we find the average weight, which is 35.46 u. This average helps us understand the typical mass of chlorine.

🎯 Exam Tip: To calculate average atomic mass, multiply the mass of each isotope by its natural abundance (as a decimal) and add the results.

 

Question 7. The relative atomic masses of hydrogen, oxygen, and carbon are 1.008 u, 16 u, and 12 u respectively. The relative molecular mass of glucose (\( C_6H_{12}O_6 \)) is
(a) 170.096 u
(b) 189.096 u
(c) 180.096 u
(d) 190.086 u
Answer: (c) 180.096 u
In simple words: To find the total weight of a glucose molecule, we add up the weights of all its carbon, hydrogen, and oxygen atoms. This sum gives us the molecular mass of glucose.

🎯 Exam Tip: Always multiply the atomic mass of each element by its subscript in the chemical formula before summing them up to find the molecular mass.

 

Question 8. The specific amount of a substance is represented in SI unit is
(a) amu
(b) mole
(c) atomic mass
(d) equivalent mass
Answer: (b) mole
In simple words: The mole is a special unit in chemistry that helps us count a very large number of atoms or molecules. It is the official way to measure the "amount of substance."

🎯 Exam Tip: Remember that the mole (mol) is the SI unit for the amount of substance, directly linking macroscopic quantities to the number of particles.

 

Question 9. Commonly used medicines for treating heart burn and acidity are called
(a) antipyretics
(b) analgesics
(c) antiseptics
(d) antacids
Answer: (d) antacids
In simple words: Medicines like antacids help calm down stomach acid when you have heartburn or acidity. They work by making the stomach less acidic.

🎯 Exam Tip: Antacids are bases that neutralize excess stomach acid, providing relief from heartburn and indigestion.

 

Question 10. The typical concentration of hydrochloric acid in gastric acid is
(a) 0.1 M
(b) 0.01 M
(c) 0.082 M
(d) 0.82 M
Answer: (c) 0.082 M
In simple words: The acid in our stomach, called gastric acid, usually has a concentration of about 0.082 M. This amount of acid helps our body digest food properly.

🎯 Exam Tip: Knowing the typical concentration of gastric acid helps understand its role in digestion and how antacids work to balance it.

 

Question 11. Antacids used to treat acidity contain mostly
(a) \( Al(OH)_3 \)
(b) \( Mg(OH)_2 \)
(c) (a) or (b)
(d) \( Ca(OH)_2 \)
Answer: (c) (a) or (b)
In simple words: Antacids often use aluminum hydroxide or magnesium hydroxide. These substances are bases that help to neutralize the extra acid in your stomach, making it feel better.

🎯 Exam Tip: Both aluminum hydroxide and magnesium hydroxide are weak bases commonly found in antacids to counteract excess stomach acid.

 

Question 13. The volume occupied by one mole of any substance in the gaseous state at 273 K and 1 atm pressure is ______ (in litres)
(a) 24.5
(b) 22.4
(e) 22.71
(d) 21.18
Answer: (b) 22.4
In simple words: At standard conditions (0°C and 1 atmosphere of pressure), one mole of any gas will always take up 22.4 litres of space. This is a very useful rule in chemistry.

🎯 Exam Tip: Remember that 22.4 L is the molar volume of an ideal gas at STP (Standard Temperature and Pressure: 0°C or 273 K and 1 atm), which is crucial for gas calculations.

 

Question 14. The equivalent mass of \( KMnO_4 \) (Molar mass 158 g \( mol^{-1} \)) based on the equation \( MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \)
(a) 158
(b) 52.66
(c) 31.6
(d) 40
Answer: (c) 31.6
In simple words: To find the equivalent mass of \( KMnO_4 \) in this reaction, we divide its total molar mass by the number of electrons it gains. Since it gains 5 electrons, the equivalent mass is 158 divided by 5, which gives 31.6.

🎯 Exam Tip: The equivalent mass of an oxidizing agent is its molar mass divided by the number of electrons gained in the reaction.

 

Question 15. The molecular formula of acetic acid is \( C_2H_4O_2 \). Its empirical formula is
(b) \( C_2H_2O \)
(c) \( CH_2O_2 \)
(d) \( CH_2O \)
Answer: (d) \( CH_2O \)
In simple words: The empirical formula is the simplest whole-number ratio of atoms in a compound. For acetic acid (\( C_2H_4O_2 \)), if you divide all the numbers by 2, you get the simplest ratio of \( CH_2O \).

🎯 Exam Tip: Always find the greatest common divisor of the subscripts in the molecular formula to determine the empirical formula.

 

Question 16. The number of moles of hydrogen required to produce 20 moles of ammonia is
(a) 10
(b) 15
(c) 30
(d) 20
Answer: (c) 30
In simple words: To make ammonia, for every 2 parts of ammonia, you need 3 parts of hydrogen. So, to make 20 parts of ammonia, you will need 30 parts of hydrogen. This is based on the balanced chemical reaction for making ammonia.

🎯 Exam Tip: To solve stoichiometry problems like this, always start with a balanced chemical equation (\( N_2 + 3H_2 \rightarrow 2NH_3 \)) and use the mole ratios.

 

Question 17. Which of the following is/are redox reactions?
(i) \( 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \)
(ii) \( H_2S + Cl_2 \rightarrow 2HCl + S \)
(iii) \( CuO + C \rightarrow Cu + CO \)
(iv) \( S +H_2 \rightarrow H_2S \)
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iv)
(d) (iii) and (iv)
Answer: (d) (iii) and (iv)
In simple words: Redox reactions involve both losing and gaining electrons. In \( CuO + C \rightarrow Cu + CO \), oxygen is removed from copper, and carbon gains oxygen. In \( S + H_2 \rightarrow H_2S \), hydrogen is added to sulfur. These are both examples where electrons move around.

🎯 Exam Tip: A redox reaction always involves a change in the oxidation states of the reacting elements, meaning one element is oxidized (loses electrons) and another is reduced (gains electrons).

 

Question 19. Choose the correct pair:

(a) A – i, B – iv, C- iii, D – ii
(b) A – iii, B – i, C – iv, D – ii
(c) A – i, B – ii, C – iv, D – iii
(d) A – iii, B – iv, C – i, D – ii
Answer: (b) A – iii, B – i, C – iv, D – ii
In simple words: To find the oxidation number, we look at how many electrons an atom seems to have lost or gained. For chromium in dichromate, it's +6. For carbon in carbon dioxide, it's +4. For carbon in difluoromethane, it's 0. For manganese in manganese sulfate, it's +2.

🎯 Exam Tip: Remember the common oxidation states for elements and apply the rules for calculating oxidation numbers in compounds, especially that the sum of oxidation numbers in a neutral compound is zero, and in an ion, it equals the ion's charge.

 

Question 20. The change in the oxidation number of Manganese in the following reaction \( 2KMnO_4 + 5FeC_2O_4 + 8H_2SO_4 \rightarrow K_2SO_4 + 5Fe_2(SO_4)_3 + 8H_2O \) is
(a) +2 to +7
(b) +2 to +5
(c) +7 to +2
(d) +5 to +2
Answer: (c) +7 to +2
In simple words: In the starting compound \( KMnO_4 \), manganese has an oxidation number of +7. After the reaction, in \( MnSO_4 \) (which is part of \( K_2SO_4 + 5Fe_2(SO_4)_3 + 8H_2O \)), manganese has an oxidation number of +2. So, its oxidation number changes from +7 to +2.

🎯 Exam Tip: To find the change in oxidation number, calculate the oxidation state of the element in the reactant and then in the product, paying attention to polyatomic ions.

 

Question 21. Which of the following statements are correct about oxidation?
(i) Removal of oxygen
(ii) Addition of hydrogen
(iii) Loss of electron
(iv) Gain of electron
(v) Addition of oxygen
(vi) Removal of hydrogen
(a) i, ii, iv
(b) iii, v, vi
(c) i, iii, v
(d) iii, iv, v
Answer: (b) iii, v, vi
In simple words: Oxidation means losing electrons. It can also mean adding oxygen to something or taking hydrogen away from it. These are all ways to describe the same chemical process.

🎯 Exam Tip: Remember the acronym OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). Also, associate oxidation with gaining oxygen or losing hydrogen.

 

Question 22. Choose the correct oxidation number of oxygen in the following compounds.

(a) A -iii, B- iv, C – i, D – ii.
(b) A – iv, B – i, C – iii, D – ii
(c) A – iii, B – i, C – iv, D – ii
(d) A – iv, B – iii, C – i, D – ii
Answer: (c) A – iii, B – i, C – iv, D – ii
In simple words: Oxygen usually has an oxidation number of -2. But in some compounds, it changes. In \( KO_2 \), it is -1/2. In \( Na_2O_2 \), it is -1. In \( H_2O_2 \), it is -1. In \( OF_2 \), it is +2 because fluorine is more electronegative.

