RBSE Solutions Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry

 

Question 1. Number of significant figures in 0.0287 is
(a) 5
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: The number 0.0287 has three important digits because we start counting from the first non-zero digit.

🎯 Exam Tip: Remember that leading zeros (zeros before the first non-zero digit) are never significant.

 

Question 3. Volume of 2 g methane at STP will be
(a) 2.8 L
(b) 5.6 L
(c) 11.2 L
(d) 22.4
Answer: (a) 2.8 L
In simple words: To find the volume, first calculate the moles of methane by dividing its mass by its molar mass, then multiply by the molar volume at STP (22.4 L/mol).

🎯 Exam Tip: For problems at STP, always remember that 1 mole of any gas occupies 22.4 liters.

 

Question 4. At STP, the number of molecules in 1 ml of ideal will be
(a) \( 6.023 \times 10^{23} \)
(b) \( 2.69 \times 10^{19} \)
(c) \( 2.69 \times 10^{23} \)
(d) \( 458 \times 10^{26} \)
Answer: (b) \( 2.69 \times 10^{19} \)
In simple words: At standard conditions, even a small amount like 1 ml of gas contains a very large number of molecules, found using Avogadro's number and the molar volume.

🎯 Exam Tip: Be careful with units and conversions (ml to L) when calculating the number of molecules at STP.

 

Question 5. Which of the following has least weight?
(a) 108 g of silver
(b) 1 mol of sulphur
(c) 1 g atom of nitrogen
(d) \( 3.011 \times 10^{23} \) atoms of carbon
Answer: (d) \( 3.011 \times 10^{23} \) atoms of carbon
In simple words: To find the least weight, convert all options to grams using their atomic or molar masses and Avogadro's number, then compare.

🎯 Exam Tip: This question tests your understanding of mole concepts and how to convert between moles, atoms, and mass for different elements.

 

Question 7. What is the difference between 8.0 g and,8.000 g weight?
Answer: The weight "8.0 g" has two significant figures, meaning its precision is up to one decimal place. On the other hand, "8.000 g" has four significant figures, showing that its precision is much higher, up to three decimal places. This difference indicates how accurately the measurement was made.
In simple words: 8.0 g is less exact than 8.000 g. More zeros after the decimal point show a more precise measurement.

🎯 Exam Tip: Remember that trailing zeros after a decimal point are always significant and indicate precision.

 

Question 8. Express 3600 g in three significant figures.
Answer: To express 3600 g in three significant figures, we write it in scientific notation as \( 3.60 \times 10^3 \) g. The zero after the 6 is added to make sure there are exactly three significant figures.
In simple words: 3600 g in three significant figures is \( 3.60 \times 10^3 \) g.

🎯 Exam Tip: When converting to scientific notation for significant figures, include trailing zeros if necessary to meet the required count.

 

Question 9. How many gram molecules of hydrogen are present in 20 g hydrogen?
Answer:
The molar mass of hydrogen molecules (\( H_2 \)) is 2 g/mol. So, 2 g of hydrogen molecules equals 1 mole. This means 1 gram molecule of hydrogen is \( \frac{1}{2} \) mol. Therefore, for 20 g of hydrogen, the number of gram molecules is \( \frac{1}{2} \times 20 = 10 \) gram molecules.
In simple words: Since 2 grams of hydrogen is one mole, 20 grams of hydrogen is equal to 10 gram molecules.

🎯 Exam Tip: Remember that a "gram molecule" refers to one mole of a molecular substance, and for hydrogen, it's \( H_2 \), not individual H atoms.

 

Question 10. How many molecules will be present in 64 g oxygen?
Answer:
The gram molecular weight of oxygen (\( O_2 \)) is 32 g. This means 32 g of \( O_2 \) contains \( 6.023 \times 10^{23} \) molecules (Avogadro's number). Therefore, 64 g of \( O_2 \) will contain \( (6.023 \times 10^{23}) \times \frac{64}{32} \). This calculates to \( 6.023 \times 10^{23} \times 2 = 12.046 \times 10^{23} \) molecules.
In simple words: Since 32g of oxygen is one mole, 64g is two moles. So, it has twice Avogadro's number of molecules.

