GSEB Class 11 Chemistry Solutions Chapter 1 રસાયણવિજ્ઞાનની કેટલીક પાયાની સંકલ્પનાઓ

 

Question 1. Calculate the molecular mass of the following :
1. H₂O
2. CO₂
3. CH₄
Answer:
1. Molecular mass of H₂O \( = 2 \times \) Atomic mass of H \( + \) Atomic mass of O \( = 2 \times 1 + 16 = 18.0 \text{ a.m.u.} \)
2. Molecular mass of CO₂ \( = \) Atomic mass of C \( + 2 \times \) Atomic mass of O \( = 12 + 2 \times 16 = 44.0 \text{ a.m.u.} \)
3. Molecular mass of CH₄ \( = \) Atomic mass of C \( + 4 \times \) Atomic mass of H \( = 12 + 4 \times 1 = 12 + 4 = 16 \text{ a.m.u.} \)
In simple words: To find the molecular mass, you add up the atomic masses of all the atoms in a molecule. Multiply the atomic mass of each element by how many times it appears in the formula, then sum those values.

Exam Tip: Remember to use the correct atomic masses for each element and count the number of atoms carefully to prevent errors.

 

Question 2. Calculate the mass percent of different elements present in sodium sulfate (Na₂SO₄).
Answer:
Molecular mass of Na₂SO₄ \( = 2 \times \) At. mass of Na \( + \) At. mass of S \( + 4 \times \) At. mass of O \( = 2 \times 23 + 32 + 4 \times 16 = 142 \)
Mass per cent of Na \( = \frac{\text{Mass of Na}}{\text{M. Mass of Na}_2\text{SO}_4} \times 100 \)
\( = \frac{46}{142} \times 100 = 32.39 \)
Mass per cent of S \( = \frac{\text{Mass of S}}{\text{M. Mass of Na}_2\text{SO}_4} \times 100 \)
\( = \frac{32}{142} \times 100 = 22.53 \)
Mass per cent of O \( = \frac{\text{Mass of oxygen}}{\text{M. Mass of Na}_2\text{SO}_4} \times 100 \)
\( = \frac{64}{142} \times 100 = 45.07 \)
In simple words: First, calculate the total mass of the molecule. Then, for each element, divide its total mass in the molecule by the total molecular mass, and multiply by 100 to get the percentage.

Exam Tip: Ensure that the sum of all mass percentages for a compound is approximately 100% (allowing for slight rounding differences) to verify your calculation accuracy.

 

Question 3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer:

The ratio between Fe and O is 2: 3.
\( \implies \) The empirical formula \( = \text{Fe}_2\text{O}_3 \).
In simple words: To find the simplest formula, first convert the mass percentages of each element into moles. Then divide all mole amounts by the smallest mole amount to get a simple whole-number ratio. This ratio gives you the empirical formula.

Exam Tip: When determining empirical formulas from percentages, remember to divide by the smallest mole ratio to get the simplest whole numbers. If ratios are close to 0.5, multiply all by 2; if close to 0.33 or 0.66, multiply by 3, and so on, to get integers.

 

Question 4. Calculate the amount of carbon dioxide that could be produced when.
1. 1 mole of carbon is burnt in the air
2. 1 mole of carbon is burnt in 16 g of dioxygen
3. 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
1. For 1 mole of carbon burnt in air:
C(s) \( + \) O₂(g) \( \longrightarrow \) CO₂(g)
1 Mole \( \quad \) air \( \quad \quad \) 1 Mole
12.0 g \( \quad \quad \quad \quad \quad \) 12 \( + \) 2 \( \times \) 16 \( = 44.0 \text{ g} \)
Amount of CO₂ produced \( = 44.0 \text{ g} \)
2. For 1 mole of carbon burnt in 16 g of dioxygen:
C(s) \( + \) O₂(g) \( \longrightarrow \) CO₂(g)
1 Mole \( \quad \) 1 Mole \( \quad \quad \) 1 Mole
12 g \( \quad \quad \) 32 g \( \quad \quad \) 44.0 g
32.0 g of O₂ produce 44.0 g of CO₂
\( \implies \) 16.0 g of O₂ produce \( = \frac{44}{32} \times 16 = 22.0 \text{ g} \) of CO₂
Amount of CO₂ produced \( = 22.0 \text{ g} \)
3. Amount of CO₂ produced when 2 moles (= 24 g) of C are burnt in 16.0 g [limited amount] of O₂ \( = 22.0 \text{ g} \)
In simple words: To figure out how much carbon dioxide forms, you need the balanced chemical equation. If one reactant is limited, that's the one you use for calculations because it runs out first.

Exam Tip: Always identify the limiting reagent in reactions to accurately calculate the product yield. The amount of product formed is always determined by the reactant that gets completely consumed first.

 

Question 5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1
Answer:
1000 mL of CH₃COONa of 1 M solution contains 82.0245 g
\( \implies \) 500 mL of it 1 M solution contains \( = \frac{82.0245}{1000} \times 500 \)
\( \implies \) 500 mL of it of 0.375 M solution \( = \frac{82.0245}{1000} \times 500 \times 0.375 = 15.38 \text{ g} \).
In simple words: To find the mass needed, first understand that 1 liter of a 1 M solution contains the molar mass in grams. Then adjust for the desired volume and the specific molarity required.

Exam Tip: Molarity (M) refers to moles per liter. Always convert volume to liters if given in milliliters before performing calculations involving molarity.