🎯 Exam Tip: Remember the exceptions for oxygen's oxidation state: -1 in peroxides, -1/2 in superoxides, and positive values when bonded to fluorine (e.g., +2 in \( OF_2 \)).

 

Question 23. The correct order of electron releasing tendency of the following elements is
(a) \( Zn > Cu > Ag \)
(b) \( Cu > Zn > Ag \)
(c) \( Ag > Zn > Cu \)
(d) \( Ag > Cu > Zn \)
Answer: (a) \( Zn > Cu > Ag \)
In simple words: Zinc likes to give away its electrons more easily than copper, and copper likes to give away electrons more easily than silver. This means zinc is the most reactive and silver is the least reactive among these three metals.

🎯 Exam Tip: The electron-releasing tendency (or reactivity) is generally found by referring to the activity series of metals, where more reactive metals are higher up and tend to lose electrons more easily.

 

Question 24. \( H_2O_2 \rightarrow 2 H_2O + O_2 \) is a
(a) Displacement reaction
(b) Combination reaction
(c) Decomposition reaction
(d) Disproportionate reaction
Answer: (d) Disproportionate reaction
In simple words: In this reaction, hydrogen peroxide both gets oxidized (forms oxygen) and reduced (forms water). When a single substance does both oxidation and reduction, it's called a disproportionation reaction.

🎯 Exam Tip: A disproportionation reaction is identified when an element in a single reactant is simultaneously oxidized and reduced to different oxidation states in the products.

 

Question 25. The molar mass and empirical formula mass of a compound are 78 and 13 respectively. The molecular formula of the compound is (Empirical formula is CH)
(a) \( C_2H_2O_2 \)
(b) \( C_2H_4 \)
(c) \( C_6H_6 \)
(d) \( C_3H_8 \)
Answer: (c) \( C_6H_6 \)
In simple words: The empirical formula is CH, which has a mass of 13. Since the actual molar mass is 78, which is 6 times 13, the molecular formula must have 6 times the atoms of the empirical formula. So, the formula becomes \( C_6H_6 \). Benzene is an example of a compound with this formula.

🎯 Exam Tip: To find the molecular formula from empirical formula and molar mass, divide the molar mass by the empirical formula mass to get a whole number, then multiply all subscripts in the empirical formula by that number.

II. Very Short Question and Answers (2 Marks):

 

Question 1. State Avogadro's Hypothesis.
Answer: Avogadro's Hypothesis states that if you have equal volumes of any two gases, and they are both at the same temperature and pressure, then they will contain the same number of molecules. This is true no matter what kind of gas it is. It helps us understand the relationship between gas volume and the number of particles.
In simple words: If you have two balloons of the same size, at the same temperature and pressure, they will both have the same number of gas particles inside, even if the gases are different.

🎯 Exam Tip: Remember the direct relationship: equal volumes of gases at identical conditions mean equal numbers of molecules.

 

Question 2. How is matter classified physically?
Answer: Matter can be classified into three main physical states: solids, liquids, and gases. Solids have a fixed shape and volume, like an ice cube. Liquids have a fixed volume but take the shape of their container, like water. Gases have no fixed shape or volume and fill their container, like steam. These states can change into one another by changing the temperature and pressure. For example, heating ice turns it into liquid water.
In simple words: Matter is grouped into solids (fixed shape), liquids (takes container's shape), and gases (no fixed shape). You can change these states by heating or pressing.

🎯 Exam Tip: Focus on the properties of shape and volume for each physical state (solid, liquid, gas) and remember that energy (temperature) and pressure influence state changes.

 

Question 3. How is matter classified chemically?
Answer: Matter is classified chemically into mixtures and pure substances. Pure substances are made of only one type of atom or molecule, like gold or water. Mixtures are made of two or more different substances that are physically mixed but not chemically joined, like salt water or air. These classifications help us understand how different materials are made and how they behave.
In simple words: Matter is split into pure substances (like gold or water) and mixtures (like salt water). Pure substances are single types of stuff, while mixtures are blends of different types.

🎯 Exam Tip: Distinguish between pure substances (elements and compounds, chemically uniform) and mixtures (homogeneous or heterogeneous, physically combined).

 

Question 4. Calculate a number of moles of carbon atoms in three moles of ethane.
Answer: Ethane has a molecular formula of \( C_2H_6 \). This means one molecule of ethane contains 2 carbon atoms. So, 1 mole of ethane contains 2 moles of carbon atoms. Therefore, 3 moles of ethane will contain \( 3 \times 2 = 6 \) moles of carbon atoms. This is \( 6 \times 6.022 \times 10^{23} \) carbon atoms, which is \( 3.6132 \times 10^{24} \) carbon atoms.
In simple words: Since each ethane molecule has two carbon atoms, three moles of ethane will have six moles of carbon atoms in total.

🎯 Exam Tip: The subscript of an element in a chemical formula tells you the number of moles of that atom present in one mole of the compound.

 

Question 5. What are pure substances? How are they classified?
Answer: Pure substances are types of matter that have a constant composition and distinct properties. They are always the same, no matter where they come from. Pure substances are classified into two groups:

• Elements: These are the simplest pure substances and cannot be broken down into simpler substances by chemical means, like gold (Au) or oxygen (\( O_2 \)). Each element has only one type of atom.
• Compounds: These are pure substances formed when two or more different elements are chemically combined in a fixed ratio, like water (\( H_2O \)) or carbon dioxide (\( CO_2 \)). Compounds have properties different from their constituent elements.

This classification helps us organize and understand all the different types of matter in the world.
In simple words: Pure substances are materials that are always the same and have fixed properties. They are divided into elements (like gold) and compounds (like water).

🎯 Exam Tip: Understand that elements are single types of atoms, while compounds are chemically bonded combinations of two or more elements in a specific ratio.

 

Question 6. Mass of one atom of an element is \( 6.66 \times 10^{-23} \) g. How many moles of the element are there in 0.320 kg?
Answer:Given: Mass of one atom = \( 6.66 \times 10^{-23} \) g. To find the molar mass, we multiply the mass of one atom by Avogadro's number: Molar mass = \( (6.66 \times 10^{-23} \text{ g/atom}) \times (6.022 \times 10^{23} \text{ atoms/mol}) \) Molar mass \( \approx 40.11 \) g/mol. Next, convert the given mass to grams: 0.320 kg = \( 0.320 \times 1000 \) g = 320 g. Now, calculate the number of moles: No. of moles = \( \frac{\text{Mass}}{\text{Molar mass}} = \frac{320 \text{ g}}{40.11 \text{ g/mol}} \approx 7.97 \text{ moles} \) Rounding to a common value, approximately 8 moles of the element are present in 0.320 kg. This calculation shows how a very small atomic mass can add up to a significant mass in a large sample.
In simple words: First, find the total weight of one mole of the element using the weight of one atom. Then, divide the total given weight (0.320 kg converted to grams) by this molar mass to find out how many moles are present.

🎯 Exam Tip: Always convert all mass units to grams for calculations involving molar mass and Avogadro's number to ensure consistency.

 

Question 7. What are compounds? Give examples.
Answer: Compounds are pure substances made up of molecules that contain two or more different types of atoms, which are chemically joined together. These atoms are always present in fixed proportions. For example, water (\( H_2O \)) is a compound where hydrogen and oxygen atoms are chemically bonded. Another example is sodium chloride (NaCl), which is common table salt. In a compound, the properties of the compound are completely different from the properties of the individual elements that make it up.
In simple words: Compounds are substances where different atoms are chemically stuck together in a set way. Water (\( H_2O \)) and salt (NaCl) are good examples.

🎯 Exam Tip: Emphasize that compounds involve chemical bonding between different elements and have unique properties distinct from their constituent elements.

 

Question 8. Calculate the weight of 0.2 moles of sodium carbonate.
Answer: The chemical formula for sodium carbonate is \( Na_2CO_3 \). First, we calculate its molecular mass: Atomic mass of Na = 23 g/mol Atomic mass of C = 12 g/mol Atomic mass of O = 16 g/mol Molecular mass of \( Na_2CO_3 = (2 \times 23) + (1 \times 12) + (3 \times 16) = 46 + 12 + 48 = 106 \) g/mol. This means 1 mole of sodium carbonate weighs 106 g. To find the weight of 0.2 moles: Weight = No. of moles \( \times \) Molecular mass = \( 0.2 \text{ mol} \times 106 \text{ g/mol} = 21.2 \) g. So, 0.2 moles of sodium carbonate weigh 21.2 g. This calculation helps determine the mass of a substance for a given number of moles.
In simple words: First, find the total weight of one molecule of sodium carbonate. Then, multiply that by 0.2 to get the weight for 0.2 moles.

🎯 Exam Tip: Always determine the molar mass of the compound correctly before calculating the mass for a given number of moles.