🎯 Exam Tip: Always use Avogadro's number to convert between moles and the number of molecules or atoms.

 

Question 12. State Avogadro's law,
Answer: Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules". This means if you have two different gases in containers of the same size, at the same temperature and pressure, they will both have the same number of gas particles.
In simple words: Same amount of gas, same space, same conditions, means same number of tiny particles.

🎯 Exam Tip: The key conditions for Avogadro's law are equal volumes, same temperature, and same pressure. Remember it's about the number of *molecules*, not necessarily mass.

 

Question 13. How many atoms of carbon will be present in 0.1 mole of \( C_6H_{12}O_6 \)?
Answer:
In 1 mole of \( C_6H_{12}O_6 \), there are 6 moles of carbon atoms. So, 1 mole of \( C_6H_{12}O_6 \) contains \( 6 \times 6.023 \times 10^{23} \) carbon atoms. Therefore, in 0.1 mole of \( C_6H_{12}O_6 \), the number of carbon atoms will be: \( 0.1 \times 6 \times 6.023 \times 10^{23} \) atoms This calculates to \( 3.6138 \times 10^{23} \) atoms of carbon.
In simple words: One molecule of \( C_6H_{12}O_6 \) has 6 carbon atoms. So, 0.1 mole of this substance has 0.1 times 6 times Avogadro's number of carbon atoms.

🎯 Exam Tip: When calculating atoms in a compound, first find the moles of the specific element within the compound, then multiply by Avogadro's number.

 

Question 14. What are significant figures?
Answer: Significant figures are all the digits in a number that are considered reliable or important for expressing its precision or accuracy. We start counting them from the first digit that is not zero. They tell us how accurate a measurement or calculation is.
In simple words: Significant figures are the important digits in a number, showing how accurate it is, starting from the first non-zero number.

🎯 Exam Tip: Understand the rules for identifying significant figures, especially regarding zeros (leading, captive, and trailing) and how they indicate precision.

 

Question 15. What is atomicity of a gas?
Answer: Atomicity of a gas is the total number of atoms of an element that are present in one molecule of that element. For example, a molecule of helium (He) has only one atom, so its atomicity is 1 (monoatomic). A molecule of hydrogen (\( H_2 \)) has two atoms, so its atomicity is 2 (diatomic).
In simple words: Atomicity tells us how many atoms make up one molecule of a gas.

🎯 Exam Tip: Remember common examples: monoatomic (He, Ne), diatomic (\( H_2, O_2, N_2, Cl_2 \)), triatomic (\( O_3 \)), and polyatomic (\( P_4, S_8 \)).

 

Question 16. Calculate the number of electrons present in 36 g water.
Answer:
First, find the number of electrons in one molecule of water (\( H_2O \)): Hydrogen (H) has 1 electron, and there are 2 H atoms: \( 2 \times 1 = 2 \) electrons. Oxygen (O) has 8 electrons, and there is 1 O atom: \( 1 \times 8 = 8 \) electrons. Total electrons in 1 molecule of \( H_2O = 2 + 8 = 10 \) electrons. Next, find the number of moles in 36 g of water: Molar mass of \( H_2O = 2(1) + 16 = 18 \) g/mol. Number of moles = \( \frac{36 \text{ g}}{18 \text{ g/mol}} = 2 \) moles. Now, calculate the number of molecules in 2 moles of water: Number of molecules = \( 2 \text{ moles} \times 6.023 \times 10^{23} \text{ molecules/mole} = 12.046 \times 10^{23} \) molecules. Finally, calculate the total number of electrons: Total electrons = Number of molecules \( \times \) electrons per molecule Total electrons = \( 12.046 \times 10^{23} \times 10 \) electrons Total electrons = \( 12.046 \times 10^{24} \) electrons.
In simple words: One water molecule has 10 electrons. 36 grams of water is 2 moles. So, we multiply 2 moles by Avogadro's number to get the total molecules, then multiply by 10 to find all the electrons.