 

Question 6. Calculate the concentration of nitric acid in moles per liter in a sample that has a density, 1.41 g mL^-1 and the mass percent of nitric acid in it being 69%.
Answer:
69% HNO₃ by mass means that 69 g of HNO₃ is present in 100 g of the solution.
Volume of 100 g of the solution \( = \frac{\text{Mass}}{\text{Density}} = \frac{100}{1.41} = 70.92 \text{ mL} \)
Molar mass of HNO₃ \( = \) 1 \( + \) 14 \( + 3 \times 16 = 63.0 \text{ g mol}^{-1} \)
Number of moles of HNO₃ in 100 g of solution \( = \frac{69.0}{63.0} \text{ moles} = 1.095 \text{ moles} \)
Thus 70.92 mL of solution contains HNO₃ \( = 1.095 \text{ moles} \)
\( \implies \) 1000 mL (= 1 L) solution contains HNO₃ \( = \frac{1.095}{70.92} \times 1000 \)
\( \implies \) Concentration of nitric acid \( = 15.44 \text{ mol L}^{-1} \).
In simple words: Start with the mass percentage to find the mass of nitric acid and the total solution mass. Use the solution's density to find its volume. Convert the mass of nitric acid to moles, then calculate moles per liter for the concentration.

Exam Tip: When given mass percent and density, assume a convenient mass (e.g., 100 g) of the solution to simplify initial calculations for mass of solute and volume of solution. Convert all units consistently, especially volume to liters for molarity.

 

Question 7. How much copper can be obtained from 100 g of copper sulfate (CuSO₄)?
Answer:
Molar mass of CuSO₄ \( = \) Atomic mass of Cu \( + \) Atomic mass of S \( + 4 \times \) Atomic mass of O \( = 63.5 + 32.0 + 4 \times 16.0 = 159.5 \text{ g mol}^{-1} \)
159.5 g of CuSO₄ \( = 63.5 \text{ g} \) of Cu
\( \implies \) 100 g of CuSO₄ \( = \frac{63.5}{159.5} \times 100 = 39.81 \text{ g} \)
\( \implies \) Mass of Cu that can be obtained \( = 39.81 \text{ g} \).
In simple words: First, find the total mass of copper sulfate. Then, see how much copper is in that total mass. Use a proportion to figure out how much copper you'd get from 100 grams of copper sulfate.

Exam Tip: Stoichiometric calculations like this require accurate molar masses. Ensure you correctly identify the molar mass of the compound and the atomic mass of the element you are extracting.

 

Question 8. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively.
Answer:
Step I: To calculate the empirical formula

Hence the empirical formula of an oxide of Iron \( = \text{Fe}_2\text{O}_3 \).
Step II: The molecular formula of the oxide of iron is the same as the empirical formula, i.e., Fe₂O₃.
Note: Molecular formula can be determined if the molecular mass of oxide of iron is given. It is not given in the question.
In simple words: First, convert percentages to moles, then find the simplest whole-number ratio of atoms to get the empirical formula. If the molecular mass is not provided, the empirical formula is considered the molecular formula for this type of problem.

Exam Tip: The empirical formula shows the simplest whole-number ratio of atoms, while the molecular formula shows the actual number of atoms. If the molecular mass is missing, the empirical formula cannot be definitively converted to a different molecular formula.

 

Question 9. Calculate the atomic mass (average) of chlorine using the following data :

Answer:
The average atomic mass of chlorine \( = \frac{75.77 \times 34.9689 + 24.23 \times 36.9659}{75.77 + 24.23} \)
\( = \frac{3545.277}{100} = 35.45 \)
In simple words: To find the average atomic mass, multiply the mass of each isotope by its natural abundance (as a decimal), then add these results together.

Exam Tip: Remember to convert percentage abundance to a decimal (divide by 100) before multiplying by the isotopic mass. The sum of abundances should be 100%.

 

Question 10. In three moles of ethane (C₂H₆), calculate the following:
1. number of moles of carbon atoms
2. no. of moles of hydrogen atoms
3. no. of molecules of ethane.
Answer:
1. 1 Mole of ethane (C₂H₆) contains 2 moles of carbon atoms
\( \implies \) 3 moles of C₂H₆ will contain \( 3 \times 2 = 6 \) moles of carbon (C) atoms.
2. 1 mole of C₂H₆ contains 6 moles of hydrogen (H) atoms
\( \implies \) 3 moles of C₂H₆ will contain \( 3 \times 6 = 18 \) moles of H atoms.
3. 1 mole of C₂H₆ contains \( 6.023 \times 10^{23} \) molecules of ethane
\( \implies \) 3 moles of C₂H₆ will contain \( 3 \times 6.023 \times 10^{23} \)
\( = 1.8069 \times 10^{24} \) molecules.
In simple words: Use the chemical formula to find out how many atoms of each element are in one molecule. Then, multiply that by the total number of moles or molecules given to get the count for the whole sample.

Exam Tip: When dealing with moles, remember that Avogadro's number (\( 6.023 \times 10^{23} \)) relates moles to the number of particles (atoms, molecules, ions).

 

Question 11. What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?
Answer:
Molar mass of sugar \( = 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \text{ g mol}^{-1} \)
No. of moles of sugar in 20 g \( = \frac{20}{342} \)
\( = \frac{20}{342 \times 2} \) [: Vol. of solution \( = 2 \text{ L} \)]
\( = 0.029 \text{ mol L}^{-1} \)
In simple words: To find concentration in moles per liter, divide the mass of sugar by its molar mass to get moles. Then, divide those moles by the total volume of the solution in liters.

Exam Tip: Molarity is defined as moles of solute per liter of solution. Always ensure mass is converted to moles and volume is in liters for this calculation.

 

Question 12. If the density of methanol is 0.793 kg L^-1 What is its volume needed for making 2.5 L of its 0.25 M solution ?
Answer:
Molar mass of methanol (CH₃OH) \( = 1 \times 12 + 4 \times 1 + 1 \times 16 = 32 \text{ g mol}^{-1} \)
Moles of methanol in 2.5 L of its 0.25 M solution \( = 2.5 \times 0.25 = 0.625 \text{ mole} \)
Mass of methanol \( = 32 \times 0.625 \text{ g} = 20.0 \text{ g} \)
Density of methanol \( = 0.793 \text{ kg L}^{-1} \)
Density of methanol \( = 0.793 \text{ g mL}^{-1} \) (since 1 kg \( = \) 1000 g and 1 L \( = \) 1000 mL)
\( \implies \) Volume of methanol required \( = \frac{20.0}{0.793} \approx 25.22 \text{ mL} \)
In simple words: First, calculate how many moles of methanol you need for the desired solution. Then, convert those moles to grams using the molar mass. Finally, use the density to find the volume of methanol in milliliters.