 

Question 9. Define relative atomic mass.
Answer: Relative atomic mass (\( A_r \)) is defined as the ratio of the average mass of an atom of an element to one-twelfth of the mass of a carbon-12 atom (which is the unified atomic mass unit). It tells us how many times heavier, on average, an atom of a particular element is compared to this standard. Since it's a ratio, relative atomic mass has no units. It helps chemists compare the masses of different atoms easily.
In simple words: Relative atomic mass compares the average weight of an atom of any element to a small piece of a carbon-12 atom. It shows how much heavier one atom is than this standard.

🎯 Exam Tip: Remember that relative atomic mass is a ratio, so it's unitless, and the standard reference is 1/12th the mass of a carbon-12 atom.

 

Question 10. What is the average atomic mass?
Answer: The average atomic mass is the weighted average of the atomic masses of all the naturally occurring isotopes of an element. Since most elements have several isotopes (atoms of the same element with different numbers of neutrons), their individual atomic masses are slightly different. The average atomic mass takes into account both the mass of each isotope and its natural abundance (how common it is) to give a single, representative mass for the element. This value is usually what you see on the periodic table.
In simple words: Average atomic mass is like a weighted average of all the different "weights" (isotopes) an element can have. It considers how much of each "weight" exists naturally to give one overall average weight for the element.

🎯 Exam Tip: Highlight that average atomic mass accounts for both the mass and the natural abundance of each isotope, making it a weighted average.

 

Question 11. Calculate the equivalent mass of barium hydroxide.
Answer: The chemical formula for barium hydroxide is \( Ba(OH)_2 \). First, calculate its molecular mass: Atomic mass of Ba = 137 g/mol Atomic mass of O = 16 g/mol Atomic mass of H = 1 g/mol Molecular mass of \( Ba(OH)_2 = 137 + (2 \times 16) + (2 \times 1) = 137 + 32 + 2 = 171 \) g/mol. Barium hydroxide is a base, and its acidity (or valency in this context for equivalent mass) is determined by the number of hydroxyl (\( OH^- \)) ions it can release. In \( Ba(OH)_2 \), there are two \( OH^- \) groups, so its acidity is 2. Equivalent mass = \( \frac{\text{Molecular mass}}{\text{Acidity}} = \frac{171 \text{ g/mol}}{2} = 85.5 \) g/eq. This means the equivalent mass of barium hydroxide is 85.5 g. Understanding equivalent mass is important for stoichiometry calculations in acid-base reactions.
In simple words: To find the equivalent mass of barium hydroxide, first find its total molecular weight. Then, divide that weight by the number of \( OH^- \) groups it has (which is 2), giving 85.5 g/eq.

🎯 Exam Tip: For bases, the equivalent mass is the molar mass divided by the number of dissociable hydroxyl ions (\( OH^- \)), often called its acidity.

 

Question 12. What is a mole?
Answer: A mole is a unit of measurement in chemistry that represents a specific amount of a substance. It is defined as the amount of substance that contains as many elementary particles (like atoms, molecules, ions, or electrons) as there are atoms in exactly 12 grams of pure carbon-12 isotope. This number of particles is called Avogadro's number, which is approximately \( 6.022 \times 10^{23} \). The mole helps chemists deal with the incredibly large numbers of particles involved in chemical reactions, making calculations much simpler. It's like having a "dozen" for atoms.
In simple words: A mole is a way to count a very large group of tiny particles, like atoms or molecules. It's exactly \( 6.022 \times 10^{23} \) of them, and it's based on how many atoms are in 12 grams of carbon-12.

🎯 Exam Tip: Define the mole as the amount of substance containing Avogadro's number of particles (\( 6.022 \times 10^{23} \)) and mention its reference to carbon-12.

 

Question 13. What do you understand by the terms empirical formula and molecular formula?
Answer:The empirical formula and molecular formula are two ways to represent the composition of a chemical compound:

• Empirical Formula: This is the simplest whole-number ratio of the atoms present in a compound. It shows the relative number of each type of atom. For example, the empirical formula of glucose (\( C_6H_{12}O_6 \)) is \( CH_2O \) because the ratio of carbon, hydrogen, and oxygen atoms is 1:2:1.
• Molecular Formula: This shows the actual number of each type of atom present in a molecule of the compound. It gives the exact composition. For example, the molecular formula of glucose is \( C_6H_{12}O_6 \), indicating exactly 6 carbon, 12 hydrogen, and 6 oxygen atoms. The molecular formula is always a whole-number multiple of the empirical formula.

These formulas are crucial for understanding the structure and properties of chemical substances.
In simple words: An empirical formula shows the simplest ratio of atoms in a compound, like \( CH_2O \) for sugar. A molecular formula shows the exact number of atoms, like \( C_6H_{12}O_6 \) for sugar.

🎯 Exam Tip: Clearly state that the empirical formula is the *simplest ratio*, while the molecular formula is the *actual count* of atoms in a molecule.

 

Question 14. What is Avogadro's number?
Answer: One mole of any substance always contains a specific number of particles, which is \( 6.022 \times 10^{23} \). This special number is known as Avogadro's number. It helps us count very tiny atoms or molecules in large quantities.
In simple words: Avogadro's number is a huge number (6.022 followed by 23 zeros) that tells you how many tiny particles (like atoms or molecules) are in one "mole" of any substance.

🎯 Exam Tip: Remember Avogadro's number as the bridge between the microscopic world of atoms and the macroscopic amounts we measure in labs.

 

Question 15. State Avogadro hypothesis.
Answer: Avogadro's hypothesis states that if you have equal volumes of different gases, and they are all at the same temperature and pressure, then they will all contain the same number of molecules. This concept shows that the volume of a gas depends on the number of particles, not their size.
In simple words: If you take two different gases but give them the same space, keep them at the same heat, and squeeze them the same amount, they will have the same number of gas particles inside.

🎯 Exam Tip: The key conditions for Avogadro's hypothesis are equal volume, same temperature, and same pressure. All three must be met.

 

Question 16. Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer: The molecular mass of sodium nitrate \( (NaNO_3) \) is calculated as: \( 23 (Na) + 14 (N) + (3 \times 16) (O) = 85 \, g/mol \).
This means \( 85 \, g \) of pure sodium nitrate contains \( 23 \, g \) of sodium.
To find the amount of sodium in \( 95 \, kg \) (or \( 95 \times 10^3 \, g \)) of pure sodium nitrate, we use a ratio:
Mass of Sodium \( = \frac{23 \, g \, Na}{85 \, g \, NaNO_3} \times 95 \times 10^3 \, g \, NaNO_3 = 25.70 \times 10^3 \, g \, Na \).
The sample is only \( 70\% \) pure. So, we calculate the actual mass of sodium:
Mass of Sodium in impure sample \( = 70\% \, \text{of} \, 25.70 \times 10^3 \, g = 0.70 \times 25.70 \times 10^3 \, g = 17.99 \times 10^3 \, g \).
This is equal to \( 17990 \, g \) or \( 17.99 \, kg \) of sodium. This calculation helps in determining the usable amount of a specific element from an impure sample.
In simple words: First, we find out how much sodium is in pure sodium nitrate. Then, we use the given amount of impure sodium nitrate and its purity (70%) to figure out the actual mass of sodium present.

🎯 Exam Tip: Always remember to account for the percentage purity of a sample; only the pure portion contributes to the reaction or elemental content.

 

Question 17. Define molar volume.
Answer: Molar volume is defined as the volume that is occupied by one mole of any substance when it is in the gaseous state, under specific conditions of temperature and pressure. For instance, at standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 liters.
In simple words: Molar volume is the space that one "mole" of any gas takes up when it's at a certain heat and pressure.

🎯 Exam Tip: Note that molar volume is specifically for substances in the gaseous state, not liquids or solids, and depends on temperature and pressure.

 

Question 18. What is gram equivalent mass?
Answer: Gram equivalent mass refers to the mass of an element, compound, or ion that can combine with or displace a fixed amount of hydrogen (1.008 g), oxygen (8 g), or chlorine (35.5 g). This concept is useful for understanding how different substances react chemically in terms of their combining power. It can also be calculated by dividing the molar mass of a substance by its equivalence factor.
Gram equivalent mass \( = \frac{\text{Molar mass}(g \, mol^{-1})}{\text{Equivalence factor}(eq \, mol^{-1})} \)
In simple words: Gram equivalent mass is how much of a substance reacts with a fixed amount of another substance, like hydrogen or oxygen. You can also find it by dividing the substance's total mass (molar mass) by how many parts it reacts with (equivalence factor).

🎯 Exam Tip: Remember that the "equivalence factor" in the formula for gram equivalent mass changes depending on the type of reaction (acid-base, redox, etc.) for a given substance.