🎯 Exam Tip: To find the number of electrons, first determine the electrons in one molecule, then convert the given mass to moles, then to molecules, and finally multiply by the electrons per molecule.

 

Question 17. What do you understand by molecular mass?
Answer: Molecular mass is a number that tells us the mass of one molecule of a substance. It is found by adding up the atomic masses of all the atoms present in that molecule. This mass is relative to the mass of a carbon-12 atom, which is set at exactly 12 atomic mass units (amu). For example, the molecular mass of water (\( H_2O \)) is 18 amu.
In simple words: Molecular mass is the total weight of all atoms in a molecule, compared to carbon-12.

🎯 Exam Tip: To calculate molecular mass, always ensure you multiply the atomic mass of each element by the number of its atoms in the chemical formula before summing them up.

 

Question 18. What is the mass of water which contains 50% heavy water (D2O)?
Answer:
If water contains 50% regular water (\( H_2O \)) and 50% heavy water (\( D_2O \)), we can find the average mass. The molecular mass of \( H_2O \) is 18 amu. The molecular mass of \( D_2O \) (where D is deuterium, mass 2) is 2(2) + 16 = 20 amu. The average mass of water in this mixture would be the sum of their masses divided by 2 (since it's a 50/50 mix): Average mass = \( \frac{(18 + 20)}{2} = \frac{38}{2} = 19 \) amu.
In simple words: To find the mass of water with 50% heavy water, just average the molecular weights of normal water (18) and heavy water (20).

🎯 Exam Tip: When dealing with mixtures of isotopes or different forms of a molecule, remember to calculate the weighted average of their masses based on their proportions.

 

Question 19. What do you understand by standard temperature and pressure?
Answer: Standard Temperature and Pressure (STP) is a set of standard conditions used for experimental measurements, making it easier to compare results. The standard temperature is 273.15 K (which is 0° Celsius or 32° Fahrenheit), and the standard pressure is 1 atmosphere (atm) or 760 mm of mercury. This temperature is actually the freezing point of pure water at sea level. At STP, one mole of any ideal gas is considered to occupy a volume of 22.4 liters.
In simple words: STP is a common reference point in science, where temperature is 0°C and pressure is 1 atmosphere.

🎯 Exam Tip: Always remember the specific values for STP (273.15 K and 1 atm) as they are frequently used in gas law calculations.

 

Question 20. At STP, x mL N2 gas reacts Completely with xmL O2 gas and forms gas A. What will be the molecular fc s unchanged after reaction?

🎯 Exam Tip: For questions that seem incomplete or malformed, focus on any parts that can be answered from the provided information, while noting the missing context.

 

Question 21. What is the weight of 1 mole of the following
(i) NaCl
(ii) CaCO3
Answer:
(i) For NaCl (Sodium Chloride): Molecular mass of NaCl = Atomic mass of Na + Atomic mass of Cl = 23 amu + 35.5 amu = 58.5 amu. The molar mass of NaCl is 58.5 g/mol. Therefore, the weight of 1 mole of NaCl is 58.5 g. (ii) For \( CaCO_3 \) (Calcium Carbonate): Molecular mass of \( CaCO_3 = \)(1 \( \times \) Atomic mass of Ca) + (1 \( \times \) Atomic mass of C) + (3 \( \times \) Atomic mass of O) Molecular mass = \( (1 \times 40) + (1 \times 12) + (3 \times 16) \) Molecular mass = \( 40 + 12 + 48 = 100 \) amu. The molar mass of \( CaCO_3 \) is 100 g/mol. Therefore, the weight of 1 mole of \( CaCO_3 \) is 100 g.
In simple words: The weight of one mole of a substance is equal to its molar mass, which is found by adding the atomic masses of all atoms in its formula.

🎯 Exam Tip: To find the weight of one mole, always calculate the molar mass by summing the atomic masses of all constituent atoms, paying attention to subscripts.

 

Question 22. Calculate number of significant figures in each of the following:
(i) 0.00468
(ii) 753
Answer:
(i) The number 0.00468 has 3 significant figures. (Leading zeros are not significant). (ii) The number 753 has 3 significant figures. (All non-zero digits are significant).
In simple words: Both numbers, 0.00468 and 753, have three significant figures.