Exam Tip: Be careful with units when converting density (e.g., kg L^-1 to g mL^-1). Make sure all units are consistent before performing calculations.

 

Question 13. The pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below :
1 Pa = 1 N m^-2
If the mass of air at sea level is 1034 g cm^-2, calculate the pressure in the pascal.
Answer:
Acceleration due to gravity (g) \( = 9.806 \text{ m s}^{-2} \)
Mass of air \( = 1034 \text{ g cm}^{-2} = 1.034 \text{ kg} (100 \text{ cm})^2 \times (1 \text{ m}^2 / 10000 \text{ cm}^2) \text{ m}^{-2} \)
\( = 1034 \text{ kg m}^{-2} \)
Force \( = \) mass \( \times \) acceleration \( = 1034 \text{ kg m}^{-2} \times 9.806 \text{ m s}^{-2} = 10139.764 \text{ N m}^{-2} \)
Pressure \( = \frac{\text{Force}}{\text{Area}} \)
\( = 10139.764 \text{ Pa} \)
The pressure of air at sea level \( = 1.0138 \times 10^5 \text{ Pa} \).
In simple words: Pressure is how much force pushes on a certain area. To calculate it, multiply the mass of air by the force of gravity, then adjust for the area to get the answer in Pascals.

Exam Tip: For pressure calculations, ensure all units are in SI base units (kilograms for mass, meters for length, seconds for time) before multiplying or dividing to get the correct answer in Pascals.

 

Question 14. What is the SI unit of mass? How is it defined?
Answer:
The SI unit of mass is Kg (kilogram). It is defined as the mass of the platinum-iridium (Pt-Ir) cylinder that is stored in an airtight jar at the International Bureau of Weights and Measures in France.
In simple words: The standard unit for mass is the kilogram. It was originally defined by a specific metal cylinder kept safe in a vault.

Exam Tip: Know the fundamental SI units for mass (kilogram), length (meter), and time (second), along with their historical definitions.

 

Question 15. Match the following prefixes with their multiples.
Prefixes Multiples
(i) micro 10⁶
(ii) deca 10⁹
(iii) mega 10^-6
(iv) giga 10^-15
(v) femto 10
Answer:
Prefixes Multiples
(i) micro 10^-6
(ii) deca 10
(iii) mega 10⁶
(iv) giga 10⁹
(v) femto 10^-15
In simple words: Match each prefix, like micro or mega, to the correct power of ten that it represents.

Exam Tip: Memorize common SI prefixes and their corresponding powers of ten (e.g., kilo = 10³, milli = 10^-3, micro = 10^-6) as they are crucial for unit conversions.

 

Question 16. What do you mean by significant figures?
Answer:
Significant figures refer to the number of digits in a number that has some importance in the magnitude of a given number. They indicate the precision of a measurement.
Example: 30.4560 has 6 significant figures.
Example: 16708.4300 has 8 significant figures.
In simple words: Significant figures are the important digits in a number that show how precise it is. They tell you which parts of a measurement are reliable.

Exam Tip: Understand the rules for determining significant figures, including non-zero digits, zeros between non-zero digits, leading zeros, and trailing zeros with and without decimal points.

 

Question 17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass)
1. Express this in percent by mass.
2. Determine the molality of chloroform in the water sample.
Answer:
1. Express this in percent by mass.
10⁶ gm of solution contains 15 g of CHCl₃
\( \implies \) 1 gm of solution contains \( = \frac{15}{10^6} \)
\( \implies \) 100 gm of solution contains \( = \frac{15}{10^6} \times 10^2 = 15 \times 10^{-4} \text{ g} \)
\( \implies \) percent by mass \( = 15 \times 10^{-4} \text{ g} \).
2. Molality of CHCl₃ :
10⁶ gm of the solution contains 15 g of CHCl₃
\( \implies \) Wt. of water \( = 1000000 - 15 = 999985 \text{ gm} \)
Now, 999985 g of water contains 15 g of CHCl₃
\( \implies \) 1000 g of water contains \( = \frac{15}{999985} \times 1000 \text{ g} \) of CHCl₃
Molar mass of CHCl₃ \( = 12 + 1 + 3 \times 35.5 = 119.5 \text{ g mol}^{-1} \)
\( \implies \) Molality \( = \frac{15}{999985} \times \frac{1000}{119.5} \text{m} = 1.25 \times 10^{-4} \text{m} \).
In simple words: To convert parts per million (ppm) to percent, remember that 1 ppm is one part per million parts, so 15 ppm means 15 parts per million. Molality is about moles of solute per kilogram of solvent, so you need to convert masses to appropriate units.

Exam Tip: Remember that ppm (parts per million) can be converted to percentage by dividing by 10,000. For molality, the denominator is the mass of the solvent in kilograms, not the total solution mass.

 

Question 18. Express the following in the scientific notation
1. 0.0048
2. 234,000
3. 8008
4. 500.0
5. 6.0012
Answer:
1. \( 0.0048 = \frac{48}{10000} = 48 \times 10^{-4} = 4.8 \times 10^{-3} \)
2. \( 234,000 = 2.34 \times 10^5 \)
3. \( 8008 = 8.008 \times 10^3 \)
4. \( 500.0 = 5.00 \times 10^2 \)
5. \( 6.0012 = 6.0012 \times 10^0 \).
In simple words: Scientific notation means writing a number as a digit between 1 and 10, multiplied by a power of 10. You move the decimal point until one non-zero digit is to its left, and the power of 10 shows how many places you moved it.

Exam Tip: When converting to scientific notation, remember that moving the decimal to the left gives a positive exponent, and moving it to the right gives a negative exponent.