 

Question 19. What is meant by Plasma state? Give an example.
Answer: The Plasma state is a special form of matter that exists at very high temperatures. In this state, gas particles become so energetic that their atoms lose electrons, forming a mixture of positively charged ions and free-moving electrons. This makes plasma an electrically conductive substance, often called the "fourth state of matter." A common example of plasma is lightning.
In simple words: Plasma is a super-hot gas where atoms have broken apart into charged pieces and free electrons. It's like a special kind of electric gas. Lightning is an example of plasma.

🎯 Exam Tip: Plasma is distinct from gas because its particles are ionized, making it highly conductive and responsive to magnetic fields.

 

Question 20. What is the acidity of a base? Give an example.
Answer: The acidity of a base tells us how many hydroxide \( (OH^-) \) ions one mole of that base can release in a solution. For instance, potassium hydroxide \( (KOH) \) has an acidity of 1 because each mole of KOH releases one mole of \( OH^- \) ions. This helps us understand how strong a base is and how it will react.
In simple words: The acidity of a base is simply how many \( OH^- \) (hydroxide) particles one molecule of the base can give away. For example, the base KOH has an acidity of 1.

🎯 Exam Tip: Acidity of a base is analogous to basicity of an acid; both refer to the number of ionizable groups per molecule.

 

Question 21. What is empirical formula of a compound?
Answer: The empirical formula of a compound shows the simplest whole-number ratio of the different types of atoms present in one molecule of that compound. It is written using atomic symbols with subscripts to indicate these ratios. For example, the empirical formula for glucose \( (C_6H_{12}O_6) \) is \( CH_2O \), showing the 1:2:1 ratio of carbon, hydrogen, and oxygen.
In simple words: An empirical formula is like a simplified recipe for a compound, showing the smallest whole number of each type of atom it contains.

🎯 Exam Tip: Always ensure the ratios in an empirical formula are the simplest whole numbers; if they can be divided further, it's not the empirical formula yet.

 

Question 22. Chlorine has a fractional average atomic mass. Justify this statement.
Answer: Chlorine has a fractional average atomic mass because it naturally exists as a mixture of two main isotopes: chlorine-35 (\( ^{35}Cl \)) and chlorine-37 (\( ^{37}Cl \)). These isotopes have different masses. Since they are found in a specific abundance ratio (77% \( ^{35}Cl \) and 23% \( ^{37}Cl \)), the average atomic mass is calculated by taking a weighted average of their masses.
The average relative atomic mass of Chlorine \( = \frac{(35 \times 77) + (37 \times 23)}{100} = \frac{2695 + 851}{100} = \frac{3546}{100} = 35.46 \, u \).
This weighted calculation results in a non-whole number, which is its fractional atomic mass.
In simple words: Chlorine has a decimal atomic mass because it's a mix of two slightly different kinds of chlorine atoms (isotopes) found in nature. We average their weights based on how much of each there is.

🎯 Exam Tip: Atomic masses on the periodic table are average atomic masses, which are fractional unless an element has only one naturally occurring isotope.

 

Question 23. What is meant by Stoichiometry?
Answer: Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a balanced chemical reaction. It allows chemists to calculate the precise amounts of substances consumed or produced in a reaction, expressed in moles, mass units, or volumes. For example, it helps predict how much product will form from a given amount of starting material.
In simple words: Stoichiometry is like chemistry's math. It helps us figure out exactly how much of each chemical we need to mix, and how much new chemical we will get, based on a balanced recipe.

🎯 Exam Tip: Stoichiometry relies heavily on balanced chemical equations, as the coefficients directly represent mole ratios.

 

Question 24. What are limiting and excess reagents?
Answer: In a chemical reaction, the limiting reagent is the reactant that is completely used up first, and it determines the maximum amount of product that can be formed. The excess reagent, on the other hand, is the reactant that is present in a greater quantity than needed, and some of it will be left over after the reaction has finished. Understanding these helps in optimizing reaction yields.
In simple words: The limiting reagent is the ingredient that runs out first in a chemical recipe, stopping the reaction. The excess reagent is the ingredient you have too much of, so some is left over.

🎯 Exam Tip: Always identify the limiting reagent first, as it dictates the theoretical yield of any product in a reaction.

 

Question 25. Define Avogadro Number.
Answer: Avogadro's number, denoted by N, is defined as the number of constituent particles (atoms, molecules, or ions) present in one mole of any substance. Its accepted value is approximately \( 6.022 \times 10^{23} \). This vast number bridges the microscopic world of atoms with the macroscopic world we can measure in labs.
In simple words: Avogadro's number is a very big number (\( 6.022 \times 10^{23} \)) that tells you how many tiny particles are in one "mole" of anything.

🎯 Exam Tip: Avogadro's number is a universal constant, meaning it is the same for one mole of any substance, whether it's water, iron, or carbon dioxide.

 

Question 26. The oxidation number is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to a set of rules.
Answer: The oxidation number is an imaginary charge assigned to an atom in a compound, assuming that all other atoms are removed in their usual oxidation states according to a specific set of rules. It represents the number of electrons an atom has gained or lost when forming a chemical bond. This concept helps us track electron transfer in redox reactions.
In simple words: An oxidation number is a made-up charge on an atom in a compound, as if all other parts were taken away. It shows how many electrons the atom might have gained or lost.

🎯 Exam Tip: Carefully follow the rules for assigning oxidation numbers, especially for elements like oxygen and hydrogen, which can have different values in specific compounds.

 

Question 27. Calculate the equivalent mass of Copper. (Atomic mass of copper = 63.5)
Answer: To calculate the equivalent mass of copper, we divide its atomic mass by its valency. Assuming copper has a valency of 2 in many compounds, the calculation is:
Equivalent mass \( = \frac{\text{Atomic mass}}{\text{Valency}} = \frac{63.5}{2} = 31.75 \, g \, eq^{-1} \).
The valency of copper is often 2 in common compounds like copper(II) sulfate.
In simple words: To find copper's equivalent mass, we take its atomic mass (63.5) and divide it by how many bonds it usually forms (its valency, which is often 2).

🎯 Exam Tip: The valency used in equivalent mass calculations can vary depending on the compound and the specific reaction copper is undergoing.

 

Question 28. Mention the types of redox reactions?
Answer: Redox reactions, which involve the transfer of electrons, can be categorized into several types. These include combination reactions (where substances combine), decomposition reactions (where a substance breaks down), displacement reactions (where one element replaces another), disproportionation reactions (where an element is both oxidized and reduced), and competitive electron transfer reactions. Each type represents a different pattern of electron exchange.
In simple words: There are different kinds of redox reactions, which are reactions where electrons move from one atom to another. Some types are when things combine, break apart, or swap places.

🎯 Exam Tip: Understanding the specific characteristics of each type of redox reaction helps in predicting products and balancing equations.

III. Short Question and Answers (3 Marks)

 

Question 1. Describe the chemical classification of matter.
Answer: Matter is defined as anything that has mass and occupies space. Essentially, anything you can touch, see, or feel is matter, and all matter is fundamentally composed of tiny particles called atoms. Matter can be broadly classified in two ways: physical classification and chemical classification.
Physical Classification: This categorizes matter based on its physical state, primarily into solids, liquids, and gases. These states are interconvertible by altering temperature and pressure. For instance, ice (solid water) can become liquid water, then water vapor (gas).
Chemical Classification: This groups matter based on its chemical composition. It divides matter into pure substances and mixtures.
* **Pure Substances:** These have a fixed chemical composition. * **Elements:** Made of only one type of atom (e.g., Gold, Au; Copper, Cu; Hydrogen, \( H_2 \); Phosphorus, \( P_4 \)). * **Compounds:** Formed when two or more different elements chemically combine in fixed ratios (e.g., carbon dioxide, \( CO_2 \); glucose, \( C_6H_{12}O_6 \)). The properties of compounds are distinct from the properties of the elements they are made from. For instance, sodium (a reactive metal) and chlorine (a poisonous gas) combine to form sodium chloride (table salt), which is a stable, edible solid essential for life. * **Mixtures:** These consist of two or more substances that are physically combined but not chemically bonded. They retain their individual properties and can be separated by physical methods. Mixtures can be homogeneous (uniform composition, e.g., green tea) or heterogeneous (non-uniform composition, e.g., oil and water).
Understanding these classifications is fundamental to chemistry.
In simple words: Matter is anything that has weight and takes up space, made of tiny atoms. We can sort matter by how it looks (solid, liquid, gas) or by what it's made of (pure stuff like elements and compounds, or mixed stuff). Pure elements are just one type of atom, while pure compounds are different atoms joined together. Mixtures are different things simply stirred or jumbled, either smoothly mixed (like tea) or visibly separate (like oil and water).

🎯 Exam Tip: Distinguishing between physical and chemical classifications is crucial. Physical changes alter state, while chemical changes alter composition.