🎯 Exam Tip: Remember the rules for significant figures: non-zero digits are always significant; leading zeros are not significant; captive zeros (between non-zeros) are significant.

 

Question 23. What will be the mass of 2 moles of following:
(i) MgSO4
(ii) KCI
Answer:
(i) For \( MgSO_4 \) (Magnesium Sulfate): Molar mass of \( MgSO_4 = \) Atomic mass of Mg + Atomic mass of S + 4 \( \times \) Atomic mass of O Molar mass = \( 24.31 + 32.07 + 4 \times 16.00 = 24.31 + 32.07 + 64.00 = 120.38 \) g/mol. Mass of 1 mole of \( MgSO_4 \) is 120.38 g. Therefore, mass of 2 moles of \( MgSO_4 = 2 \times 120.38 = 240.76 \) g. (ii) For KCl (Potassium Chloride): Molar mass of KCl = Atomic mass of K + Atomic mass of Cl = \( 39.10 + 35.45 = 74.55 \) g/mol. Mass of 1 mole of KCl is 74.55 g. Therefore, mass of 2 moles of KCl = \( 2 \times 74.55 = 149.10 \) g.
In simple words: To find the mass of 2 moles, first calculate the molar mass of each compound, then multiply that molar mass by 2.

🎯 Exam Tip: Always calculate the molar mass correctly for each compound before multiplying by the number of moles required.

 

Question 24. Calculate number of significant figures, in each of the following:
(i) 0.868
(ii) \( 3.865 \times 10^4 \)
Answer:
(i) In 0.868, all non-zero digits (8, 6, 8) are significant. So, there are 3 significant figures. (ii) In \( 3.865 \times 10^4 \), the significant figures are determined by the number before the power of 10. The digits 3, 8, 6, 5 are all non-zero and therefore significant. So, there are 4 significant figures.
In simple words: The number 0.868 has three important digits, and \( 3.865 \times 10^4 \) has four important digits.

🎯 Exam Tip: When a number is in scientific notation, only the digits in the coefficient (the part before \( \times 10^n \)) are counted as significant figures.

 

Question 25. What is mole? Explain.
Answer: A mole is a unit of measurement used in chemistry to express amounts of a chemical substance. One mole of any substance contains the same number of "entities" (like atoms, molecules, or ions) as there are atoms in exactly 12 grams of the carbon-12 isotope. This specific number is known as Avogadro's number (\( N_A \)), which is approximately \( 6.023 \times 10^{23} \) entities per mole. The term 'mole' was named after Amedeo Avogadro, which is why Avogadro's number is also commonly used.
In simple words: A mole is a count of a very large number of tiny particles, like atoms or molecules, similar to how a 'dozen' counts 12 items.

🎯 Exam Tip: Understand that the mole provides a link between the microscopic world (atoms/molecules) and the macroscopic world (grams), using Avogadro's number.

 

Question 26. What is the Limiting Reagent? Explain with the help of example.
Answer: A limiting reagent is a reactant in a chemical reaction that is completely used up first. Because it runs out, it limits the amount of product that can be formed, even if other reactants are still present. It essentially determines how much of the final product you can make. Let's consider an example: the reaction between hydrogen (\( H_2 \)) and oxygen (\( O_2 \)) to form water (\( H_2O \)): \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \) Suppose we start with 10 moles of \( H_2 \) and 7 moles of \( O_2 \).

From the balanced equation, 2 moles of \( H_2 \) react with 1 mole of \( O_2 \). If we have 10 moles of \( H_2 \), it would need 5 moles of \( O_2 \) to react completely (\( 10 \text{ moles } H_2 \times \frac{1 \text{ mole } O_2}{2 \text{ moles } H_2} = 5 \text{ moles } O_2 \)). Since we started with 7 moles of \( O_2 \), and only 5 moles are needed, \( O_2 \) is in excess (7-5=2 moles of \( O_2 \) are left over). Hydrogen (\( H_2 \)) is completely used up, so \( H_2 \) is the limiting reagent in this reaction. It stops the reaction once all 10 moles are consumed.
In simple words: A limiting reagent is the ingredient that runs out first in a chemical recipe, stopping the reaction and deciding how much product can be made. Hydrogen is the limiting reagent here.