 

Question 19. How many significant figures are present in the following?
1. 0.0025
2. 208
3. 5005
4. 126,000
5. 500.0
6. 2.0034
Answer:
1. 0.0025 has 2 significant figures.
2. 208 has 3 significant figures.
3. 5005 has 4 significant figures.
4. 126,000 has 3 significant figures (trailing zeros without a decimal are not significant).
5. 500.0 has 4 significant figures (trailing zeros with a decimal are significant).
6. 2.0034 has 5 significant figures.
In simple words: Count all digits that are not leading zeros. Zeros between non-zero digits are significant. Trailing zeros are significant only if there is a decimal point in the number.

Exam Tip: Remember the rules for counting significant figures: non-zero digits are always significant; zeros between non-zero digits are significant; leading zeros are never significant; trailing zeros are significant only if a decimal point is present.

 

Question 20. Round up the following up to three significant figures :
1. 34.216
2. 10.4107
3. 0.04597
4. 280.8
Answer:
1. \( 34.216 \sim 34.2 \)
2. \( 10.4107 \sim 10.4 \)
3. \( 0.04597 \sim 0.0460 \)
4. \( 280.8 \sim 281 \)
In simple words: To round a number to three significant figures, look at the fourth digit. If it's 5 or more, round up the third digit. If it's less than 5, keep the third digit the same. Remember to keep the place value correct using zeros if needed.

Exam Tip: When rounding, ensure you maintain the place value of the digits. For example, rounding 280.8 to three significant figures means 281, not 281.0, and 0.04597 to 0.0460 correctly indicates the trailing zero as significant after rounding.

 

Question 21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds :

(1) Which law of chemical combination is obeyed by the above experimental data? Give
(2) Fill in the blanks in the following conversion :
1. 1 km \( = \) ____ mm \( = \) ____ pm
2. 1 mg \( = \) ____ kg \( = \) ____ ng
3. 1 mL \( = \) ____ L \( = \) ____ dm³
Answer:
(1) The given data is in accordance with the law of multiple proportions, which states; When two elements combine to form two or more than two compounds, the weights of one of the two elements which combine with the fixed weight of the other bears a simple ratio to one another.
In the said question, if we fix the weight of dinitrogen at 14 g, then the weights of dioxygen which combines with the fixed weight (=14 g) of dinitrogen will be 16, 32, 16, 40 which are in the simple whole-number ratio of 1: 2: 1: 2.5 or 2: 4: 2: 5.
(2)
1. 1 km \( = 10^6 \) mm \( = 10^{15} \) pm
2. 1 mg \( = 10^{-6} \) kg \( = 10^6 \) ng
3. 1 mL \( = 10^{-3} \) L \( = 10^{-3} \) dm³
In simple words: The data follows the law of multiple proportions, meaning that when two elements combine to make different compounds, one element's mass will combine with a fixed mass of the other in simple whole number ratios. For conversions, know the standard scientific prefixes and their associated powers of ten.

Exam Tip: Understand the Law of Multiple Proportions clearly. For unit conversions, memorize the SI prefixes (e.g., mega, kilo, centi, milli, micro, nano, pico, femto) and their respective powers of 10 to convert between units accurately.

 

Question 22. If the speed of light is \( 3.0 \times 10^8 \text{ ms}^{-1} \), calculate the distance covered by light in 2.00 ns.
Answer:
Distance covered by light in one second \( = 3.0 \times 10^8 \text{ m} \)
\( \implies \) Distance covered by light in 2.00 ns or \( 2 \times 10^{-9} \) second \( = 3.0 \times 10^8 \times 2 \times 10^{-9} \text{ m} \)
\( \implies \) Distance covered in 2.00 ns \( = 6.0 \times 10^{-1} \text{ m} = 0.6 \text{ m} \).
In simple words: To find the distance light travels, multiply its speed by the time taken. Remember to convert nanoseconds to seconds first.

Exam Tip: Ensure that all units are consistent (e.g., convert nanoseconds to seconds) before performing calculations. Distance = Speed × Time.

 

Question 23. In a reaction, A \( + \) B₂ \( \longrightarrow \) AB₂ Identify the limiting reagent, if any, in the following reaction mixtures :
1. 300 atoms of A \( + \) 200 molecules of B
2. 2 mol A \( + \) 3 mol B
3. 100 atoms of A \( + \) 100 molecules of B
4. 5 mol A \( + \) 2.5 mol B
5. 2.5 mol A \( + \) 5 mol B.
Answer:
1. The given reaction is A \( + \) B₂ \( \longrightarrow \) AB₂
Here 300 atoms of A requires 300 molecules of B. Since there are only 200 molecules of B provided, B is the limiting reagent.
2. 3 mol B requires 3 mol of A (from 1:1 ratio). Since only 2 mol of A is provided, A is the limiting reagent.
3. 100 atoms of A \( + \) 100 molecules of B constitute a stoichiometric mixture. Neither A nor B is the limiting reagent.
4. B is the limiting reagent as 5 mol A requires 5 mol B but only 2.5 mol B is given.
5. A is the limiting reagent as 5 mol B requires 5 mol A, but only 2.5 mol A is provided.
In simple words: The limiting reagent is the reactant that runs out first in a chemical reaction. It determines how much product can be formed. Compare the moles or atoms of each reactant needed versus what's available based on the reaction's ratio.

Exam Tip: To identify the limiting reagent, calculate the amount of product each reactant could form. The reactant that yields the least product is the limiting reagent.