 

Question 2. Distinguish between element and compound.
Answer: An element is a pure substance made up of only one type of atom. These atoms can exist alone (monoatomic, like Gold or Copper) or bonded together in small groups (polyatomic molecules, like Hydrogen or Phosphorus). A compound, however, is a pure substance formed when two or more different elements are chemically combined in a fixed ratio. For example, carbon dioxide (\( CO_2 \)) and glucose (\( C_6H_{12}O_6 \)) are compounds. Elements are the basic building blocks, while compounds are more complex substances formed from them.
In simple words: An element is made of only one kind of atom, like pure gold. A compound is made when two or more different kinds of atoms stick together, like water (\( H_2O \)) where hydrogen and oxygen join.

🎯 Exam Tip: Remember that elements cannot be broken down into simpler substances by chemical means, but compounds can be.

 

Question 3. What is average atomic mass? How is average atomic mass of chlorine calculated?
Answer: Average atomic mass is the weighted average of the atomic masses of all naturally occurring isotopes of an element. It accounts for the relative abundance of each isotope. For chlorine, which exists as two main isotopes, \( ^{35}Cl \) (77% abundant) and \( ^{37}Cl \) (23% abundant), the average atomic mass is calculated as:
Average atomic mass \( = \frac{(35 \times 77) + (37 \times 23)}{100} = \frac{2695 + 851}{100} = \frac{3546}{100} = 35.46 \, u \).
This weighted average explains why atomic masses often have decimal values on the periodic table.
In simple words: Average atomic mass is the average weight of all the different types of atoms (isotopes) of an element, taking into account how common each type is. For chlorine, we add up the weights of its two types of atoms, multiplied by how often they appear, and then divide by 100 to get the average.

🎯 Exam Tip: Always multiply each isotope's mass by its fractional abundance (percentage/100) before summing them up to get the weighted average atomic mass.

 

Question 4. What is mass of one \( ^{12}C \) atom in g?
Answer: The molar mass of carbon-12 \( (C^{12}) \) is \( 12.00 \, g \, mol^{-1} \). This means that one mole of carbon-12, which contains Avogadro's number of atoms (\( 6.022 \times 10^{23} \) atoms), has a mass of \( 12.0 \, g \). To find the mass of a single carbon-12 atom, we divide the molar mass by Avogadro's number:
Mass of one \( ^{12}C \) atom \( = \frac{12.0 \, g}{6.023 \times 10^{23} \, \text{atoms/mol}} \approx 1.992 \times 10^{-23} \, g \).
This incredibly small mass highlights the tiny nature of individual atoms.
In simple words: To find the mass of one carbon-12 atom, we take the mass of one mole of carbon-12 (12 grams) and divide it by Avogadro's number (the total number of atoms in a mole). This gives us a very, very small number for one atom's weight.

🎯 Exam Tip: Remember to use Avogadro's number when converting between macroscopic amounts (moles, grams) and microscopic quantities (number of atoms/molecules).

 

Question 5. Discuss the role of antacids.
Answer: Antacids are medications used to relieve heartburn and indigestion, which are often caused by an excess of hydrochloric acid in the stomach. Gastric acid normally helps in digestion, but if its concentration rises above 0.1M, it can lead to discomfort. Antacids work by containing weak bases, such as magnesium hydroxide (\( Mg(OH)_2 \)) or aluminum hydroxide (\( Al(OH)_3 \)), which neutralize the excess stomach acid through chemical reactions like:
\( 3HCl + Al(OH)_3 \rightarrow AlCl_3 + 3H_2O \)
\( 2HCl + Mg(OH)_2 \rightarrow MgCl_2 + 2H_2O \)
These reactions help to restore the stomach's pH balance, providing relief.
In simple words: Antacids help stop heartburn by neutralizing extra stomach acid. They have weak bases in them, like aluminum or magnesium hydroxide, which react with the acid to calm your stomach.

🎯 Exam Tip: Antacids provide temporary relief by neutralizing excess acid, but they don't treat the underlying cause of acid reflux.

 

Question 6. Explain gram equivalent mass.
Answer: Gram equivalent mass is the mass of a substance that can react with or displace a standard amount of another substance, specifically 1.008 grams of hydrogen, 8 grams of oxygen, or 35.5 grams of chlorine. For example, in the reaction \( Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2 \), 65.38 grams of zinc displace 2.016 grams of hydrogen. The mass of zinc required to displace 1.008 grams of hydrogen (which is half of 2.016 g) is its equivalent mass.
Mass of zinc required to displace 1.008 g hydrogen \( = \frac{65.38 \, g}{2.016 \, g} \times 1.008 \, g = 32.69 \, g \).
So, the equivalent mass of zinc is \( 32.69 \, g \, eq^{-1} \). This concept is vital for understanding stoichiometry and redox reactions.
Gram equivalent mass \( = \frac{\text{Molar mass}(g \, mol^{-1})}{\text{Equivalence factor}(eq \, mol^{-1})} \)
In simple words: Gram equivalent mass is how much of a substance (like an element or a compound) can combine with or replace a fixed amount of other common elements like hydrogen or oxygen. It helps us know how much of one chemical will react with another.

🎯 Exam Tip: The equivalence factor for a metal in a displacement reaction is typically its valency.

 

Question 7. Calculate the gram equivalent mass of sulphuric acid.
Answer: To calculate the gram equivalent mass of sulphuric acid \( (H_2SO_4) \), we first determine its basicity and molar mass. Sulphuric acid is a diprotic acid, so its basicity is 2, meaning it can donate two hydrogen ions. Its molar mass is calculated as: \( (2 \times 1) + (1 \times 32) + (4 \times 16) = 2 + 32 + 64 = 98 \, g/mol \).
The gram equivalent mass is then:
Equivalent mass \( = \frac{\text{Molar mass}}{\text{Basicity}} = \frac{98 \, g/mol}{2 \, eq/mol} = 49 \, g \, eq^{-1} \).
This value is important for acid-base titration calculations.
In simple words: For sulphuric acid, we divide its total weight (molar mass) by how many hydrogen atoms it can give away (its basicity, which is 2). This gives us its gram equivalent mass, 49 grams per equivalent.

🎯 Exam Tip: For acids, the equivalence factor is its basicity (number of replaceable hydrogen ions), while for bases, it's its acidity (number of replaceable hydroxide ions).

 

Question 8. Calculate the gram equivalent mass of potassium hydroxide.
Answer: To find the gram equivalent mass of potassium hydroxide \( (KOH) \), we need its acidity and molar mass. Potassium hydroxide is a monoprotic base, so its acidity is 1, as it releases one hydroxide ion. Its molar mass is calculated as: \( (1 \times 39) + (1 \times 16) + (1 \times 1) = 39 + 16 + 1 = 56 \, g/mol \).
The gram equivalent mass is then:
Equivalent mass \( = \frac{\text{Molar mass}}{\text{Acidity}} = \frac{56 \, g/mol}{1 \, eq/mol} = 56 \, g \, eq^{-1} \).
This value indicates the amount of \( KOH \) required to neutralize one equivalent of an acid.
In simple words: For potassium hydroxide, we divide its total weight (molar mass) by how many hydroxide groups it can give away (its acidity, which is 1). This results in a gram equivalent mass of 56 grams per equivalent.

🎯 Exam Tip: Strong bases like KOH are good examples for understanding basicity, as they fully dissociate in solution.

 

Question 9. Calculate the gram equivalent mass of Potassium permanganate.
Answer: Potassium permanganate \( (KMnO_4) \) acts as a strong oxidizing agent, and its equivalent mass depends on the reaction medium. In an acidic medium, \( KMnO_4 \) gains 5 electrons during reduction. Its molar mass is calculated as: \( (1 \times 39) + (1 \times 55) + (4 \times 16) = 39 + 55 + 64 = 158 \, g/mol \).
In an acidic medium, the reduction of permanganate involves the gain of 5 electrons, as shown:
\( MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \)
Since 5 electrons are gained, the equivalence factor (n) is 5. Thus, the gram equivalent mass is:
Gram equivalent mass \( = \frac{\text{Molar mass}}{\text{n}} = \frac{158 \, g/mol}{5 \, eq/mol} = 31.6 \, g \, eq^{-1} \).
The equivalent mass changes if the reaction medium is neutral or basic.
In simple words: In a specific type of reaction (acidic), potassium permanganate takes in 5 electrons. So, we divide its total weight (158 grams) by 5 to get its equivalent weight, which is 31.6 grams.

🎯 Exam Tip: The equivalent mass of oxidizing/reducing agents depends on the number of electrons gained or lost, which can vary with the reaction conditions (acidic, neutral, basic).

 

Question 10. An acid found in tamarind on analysis shows the following percentage composition: 32 % Carbon; 4% Hydrogen; 64 % Oxygen. Find the empirical formula of the compound.
Answer: To find the empirical formula, we follow these steps:
1. **Convert percentages to mass:** Assume 100g of compound, so 32g C, 4g H, 64g O. 2. **Calculate moles:** * C: \( \frac{32}{12} = 2.66 \, mol \) * H: \( \frac{4}{1} = 4 \, mol \) * O: \( \frac{64}{16} = 4 \, mol \) 3. **Find simplest mole ratio (divide by smallest, 2.66):** * C: \( \frac{2.66}{2.66} = 1 \) * H: \( \frac{4}{2.66} \approx 1.5 \) * O: \( \frac{4}{2.66} \approx 1.5 \) 4. **Convert to whole numbers (multiply by 2):** * C: \( 1 \times 2 = 2 \) * H: \( 1.5 \times 2 = 3 \) * O: \( 1.5 \times 2 = 3 \) Thus, the empirical formula is \( C_2H_3O_3 \). This formula gives the simplest ratio of elements in the compound.
In simple words: To find the empirical formula, we change percentages to grams, then to moles for each element. Next, we find the simplest whole-number ratio of these moles. For this acid, the formula is \( C_2H_3O_3 \).

🎯 Exam Tip: When converting mole ratios to whole numbers, if you get a ratio like 1.5, multiply all ratios by 2; if 1.33 or 1.66, multiply by 3 to get integers.

 

Question 11. An organic compound present in vinegar has 40 % Carbon, 6.6% Hydrogen: and 53.4% Oxygen. Find the empirical formula of the compound.
Answer: To determine the empirical formula, we convert the given percentages to moles and find their simplest whole-number ratio:
1. **Assume 100g sample:** 40g C, 6.6g H, 53.4g O. 2. **Calculate moles:** * C: \( \frac{40}{12} \approx 3.33 \, mol \) * H: \( \frac{6.6}{1} = 6.6 \, mol \) * O: \( \frac{53.4}{16} \approx 3.33 \, mol \) 3. **Find simplest mole ratio (divide by smallest, 3.33):** * C: \( \frac{3.33}{3.33} = 1 \) * H: \( \frac{6.6}{3.33} \approx 2 \) * O: \( \frac{3.33}{3.33} = 1 \) The ratio of C:H:O is 1:2:1, leading to an empirical formula of \( CH_2O \). This formula indicates the simplest building block unit of the compound.
In simple words: We convert the percentages of carbon, hydrogen, and oxygen into moles. Then we find the simplest whole-number ratio of these moles. For this compound, the simplest recipe is \( CH_2O \).

🎯 Exam Tip: Always double-check your arithmetic, especially when dividing by the smallest mole value, to avoid errors in the empirical formula.

 

Question 12. How much copper can be obtained from 100 g of anhydrous copper sulphate?
Answer: To determine the amount of copper obtainable from 100 g of anhydrous copper sulfate \( (CuSO_4) \), we first calculate the molar mass of \( CuSO_4 \):
Molar mass of \( CuSO_4 = (1 \times 63.5) + (1 \times 32) + (4 \times 16) = 63.5 + 32 + 64 = 159.5 \, g/mol \).
This means that \( 159.5 \, g \) of \( CuSO_4 \) contains \( 63.5 \, g \) of copper.
Now, to find the copper content in \( 100 \, g \) of \( CuSO_4 \):
Mass of copper \( = \frac{63.5 \, g \, Cu}{159.5 \, g \, CuSO_4} \times 100 \, g \, CuSO_4 \approx 39.81 \, g \).
This shows that nearly 40% of copper sulfate is actually copper.
In simple words: We find the total weight of copper sulfate. Then we see how much copper is in that total weight. Using these numbers, we can figure out that from 100 grams of copper sulfate, we can get about 39.81 grams of pure copper.

🎯 Exam Tip: When dealing with mass percentages, ensure you use the correct molar masses for all elements in the compound.

 

Question 13. Explain the classical concept of oxidation and reduction.
Answer: In the classical concept, oxidation is defined as a chemical process that involves the addition of oxygen to a substance or the removal of hydrogen from it. An example is the rusting of iron: \( 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \), where iron gains oxygen. Reduction, conversely, is the process of removing oxygen from a substance or adding hydrogen to it. For instance, in the reaction \( H_2S + Cl_2 \rightarrow 2HCl + S \), hydrogen is removed from \( H_2S \). Another example is \( CuO + C \rightarrow Cu + CO \), where oxygen is removed from copper oxide, and \( S + H_2 \rightarrow H_2S \), where hydrogen is added to sulfur. These definitions were among the earliest ways to understand chemical changes.
In simple words: Classically, oxidation means adding oxygen or taking away hydrogen from something. Reduction means taking away oxygen or adding hydrogen.

🎯 Exam Tip: It's helpful to remember that oxidation and reduction always occur simultaneously in a chemical reaction (redox reaction).

 

Question 14. Describe the electron concept of oxidation and reduction.
Answer: The electron concept provides a more fundamental understanding of oxidation and reduction, defining them in terms of electron transfer. Oxidation is the process where a substance loses electrons, causing its oxidation state to increase. For example: \( Fe^{2+} \rightarrow Fe^{3+} + e^- \). Reduction, on the other hand, is the process where a substance gains electrons, leading to a decrease in its oxidation state. An example is: \( Cu^{2+} + 2e^- \rightarrow Cu \). These two processes always occur together in what are known as redox reactions.
In simple words: In simple terms, oxidation is when an atom loses electrons, and reduction is when an atom gains electrons. These electron swaps always happen together in a chemical reaction.

🎯 Exam Tip: A useful mnemonic is OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

 

Question 15. Describe the oxidation number concept of oxidation and reduction.
Answer: The oxidation number concept defines oxidation as an increase in the oxidation number of an element during a chemical reaction, while reduction is defined as a decrease in the oxidation number of an element. For instance, in the reaction \( 2KMnO_4 + 10FeSO_4 + 8H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 8H_2O \), manganese changes from \( +7 \) to \( +2 \) (reduction), and iron changes from \( +2 \) to \( +3 \) (oxidation). Tracking these changes helps identify which species is oxidized and which is reduced, even in complex reactions.
In simple words: When an atom's oxidation number (a special assigned value) goes up, it's oxidation. When it goes down, it's reduction. This helps us see which atoms are gaining or losing electrons.

🎯 Exam Tip: Practice assigning oxidation numbers in various compounds to quickly identify oxidation and reduction steps in a reaction.

 

Question 16. Write notes on displacement reaction.
Answer: A displacement reaction is a type of redox reaction where a more reactive element replaces a less reactive element in a compound. These reactions can be broadly categorized into metal displacement reactions and non-metal displacement reactions. In such reactions, atoms are typically transferred from one compound to another, leading to a new set of compounds.
1. **Metal Displacement Reactions:** A more reactive metal displaces a less reactive metal from its compound. For example, when a zinc strip is placed in copper sulfate solution, zinc displaces copper: \( CuSO_4(aq) + Zn(s) \rightarrow Cu(s) + ZnSO_4(aq) \). The blue color of the solution diminishes, and a brown copper coating forms on the zinc. 2. **Non-metal Displacement Reactions:** A more reactive non-metal displaces a less reactive non-metal. For example, zinc can displace hydrogen from hydrochloric acid: \( Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g) \). These reactions are common in metallurgy and chemical industries.
In simple words: A displacement reaction is when one element kicks out another element from a chemical compound because it's stronger or more reactive. It's like one team member replacing another. In metal displacement, a metal like zinc can push out a weaker metal like copper from its salt solution. In non-metal displacement, a non-metal like zinc can push out hydrogen from an acid.

🎯 Exam Tip: Use the activity series of metals (or non-metals) to predict whether a displacement reaction will occur, as a more reactive element is needed to displace a less reactive one.

IV. Long Question and Answers

 

Question 1. What is a matter? Explain its classification.
Answer: Matter is defined as anything that has mass and occupies space. Essentially, anything you can touch, see, or feel is matter, and all matter is fundamentally composed of tiny particles called atoms. Matter can be broadly classified in two ways: physical classification and chemical classification.
Physical Classification: This categorizes matter based on its physical state, primarily into solids, liquids, and gases. These states are interconvertible by altering temperature and pressure. For instance, ice (solid water) can become liquid water, then water vapor (gas).
Chemical Classification: This groups matter based on its chemical composition. It divides matter into pure substances and mixtures. This system helps us understand the composition and reactivity of materials.
* **Pure Substances:** These have a fixed chemical composition. * **Elements:** Made of only one type of atom (e.g., Gold, Au; Copper, Cu; Hydrogen, \( H_2 \); Phosphorus, \( P_4 \)). * **Compounds:** Formed when two or more different elements chemically combine in fixed ratios (e.g., carbon dioxide, \( CO_2 \); glucose, \( C_6H_{12}O_6 \)). The properties of compounds are distinct from the properties of the elements they are made from. For instance, sodium (a reactive metal) and chlorine (a poisonous gas) combine to form sodium chloride (table salt), which is a stable, edible solid essential for life. * **Mixtures:** These consist of two or more substances that are physically combined but not chemically bonded. They retain their individual properties and can be separated by physical methods. Mixtures can be homogeneous (uniform composition, e.g., green tea) or heterogeneous (non-uniform composition, e.g., oil and water).
Understanding these classifications is fundamental to chemistry.
In simple words: Matter is anything that has weight and takes up space, made of tiny atoms. We can sort matter by how it looks (solid, liquid, gas) or by what it's made of (pure stuff like elements and compounds, or mixed stuff). Pure elements are just one type of atom, while pure compounds are different atoms joined together. Mixtures are different things simply stirred or jumbled, either smoothly mixed (like tea) or visibly separate (like oil and water).