 

Question 24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation :
N₂(g) \( + \) 3H₂(g) \( \longrightarrow \) 2NH₃(g)
1. Calculate the mass of ammonia produced if \( 2.00 \times 10^3 \) g dinitrogen reacts with \( 1.00 \times 10^3 \) g of dihydrogen.
2. Will any of the two reactants remain unreacted?
3. If yes, which one and what would be its mass?
Answer:
The balanced equation shows the molar masses:
N₂(g) \( + \) 3H₂(g) \( \longrightarrow \) 2NH₃(g)
1 mol \( \quad \) 3 mol \( \quad \quad \quad \) 2 mol
\( = 28.0 \text{ g} \quad = 6.0 \text{ g} \quad \quad = 34.0 \text{ g} \)
1. Calculate the mass of ammonia produced:
28.0 g of N₂ require 6.0 g of H₂ to produce 34.0 g of NH₃
Given: \( 2.00 \times 10^3 \text{ g} \) of N₂ and \( 1.00 \times 10^3 \text{ g} \) of H₂.
Mass of H₂ required for \( 2.00 \times 10^3 \text{ g} \) of N₂ \( = \frac{6.0}{28.0} \times 2.00 \times 10^3 \text{ g} \) of H₂ \( = 428.57 \text{ g} \) of H₂.
Since \( 1.00 \times 10^3 \text{ g} \) (1000 g) of H₂ is available, which is more than 428.57 g, N₂ is the limiting reagent.
Mass of NH₃ produced by \( 2.00 \times 10^3 \text{ g} \) of N₂ \( = \frac{34.0}{28.0} \times 2.00 \times 10^3 \text{ g} \) of NH₃ \( = 2428.57 \text{ g} \text{ NH}_3 \approx 2.43 \times 10^3 \text{ g NH}_3 \).
2. Yes, Dihydrogen (H₂) will remain unreacted to some extent.
3. Amount of hydrogen that remains unreacted \( = (1.00 \times 10^3 - 428.57) \text{ g} = 571.43 \text{ g} \).
In simple words: First, use the balanced equation to find out which reactant is the limiting one, meaning it will run out first. That reactant determines how much ammonia can be made. Then, subtract the amount of the excess reactant used from the starting amount to find out how much is left over.

Exam Tip: Always begin by writing a balanced chemical equation and calculating molar masses. Identifying the limiting reagent is critical for accurately predicting product yield and unreacted amounts of other reactants.

 

Question 25. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Answer:
0.50 mol Na₂CO₃ means half the molar mass of Na₂CO₃.
Molar mass of Na₂CO₃ \( = 2 \times 23 + 12 + 3 \times 16 = 46 + 12 + 48 = 106 \text{ g mol}^{-1} \)
So, 0.50 mol Na₂CO₃ represents the mass \( = 0.50 \times 106 \text{ g} = 53.0 \text{ g} \).
Whereas 0.50 M Na₂CO₃ represents its molarity in solution. 0.50 M Na₂CO₃ indicates that 53.0 g of Na₂CO₃ have been dissolved in 1 L of its solution.
In simple words: 0.50 mol tells you the actual amount of a substance in terms of its mass. 0.50 M tells you the concentration of a solution, specifically how many moles of a substance are dissolved in each liter of that solution.

Exam Tip: "Mol" refers to a quantity of substance (like a dozen for eggs), representing a specific mass. "M" (Molar) refers to concentration, indicating how much substance is dissolved in a specific volume of solution.

 

Question 26. If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapor would be produced?
Answer:
The balanced chemical equation is:
2H₂(g) \( + \) O₂(g) \( \longrightarrow \) 2 H₂O(g)
2 volumes \( \quad \) 1 volume \( \quad \) 2 volumes (experimentally)
The ratio by volumes is 2: 1: 2.
\( \implies \) 10 volumes of dihydrogen will react with 5 volumes of dioxygen to produce 10 volumes of water vapors.
In simple words: Based on the chemical equation, two parts of hydrogen gas react with one part of oxygen gas to make two parts of water vapor. So if you have ten parts of hydrogen and five parts of oxygen, they will all react to create ten parts of water vapor.

Exam Tip: For reactions involving gases at constant temperature and pressure, the volume ratios are directly proportional to the mole ratios from the balanced chemical equation (Gay-Lussac's Law of Gaseous Volumes).

 

Question 27. Convert the following into basic units :
1. 28.7 pm
2. 1515 ps
3. 25365 mg
Answer:
1. \( 28.7 \text{ pm} = 2.87 \times 10^{-11} \text{ m} \)
2. \( 15.15 \text{ ps} = 15.15 \times 10^{-12} \text{ s} \approx 1.515 \times 10^{-11} \text{ s} \)
3. \( 25365 \text{ mg} = 2.5365 \times 10^{-2} \text{ kg} \)
In simple words: To change units to their basic forms, remember the conversion factor for each prefix. For example, pico (p) means \( 10^{-12} \), and milli (m) means \( 10^{-3} \). Kilograms are already a basic unit for mass.

Exam Tip: Memorize the standard SI prefixes and their corresponding powers of ten (e.g., pico = 10^-12, milli = 10^-3, kilo = 10³) to perform conversions accurately.

 

Question 28. Which one of the following will have the largest number of atoms?
1. 1 g Au(s)
2. 1 g Na(s)
3. 1 g Li(s)
4. 1 g Cl₂(g)
Answer:
1. For 1 g Au(s): Atomic mass of Au \( = 197 \text{ g mol}^{-1} \).
Number of atoms \( = \frac{1 \text{ g}}{197 \text{ g/mol}} \times 6.023 \times 10^{23} \text{ atoms/mol} = 3.057 \times 10^{21} \text{ atoms} \).
2. For 1 g Na(s): Atomic mass of Na \( = 23.0 \text{ g mol}^{-1} \).
Number of atoms \( = \frac{1 \text{ g}}{23.0 \text{ g/mol}} \times 6.023 \times 10^{23} \text{ atoms/mol} = 2.618 \times 10^{22} \text{ atoms} \).
3. For 1 g Li(s): Atomic mass of Li \( = 7.0 \text{ g mol}^{-1} \).
Number of atoms \( = \frac{1 \text{ g}}{7.0 \text{ g/mol}} \times 6.023 \times 10^{23} \text{ atoms/mol} = 8.604 \times 10^{22} \text{ atoms} \).
4. For 1 g Cl₂(g): Molar mass of Cl₂ \( = 2 \times 35.5 = 71.0 \text{ g mol}^{-1} \).
Number of molecules \( = \frac{1 \text{ g}}{71.0 \text{ g/mol}} \times 6.023 \times 10^{23} \text{ molecules/mol} = 8.483 \times 10^{21} \text{ molecules} \).
Number of atoms \( = 8.483 \times 10^{21} \text{ molecules} \times 2 \text{ atoms/molecule} = 1.697 \times 10^{22} \text{ atoms} \).
By comparison (iii) i.e., 1 g Li(s) will have the maximum number of atoms, which is \( 8.604 \times 10^{22} \) atoms.
In simple words: To find out which sample has the most atoms, divide the given mass (1g) by the atomic mass of each element to get moles. The element with the smallest atomic mass will have the most moles, and thus the most atoms, for the same mass. Remember to account for diatomic molecules like Cl₂.