🎯 Exam Tip: Be able to give examples for each classification type, such as solid (ice), liquid (water), gas (water vapor), element (gold), compound (salt), homogeneous mixture (sugar solution), and heterogeneous mixture (sand in water).

 

Question 2. How is empirical formula of a compound determined from the elemental analysis?
Answer: The empirical formula of a compound, which shows the simplest whole-number ratio of its constituent atoms, is determined from elemental analysis through a systematic four-step process:
1. **Percentage to Mass:** Convert the given percentage composition of each element into its mass by assuming a 100-gram sample of the compound (e.g., 20% carbon means 20g carbon). 2. **Mass to Moles:** Divide the mass of each element by its respective atomic mass to find the number of moles for each element. 3. **Simplest Mole Ratio:** Divide all the mole values obtained in step 2 by the smallest number of moles among them. This yields the simplest ratio of elements. 4. **Whole Numbers:** If the ratios from step 3 are not whole numbers, multiply all the ratios by the smallest possible integer to convert them into whole numbers. These whole numbers become the subscripts in the empirical formula. This methodical approach ensures accurate representation of elemental proportions.
In simple words: To find the empirical formula from elemental analysis, you first change percentages to grams, then to moles. Next, you find the simplest whole-number ratio of these moles by dividing them all by the smallest mole number. If needed, you multiply to make them all whole numbers.

🎯 Exam Tip: Make sure to use the correct atomic masses for each element when converting mass to moles, as small errors here can lead to incorrect ratios.

 

Question 3. An organic compound present in vinegar has 40% carbon, 6.6% hydrogen, and 53.4% oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 60 g mol-1).
Answer: To determine both the empirical and molecular formulas for this organic compound, we follow these steps:
1. **Calculate Moles (from percentages):** * Carbon: \( \frac{40}{12} \approx 3.33 \, mol \) * Hydrogen: \( \frac{6.6}{1} = 6.6 \, mol \) * Oxygen: \( \frac{53.4}{16} \approx 3.33 \, mol \) 2. **Find Simplest Mole Ratio:** Divide by the smallest mole value (3.33). * Carbon: \( \frac{3.33}{3.33} = 1 \) * Hydrogen: \( \frac{6.6}{3.33} \approx 2 \) * Oxygen: \( \frac{3.33}{3.33} = 1 \) Thus, the empirical formula is \( CH_2O \). 3. **Calculate Empirical Formula Mass:** For \( CH_2O \), the mass is \( (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30 \, g/mol \). 4. **Determine 'n' (Molecular Formula Multiplier):** Given molar mass is \( 60 \, g/mol \). \( n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2 \). 5. **Calculate Molecular Formula:** Multiply the subscripts in the empirical formula by \( n \). Molecular formula \( = (CH_2O)_2 = C_2H_4O_2 \).
This specific compound is acetic acid, which is the main component of vinegar.
In simple words: We find the simple recipe (empirical formula) like before, which is \( CH_2O \). Then, we use the compound's total weight (molar mass = 60) and the weight of our simple recipe (empirical formula mass = 30) to see how many times the simple recipe fits into the full one. Here it's 2 times, so the full recipe (molecular formula) is \( C_2H_4O_2 \).

🎯 Exam Tip: Always calculate the empirical formula mass correctly before finding 'n' (the molecular formula multiplier), as any error will propagate to the molecular formula.

 

Question 4. An organic compound present in vinegar has 40% carbon, 6.6 % hydrogen, and 53.4% oxygen, Find the empirical formula and molecular formula of the compound. (Given, Molar mass: 90 g mol-1).
Answer: For this organic compound, the steps to find its empirical and molecular formulas are:
1. **Calculate Moles (from percentages):** * Carbon: \( \frac{40}{12} \approx 3.33 \, mol \) * Hydrogen: \( \frac{6.6}{1} = 6.6 \, mol \) * Oxygen: \( \frac{53.4}{16} \approx 3.33 \, mol \) 2. **Find Simplest Mole Ratio:** Divide by the smallest mole value (3.33). * Carbon: \( \frac{3.33}{3.33} = 1 \) * Hydrogen: \( \frac{6.6}{3.33} \approx 2 \) * Oxygen: \( \frac{3.33}{3.33} = 1 \) So, the empirical formula is \( CH_2O \). 3. **Calculate Empirical Formula Mass:** For \( CH_2O \), the mass is \( (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30 \, g/mol \). 4. **Determine 'n' (Molecular Formula Multiplier):** Given molar mass is \( 90 \, g/mol \). \( n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{90}{30} = 3 \). 5. **Calculate Molecular Formula:** Multiply the subscripts in the empirical formula by \( n \). Molecular formula \( = (CH_2O)_3 = C_3H_6O_3 \).
This compound could be lactic acid or glyceraldehyde, both sharing this molecular formula.
In simple words: Like the last question, the simple recipe (empirical formula) is \( CH_2O \). This time, the compound's total weight (molar mass = 90) is 3 times the weight of the simple recipe (empirical formula mass = 30). So, the full recipe (molecular formula) is \( C_3H_6O_3 \).

🎯 Exam Tip: When given percentages for elemental composition, it's a good practice to sum them up to 100% to ensure all elements are accounted for.

 

Question 5. What is a redox reaction? Explain the different concepts of redox reaction.
Answer: A redox reaction involves both oxidation and reduction happening at the same time. Oxidation is when a substance loses electrons, and reduction is when it gains electrons. These two processes are always linked in a chemical reaction, meaning one cannot happen without the other. Understanding redox reactions helps in studying many natural processes, including how batteries work.
Classical concept of oxidation and reduction:
According to the older, classical idea, oxidation means adding oxygen to a substance or removing hydrogen from it. Here are some examples:
\( 4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 \)
\( \text{H}_2\text{S} + \text{Cl}_2 \rightarrow 2\text{HCl} + \text{S} \)
In the first example, oxygen is added to iron, which causes rusting. In the second example, hydrogen is taken away from hydrogen sulfide.
Reduction, in the classical sense, means adding hydrogen to a substance or removing oxygen from it. Here are some examples:
\( \text{CuO} + \text{C} \rightarrow \text{Cu} + \text{CO} \)
In the first reaction, oxygen is removed from copper oxide. In the second, hydrogen is added to sulfur.
Electron concept of oxidation and reduction:
The electron concept defines oxidation as the loss of electrons and reduction as the gain of electrons. Both processes occur together in a redox reaction.
\( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^{-} \) (This is oxidation, as iron loses an electron)
\( \text{Cu}^{2+} + 2\text{e}^{-} \rightarrow \text{Cu} \) (This is reduction, as copper gains electrons)
Oxidation number concept of oxidation and reduction:
In redox reactions, the oxidation number of elements changes. A reaction is called oxidation if the oxidation number of an element increases. It is called reduction if the oxidation number decreases. This concept is very useful for balancing complex chemical equations.
In simple words: A redox reaction is when chemicals swap electrons. Oxidation means losing electrons or gaining oxygen. Reduction means gaining electrons or gaining hydrogen. These two processes always happen together in a reaction.

🎯 Exam Tip: Remember the mnemonic "OIL RIG" - Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons) - to easily recall the electron concept of redox reactions.

 

Question 6. What is an oxidation number? State the rules to find the oxidation number.
Answer: An oxidation number is a number given to an atom in a compound that shows how many electrons it has lost or gained. It is like an imaginary charge an atom would have if all its bonds were ionic. It is a very helpful concept for balancing redox equations.
Rules to find the oxidation number:
1. For a free element (one that is not combined with other elements), the oxidation state is always zero. For example, in \( \text{H}_2 \), \( \text{Cl}_2 \), \( \text{Na} \), or \( \text{S}_8 \), the oxidation number is 0.
2. For a monoatomic ion (an ion made of only one atom), the oxidation state is the same as its charge. For instance, the oxidation number of sodium in \( \text{Na}^+ \) is +1, and for chlorine in \( \text{Cl}^- \) is -1.
3. The sum of all oxidation states in a neutral molecule must be zero. In an ion, the sum of oxidation states must equal the ion's net charge.
4. Hydrogen usually has an oxidation number of +1 in its compounds. The only exception is in metal hydrides (like \( \text{NaH} \)), where it has an oxidation number of -1.
5. Fluorine always has an oxidation state of -1 in all its compounds because it is the most electronegative element.
6. Oxygen usually has an oxidation state of -2 in most compounds. There are a few exceptions: in peroxides (like \( \text{H}_2\text{O}_2 \)), it is -1; in superoxides (like \( \text{KO}_2 \)), it is -1/2; and in oxygen difluoride (OF2), it is +2.
7. Alkali metals (Group 1 elements like Li, Na, K) always have an oxidation state of +1. Alkaline earth metals (Group 2 elements like Be, Mg, Ca) always have an oxidation state of +2 in all their compounds. These rules simplify finding the oxidation states for different atoms within a molecule.
In simple words: An oxidation number is like a pretend charge that tells you if an atom has gained or lost electrons when it joins with other atoms. There are simple rules to find this number, like elements on their own have zero, and single-atom ions have a charge equal to their oxidation number.

🎯 Exam Tip: When applying oxidation number rules, always remember the hierarchy: Group 1 & 2 metals are always +1/+2, then Fluorine is always -1, then Hydrogen is +1 (unless with a metal), then Oxygen is -2 (unless in peroxides/superoxides/with Fluorine). The sum must always equal the overall charge.

 

Question 7. Balance the following chemical equation by oxidation number method.
\( \text{KMnO}_4 + \text{FeSO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Fe}_2(\text{SO}_4)_3 + \text{8H}_2\text{O} \)
Answer: The oxidation number method helps balance complex redox reactions by ensuring that the total increase in oxidation numbers equals the total decrease. This method makes sure electrons are conserved in the reaction.
The given reaction is:
\( \text{KMnO}_4 + \text{FeSO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{MnSO}_4 + \text{Fe}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \)
First, we identify the oxidation numbers for the changing elements:
In \( \text{KMnO}_4 \), Mn is +7. In \( \text{MnSO}_4 \), Mn is +2. (Gain of 5 electrons, so reduction)
In \( \text{FeSO}_4 \), Fe is +2. In \( \text{Fe}_2(\text{SO}_4)_3 \), Fe is +3. (Loss of 1 electron, so oxidation)

Reduction half-reaction: \( \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \)
Oxidation half-reaction: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} \)

Balance electrons:
\( \text{MnO}_4^- \) gains 5 electrons to become \( \text{Mn}^{2+} \).
\( \text{Fe}^{2+} \) loses 1 electron to become \( \text{Fe}^{3+} \).
To balance the electrons, we multiply the oxidation half-reaction by 5.
\( \text{MnO}_4^- + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} \)

Now, balance the other atoms and charges (in acidic medium, use \( \text{H}^+ \) and \( \text{H}_2\text{O} \)):
Balance oxygen atoms in \( \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \) by adding \( \text{H}_2\text{O} \) to the right side:
\( \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)
Balance hydrogen atoms by adding \( \text{H}^+ \) to the left side:
\( \text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)
Now, combine this with the balanced oxidation step:
\( \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \)

Finally, add the spectator ions (K, SO4) back into the equation:
\( 2\text{KMnO}_4 + 10\text{FeSO}_4 + 8\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3 + 8\text{H}_2\text{O} \)
This equation is now fully balanced.
In simple words: We balance the equation by looking at how the oxidation numbers of manganese and iron change. Manganese goes from +7 to +2 (gains 5 electrons), and iron goes from +2 to +3 (loses 1 electron). We make sure the electrons gained and lost are equal by multiplying, then add hydrogen and oxygen as water to balance the rest of the atoms.

🎯 Exam Tip: Always double-check that the total number of atoms for each element and the total charge on both sides of the final equation are equal. This ensures a perfectly balanced equation.

 

Question 8. How is the following equation is balanced by lon electron method?
\( \text{MnO}_4^- + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_2\text{O} \)
Answer: The ion-electron method (also known as the half-reaction method) is a structured way to balance redox equations by splitting the overall reaction into two half-reactions: one for oxidation and one for reduction. This helps in balancing both atoms and charges separately.
Given unbalanced ionic equation:
\( \text{MnO}_4^- + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_2\text{O} \)

Step 1: Write the unbalanced half-reactions.
Oxidation half-reaction: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} \)
Reduction half-reaction: \( \text{MnO}_4^- \rightarrow \text{Mn}^{2+} \)

Step 2: Balance atoms other than oxygen and hydrogen.
For oxidation: Fe atoms are already balanced.
For reduction: Mn atoms are already balanced.

Step 3: Balance oxygen atoms by adding \( \text{H}_2\text{O} \) molecules (in acidic medium).
For oxidation: No oxygen, so no change.
For reduction: \( \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

Step 4: Balance hydrogen atoms by adding \( \text{H}^+ \) ions (in acidic medium).
For oxidation: No hydrogen, so no change.
For reduction: \( \text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

Step 5: Balance the charges by adding electrons.
For oxidation: \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^{-} \) (Charge: +2 on left, +3 - 1 = +2 on right. Balanced.)
For reduction: \( \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \) (Charge: -1 + 8 = +7 on left, +2 on right. Add 5 electrons to the left. +7 - 5 = +2. Balanced.)

Step 6: Make the number of electrons equal in both half-reactions.
Multiply the oxidation half-reaction by 5:
\( 5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5\text{e}^{-} \)
The reduction half-reaction remains:
\( \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \)

Step 7: Add the two balanced half-reactions and simplify.
\( \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+} + 5\text{e}^{-} \)
Cancel out the electrons from both sides:
\( \text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \)
This is the balanced ionic equation.
In simple words: To balance this reaction, we first split it into two parts: iron changing (oxidation) and manganese changing (reduction). We add water to balance oxygen and \( \text{H}^+ \) to balance hydrogen. Then, we add electrons to balance the charge in each part. Finally, we multiply one part so both parts use the same number of electrons, and then we put them back together to get the complete balanced equation.

🎯 Exam Tip: Always make sure to state whether the reaction is in acidic or basic medium as it dictates how hydrogen and oxygen atoms are balanced (using \( \text{H}^+ \) and \( \text{H}_2\text{O} \) for acidic, or \( \text{OH}^- \) and \( \text{H}_2\text{O} \) for basic).

 

Question 9. Calculate the percentage composition of the elements present in lead nitrate. How many Kg of O2 can be obtained from 50 kg of 70% pure lead nitrate?
Answer: To find the percentage composition, we first need the molar mass of lead nitrate \( \text{Pb}(\text{NO}_3)_2 \). This helps determine the amount of each element within the compound. Then, we can calculate the oxygen obtained from a given impure sample.
Lead nitrate formula: \( \text{Pb}(\text{NO}_3)_2 \)
Atomic masses:
Pb = 207 g/mol
N = 14 g/mol
O = 16 g/mol

Molar mass of \( \text{Pb}(\text{NO}_3)_2 = 207 + (2 \times 14) + (6 \times 16) = 207 + 28 + 96 = 331 \text{ g/mol} \)

Percentage composition:
Percentage of Pb \( = \frac{207}{331} \times 100\% = 62.54\% \)
Percentage of N \( = \frac{2 \times 14}{331} \times 100\% = \frac{28}{331} \times 100\% = 8.46\% \)
Percentage of O \( = \frac{6 \times 16}{331} \times 100\% = \frac{96}{331} \times 100\% = 28.99\% \)

Next, we calculate the amount of oxygen that can be obtained from 50 kg of 70% pure lead nitrate:
Mass of pure lead nitrate in the sample \( = 50 \text{ kg} \times 70\% = 50 \text{ kg} \times 0.70 = 35 \text{ kg} \)
From the molar mass, 331 g of \( \text{Pb}(\text{NO}_3)_2 \) contains 96 g of oxygen.
So, 331 kg of \( \text{Pb}(\text{NO}_3)_2 \) contains 96 kg of oxygen.

Amount of oxygen from 35 kg of pure \( \text{Pb}(\text{NO}_3)_2 \):
\( = \frac{96 \text{ kg O}}{331 \text{ kg Pb(NO}_3)_2} \times 35 \text{ kg Pb(NO}_3)_2 = \frac{96 \times 35}{331} \text{ kg} \)
\( = \frac{3360}{331} \text{ kg} \approx 10.15 \text{ kg} \)
Therefore, approximately 10.15 Kg of oxygen can be obtained from 50 kg of 70% pure lead nitrate.
In simple words: First, we find out how much lead, nitrogen, and oxygen are in one unit of lead nitrate, as a percentage. Then, we figure out how much of the 50 kg sample is actually pure lead nitrate (70% of it). Since we know how much oxygen is in pure lead nitrate, we can then calculate the total amount of oxygen from the pure part of the sample.

🎯 Exam Tip: When dealing with impure samples, always calculate the mass of the pure substance first before proceeding with stoichiometric calculations to avoid errors.