Exam Tip: A gram of a lighter element will always contain more atoms than a gram of a heavier element. For molecular substances like Cl₂, remember to multiply the number of molecules by the number of atoms per molecule.

 

Question 29. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040?
Answer:
Let us calculate the mass of ethanol in which its mole fraction is 0.040.
Mole fraction of ethanol \( = \frac{\text{no. of moles of ethanol}}{\text{no. of moles of water} + \text{no. of moles of ethanol}} \)
Let \( w \) be the mass of ethanol in grams.
Moles of ethanol \( = \frac{w}{46} \) (Molar mass of ethanol is 46 g/mol)
Assuming 1000 g of water, moles of water \( = \frac{1000}{18} \) (Molar mass of water is 18 g/mol)
\( 0.040 = \frac{\frac{w}{46}}{\frac{1000}{18} + \frac{w}{46}} \)
\( 0.040 = \frac{\frac{w}{46}}{55.55 + \frac{w}{46}} \)
\( 0.040 \left(55.55 + \frac{w}{46}\right) = \frac{w}{46} \)
\( 2.222 + 0.040 \frac{w}{46} = \frac{w}{46} \)
\( 2.222 = \frac{w}{46} - 0.040 \frac{w}{46} \)
\( 2.222 = \frac{w}{46} (1 - 0.040) \)
\( 2.222 = \frac{w}{46} (0.960) \)
\( w = \frac{2.222 \times 46}{0.960} \approx 106.48 \text{ g} \)
Density of solution is approximated as \( 1 \text{ g cm}^{-3} \) (assuming dilute aqueous solution). So volume of 106.48 g ethanol \( = \frac{106.48 \text{ g}}{0.789 \text{ g/mL}} \approx 134.9 \text{ mL} \). This is the volume of ethanol. Total volume of solution is not simply sum of volumes, and water is 1000g approx 1L. It implies total volume of solution \( \approx \) volume of water. So assume volume of solution \( \approx 1 \text{ L} \).
Moles of ethanol \( = \frac{106.48}{46} = 2.315 \text{ mol} \)
\( \implies \) Molarity of the solution \( = \frac{2.315 \text{ mol}}{1 \text{ L}} = 2.315 \text{ M} \).
In simple words: First, use the mole fraction and the molar masses of ethanol and water to find the mass of ethanol in a given amount of water. Then, convert the mass of ethanol to moles. Assuming the solution's volume is close to the water's volume for dilute solutions (1000g water is approximately 1L), divide the moles of ethanol by the total volume to get the molarity.

Exam Tip: When calculating molarity from mole fraction in aqueous solutions, it is common to assume 1 kg (or 1000 g) of water (solvent) to simplify the calculations, which approximates the solution volume to 1 L for dilute solutions. Ensure molar masses are correct.

 

Question 30. What will be the mass of one ¹²C atom in g?
Answer:
We want to calculate the mass of one atom of ¹²C.
1 gm atom of ¹²C \( = 12.0 \text{ g} \)
\( 6.023 \times 10^{23} \) atoms of ¹²Carbon weigh \( = 12.0 \text{ g} \)
\( \implies \) 1 atom of ¹²Carbon weight \( = \frac{12.0}{6.023 \times 10^{23}} \text{ g} \)
\( \implies \) Weight of 1 atom of ¹²C \( = 1.99265 \times 10^{-23} \text{ g} \).
In simple words: To find the mass of a single atom, divide the atomic mass of the element by Avogadro's number, which tells you how many atoms are in one mole.

Exam Tip: Remember Avogadro's number (\( 6.023 \times 10^{23} \text{ particles/mol} \)) is the key conversion factor between moles and the number of atoms or molecules.

 

Question 31. How many significant figures should be present in the answer to the following calculations?
1. \( \frac{0.02856 \times 298.15 \times 0.112}{0.5785} \)
2. \( 5 \times 5.364 \)
3. \( 0.0125 + 0.7864 + 0.0215 \)
Answer:
1. For multiplication/division, the answer should have the same number of significant figures as the measurement with the fewest significant figures.
\( 0.02856 \) (5 sig figs)
\( 298.15 \) (5 sig figs)
\( 0.112 \) (3 sig figs) - This is the limiting factor.
\( 0.5785 \) (4 sig figs)
So the answer should have 3 significant figures.
\( \frac{0.02856 \times 298.15 \times 0.112}{0.5785} = \frac{0.953698}{0.5785} = 1.64857 \approx 1.65 \)
2. For multiplication/division, the answer should have the same number of significant figures as the measurement with the fewest significant figures.
\( 5 \) (1 sig fig, assuming it's an exact number for counting or the context implies otherwise. If it's a measured value, it's 1 sig fig. However, usually, a single digit like 5 is considered exact or has unlimited sig figs in introductory problems. If 5 is an exact count, then the answer is limited by 5.364). Assuming 5 is a measured value with 1 sig fig, then the result should be 1 sig fig.
\( 5.364 \) (4 sig figs)
If 5 is an exact number, the answer should have 4 significant figures. If 5 is a measurement, it has 1 significant figure.
Here, the solution indicates 4 significant figures, implying '5' is treated as an exact number.
\( 5 \times 5.364 = 26.82 \). It should have 4 significant figures if 5 is exact.
3. For addition/subtraction, the answer should have the same number of decimal places as the measurement with the fewest decimal places.
\( 0.0125 \) (4 decimal places)
\( 0.7864 \) (4 decimal places)
\( 0.0215 \) (4 decimal places)
The answer should have 4 decimal places.
\( 0.0125 + 0.7864 + 0.0215 = 0.8204 \). It should have 4 significant figures.
In simple words: For multiplying and dividing, the answer should only have as many important digits as the number in your calculation with the fewest important digits. For adding and subtracting, the answer should only have as many digits after the decimal point as the number in your calculation with the fewest digits after the decimal.

Exam Tip: Master the rules for significant figures in both multiplication/division (limited by the fewest significant figures) and addition/subtraction (limited by the fewest decimal places) to ensure correct precision in your final answers.

 

Question 32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Answer: The molar mass of naturally occurring Argon is calculated as follows:
\[ = \frac{ (35.96755 \times 0.337) + (37.96272 \times 0.063) + (39.9624 \times 99.600) }{ 100.00 } \]
\[ = \frac{ 12.12106 + 2.39165 + 3980.255 }{ 100 } \]
\[ = \frac{ 3994.76 }{ 100 } = 39.948 \text{ g mol}^{-1} \]In simple words: To find the average molar mass, we multiply each isotope's mass by its abundance percentage. Then we add these results together and divide by 100. This gives us the overall molar mass for argon as it appears in nature.

Exam Tip: Remember to use the precise isotopic molar masses and their respective abundances. Ensure you divide the sum of (mass × abundance) products by the total abundance (which is 100%) to get the accurate average molar mass.

 

Question 33. Calculate the number of atoms in each of the following:
1. 52 moles of Ar
2. 4 u of He
3. 52 g of He.
Answer:
(1) We know that \( 1 \) mole of Argon (\( \text{Ar} \)) contains \( 6.023 \times 10^{23} \) atoms.
\( \implies \) So, \( 52 \) moles of \( \text{Ar} \) will contain \( 6.023 \times 10^{23} \times 52 \) atoms \( = 3.13 \times 10^{25} \) atoms.
(2) We know that \( 4 \) u of Helium (\( \text{He} \)) is equal to \( 1 \) atom of \( \text{He} \).
\( \implies \) Therefore, \( 52 \) u of \( \text{He} \) will be \( \frac{ 1 }{ 4 } \times 52 = 13 \) atoms.
(3) We know that \( 4 \) g of \( \text{He} \) contains \( 6.023 \times 10^{23} \) atoms.
\( \implies \) So, \( 52 \) g of \( \text{He} \) will contain \( \frac{ 6.023 \times 10^{23} \times 52 }{ 4 } = 7.8286 \times 10^{24} \) atoms.
In simple words: To find the number of atoms, use Avogadro's number for moles or grams. For atomic mass units (u), remember that the atomic mass of an element in u represents one atom.

Exam Tip: Pay close attention to the units given (moles, u, or grams) as each requires a different approach using Avogadro's number or the atomic mass in u.

 

Question 34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives \( 3.38 \text{ g} \) carbon dioxide, \( 0.690 \text{ g} \) of water, and no other products. A volume of \( 10.0 \text{ L} \) (measured at STP) of this welding gas is found to weigh \( 11.6 \text{ g} \). Calculate
1. empirical formula
2. the molar mass of the gas, and
3. molecular formula.
Answer: The welding fuel gas consists only of carbon (\( \text{C} \)) and hydrogen (\( \text{H} \)), which can be written as \( \text{C}_x\text{H}_y \).
First, let's find the molar mass of the gas.
A \( 10.0 \text{ L} \) sample of this gas at STP has a weight of \( 11.6 \text{ g} \).
Since \( 22.4 \text{ L} \) of any gas at STP corresponds to its molecular mass:
Weight of \( 22.4 \text{ L} \) of this gas at STP \( = \frac{ 11.6 }{ 10.0 } \times 22.4 \text{ g} = 25.98 \text{ g} \).
Thus, the molar mass of the welding fuel gas is approximately \( 26.0 \text{ g/mol} \) (rounding to a whole number).
From the burning data, we need to find the amounts of C and H.
Mass of \( \text{C} \) in \( 3.38 \text{ g CO}_2 \): \( \text{Molar mass of CO}_2 = 12 + 2 \times 16 = 44 \text{ g/mol} \).
Mass of \( \text{C} = 3.38 \text{ g CO}_2 \times \frac{ 12 \text{ g C} }{ 44 \text{ g CO}_2 } = 0.9218 \text{ g C} \).
Moles of \( \text{C} = \frac{ 0.9218 \text{ g} }{ 12.01 \text{ g/mol} } = 0.07675 \text{ mol} \).
Mass of \( \text{H} \) in \( 0.690 \text{ g H}_2\text{O} \): \( \text{Molar mass of H}_2\text{O} = 2 \times 1 + 16 = 18 \text{ g/mol} \).
Mass of \( \text{H} = 0.690 \text{ g H}_2\text{O} \times \frac{ 2 \text{ g H} }{ 18 \text{ g H}_2\text{O} } = 0.07667 \text{ g H} \).
Moles of \( \text{H} = \frac{ 0.07667 \text{ g} }{ 1.008 \text{ g/mol} } = 0.07606 \text{ mol} \).
To find the empirical formula, divide by the smallest number of moles:
Ratio of \( \text{C} : \text{H} = \frac{ 0.07675 }{ 0.07606 } : \frac{ 0.07606 }{ 0.07606 } \approx 1 : 1 \).
1. The empirical formula is \( \text{CH} \).
Empirical formula mass of \( \text{CH} = 12.01 + 1.008 = 13.018 \text{ g/mol} \).
2. The molar mass of the gas is \( 26.0 \text{ g/mol} \).
To find the molecular formula, calculate \( n = \frac{ \text{Molar mass} }{ \text{Empirical formula mass} } \).
\( n = \frac{ 26.0 }{ 13.018 } \approx 2 \).
3. The molecular formula is \( (\text{CH})_2 = \text{C}_2\text{H}_2 \).
In simple words: We first find the total weight of the gas for a standard volume to get its molar mass. Then, by looking at how much carbon dioxide and water are made when the gas burns, we figure out the amounts of carbon and hydrogen. This helps us find the simplest formula (empirical) and then the real formula (molecular).

Exam Tip: Remember to calculate the moles of carbon from \( \text{CO}_2 \) and hydrogen from \( \text{H}_2\text{O} \) when given combustion data. The molar volume of any gas at STP is \( 22.4 \text{ L} \), which is key for finding molar mass.

 

Question 35. Calcium carbonate reacts with aqueous \( \text{HCl} \) to give \( \text{CaCl}_2 \) and \( \text{CO}_2 \) according to the reaction, \( \text{CaCO}_3\text{(s)} + 2\text{ HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} \). What mass of \( \text{CaCO}_3 \) is required to react completely with \( 25 \text{ mL} \) of \( 0.75 \text{ M HCl} \)?
Answer: The balanced chemical reaction is:
\[ \text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} \] First, we need to find the moles of \( \text{HCl} \) present in \( 25 \text{ mL} \) of \( 0.75 \text{ M HCl} \).
Molarity \( = \frac{ \text{Moles of solute} }{ \text{Volume of solution in L} } \)
Moles of \( \text{HCl} = \text{Molarity} \times \text{Volume (L)} = 0.75 \text{ mol/L} \times \frac{ 25 }{ 1000 } \text{ L} = 0.01875 \text{ mol HCl} \).
From the balanced equation, \( 1 \) mole of \( \text{CaCO}_3 \) reacts with \( 2 \) moles of \( \text{HCl} \).
So, moles of \( \text{CaCO}_3 \) required \( = \frac{ 0.01875 \text{ mol HCl} }{ 2 } = 0.009375 \text{ mol CaCO}_3 \).
Now, calculate the mass of \( \text{CaCO}_3 \) required.
Molar mass of \( \text{CaCO}_3 = 40.08 (\text{Ca}) + 12.01 (\text{C}) + 3 \times 16.00 (\text{O}) = 100.09 \text{ g/mol} \).
Mass of \( \text{CaCO}_3 = \text{Moles} \times \text{Molar mass} = 0.009375 \text{ mol} \times 100.09 \text{ g/mol} = 0.9383 \text{ g CaCO}_3 \).
Let's recheck the calculation based on the provided solution's approach:
The source calculates mass of HCl first:
\( 1000 \text{ mL} \) of \( 1.0 \text{ M HCl} \) contains \( 36.5 \text{ g HCl} \).
Mass of \( \text{HCl} \) in \( 25 \text{ mL} \) of \( 0.75 \text{ M HCl} = \frac{ 36.5 }{ 1000 } \times 25 \times 0.75 = 0.6844 \text{ g HCl} \).
According to the equation, \( 2 \) moles of \( \text{HCl} \) (i.e., \( 2 \times (1 + 35.5) = 73 \text{ g HCl} \)) react with \( 1 \) mole of \( \text{CaCO}_3 \) (i.e., \( 100.0 \text{ g CaCO}_3 \)).
So, \( 73 \text{ g HCl} \) reacts with \( 100.0 \text{ g CaCO}_3 \).
Therefore, \( 0.6844 \text{ g HCl} \) will react with \( = \frac{ 100 }{ 73 } \times 0.6844 = 0.94 \text{ g CaCO}_3 \).
In simple words: We first find out how much \( \text{HCl} \) we have. Then, using the chemical equation, which tells us how \( \text{HCl} \) and \( \text{CaCO}_3 \) react, we can figure out the exact amount of \( \text{CaCO}_3 \) needed to use up all that \( \text{HCl} \).

Exam Tip: Always write a balanced chemical equation first. Convert given quantities to moles to use stoichiometry, and then convert back to mass if needed. Pay attention to units and significant figures throughout the calculation.

 

Question 36. Chlorine is prepared in the laboratory by treating manganese dioxide (\( \text{MnO}_2 \)) with hydrochloric acid according to the reaction \( 4 \text{ HCl(aq)} + \text{MnO}_2\text{(s)} \rightarrow 2\text{H}_2\text{O(l)} + \text{MnCl}_2\text{(aq)} + \text{Cl}_2\text{(g)} \). How many grams of \( \text{HCl} \) react with \( 5.0 \text{ g} \) of manganese dioxide?
Answer: The chemical equation for the reaction is:
\[ 4 \text{ HCl(aq)} + \text{MnO}_2\text{(s)} \rightarrow 2\text{H}_2\text{O(l)} + \text{MnCl}_2\text{(aq)} + \text{Cl}_2\text{(g)} \] First, we need to find the molar masses of \( \text{MnO}_2 \) and \( \text{HCl} \).
Molar mass of \( \text{MnO}_2 = \text{Atomic mass of Mn} + (2 \times \text{Atomic mass of O}) = 55 + (2 \times 16) = 55 + 32 = 87 \text{ g/mol} \).
Molar mass of \( \text{HCl} = \text{Atomic mass of H} + \text{Atomic mass of Cl} = 1 + 35.5 = 36.5 \text{ g/mol} \).
From the balanced equation, \( 1 \) mole of \( \text{MnO}_2 \) reacts with \( 4 \) moles of \( \text{HCl} \).
In terms of mass:
\( 87 \text{ g} \) of \( \text{MnO}_2 \) reacts with \( 4 \times 36.5 \text{ g HCl} = 146 \text{ g HCl} \).
Now, we need to calculate how many grams of \( \text{HCl} \) will react with \( 5.0 \text{ g} \) of \( \text{MnO}_2 \):
Mass of \( \text{HCl} \) reacting \( = \frac{ 146 \text{ g HCl} }{ 87 \text{ g MnO}_2 } \times 5.0 \text{ g MnO}_2 = 8.40 \text{ g HCl} \).
Thus, the amount of \( \text{HCl} \) in grams that will react with \( 5.0 \text{ g} \) of manganese dioxide is \( 8.40 \text{ g} \).
In simple words: To find out how much hydrochloric acid is needed, we look at the balanced chemical recipe. It tells us how many grams of manganese dioxide react with how many grams of hydrochloric acid. We then use this ratio to calculate the exact amount for 5 grams of manganese dioxide.

Exam Tip: For stoichiometry problems, it's crucial to first balance the chemical equation. Then, use the molar masses to convert between mass and moles, and apply the mole ratios from the balanced equation correctly.