🎯 Exam Tip: To identify a limiting reagent, calculate how much product each reactant could form if it were completely consumed. The reactant that yields the least product is the limiting reagent.

 

Question 28. Define molarity, molality and normality of a solution.
Answer:
Molarity (M): Molarity is a measure of the concentration of a solution, defined as the number of moles of solute dissolved per liter of solution. Formula: \( \text{M} = \frac{\text{Moles of solute}}{\text{Volume of solution (in liters)}} \). The unit for Molarity is mol/L.Molality (m): Molality is another measure of concentration, defined as the number of moles of solute dissolved per kilogram of solvent. Formula: \( \text{m} = \frac{\text{Moles of solute}}{\text{Weight of solvent (in kg)}} \). The unit for Molality is mol/kg.Normality (N): Normality is defined as the number of gram equivalents of solute dissolved per liter of solution. It is often used in acid-base reactions and redox reactions. Formula: \( \text{N} = \frac{\text{No. of gram equivalent of solute}}{\text{Volume of solution (in liters)}} \). The unit for Normality is gram equivalent per liter.
In simple words: Molarity is moles per liter of solution; Molality is moles per kilogram of solvent; Normality is gram equivalents per liter of solution. They all describe how much substance is dissolved.

🎯 Exam Tip: Clearly differentiate between molarity (volume of solution) and molality (mass of solvent). Normality involves equivalents, which depend on the reaction type.

 

Question 29. Write the formula of the following compounds and calculate molecular mass :
(i) Calcium Carbonate
(ii) Magnesium Phosphate
(iii) Ferric chloride,
Answer:
(i) Calcium Carbonate: Formula: \( CaCO_3 \) Molecular mass = (Atomic mass of Ca) + (Atomic mass of C) + (3 \( \times \) Atomic mass of O) Molecular mass = \( 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \) g/mol. (ii) Magnesium Phosphate: Formula: \( Mg_3(PO_4)_2 \) Molecular mass = (3 \( \times \) Atomic mass of Mg) + (2 \( \times \) Atomic mass of P) + (8 \( \times \) Atomic mass of O) Molecular mass = \( (3 \times 24) + (2 \times 31) + (8 \times 16) \) Molecular mass = \( 72 + 62 + 128 = 262 \) g/mol. (iii) Ferric Chloride: Formula: \( FeCl_3 \) Molecular mass = (Atomic mass of Fe) + (3 \( \times \) Atomic mass of Cl) Molecular mass = \( 55.8 + (3 \times 35.5) \) Molecular mass = \( 55.8 + 106.5 = 162.3 \) g/mol.
In simple words: First, write the correct chemical formula for each compound. Then, add up the atomic masses of all atoms in the formula to find its molecular mass.

🎯 Exam Tip: Pay close attention to subscripts in chemical formulas, especially when they are outside parentheses, as they indicate the number of atoms for each element.

 

Question 30. Calculate the number of moles in the following
(i) 100 g of CaCO3
(ii) 80 g of Oxygen
(iii) 10 g of C12H22O11
Answer:
(i) For 100 g of \( CaCO_3 \): The molar mass of \( CaCO_3 \) is 100 g/mol. So, 100 g of \( CaCO_3 \) contains 1 mole. (ii) For 80 g of Oxygen (\( O_2 \)): The molar mass of Oxygen (\( O_2 \)) is 2 \( \times \) 16 = 32 g/mol. So, 32 g of Oxygen contains 1 mole. Therefore, 80 g of Oxygen will contain \( \frac{80}{32} = 2.5 \) moles. (iii) For 10 g of \( C_{12}H_{22}O_{11} \) (Sucrose): The molar mass of Sucrose (\( C_{12}H_{22}O_{11} \)) is \( (12 \times 12) + (22 \times 1) + (11 \times 16) = 144 + 22 + 176 = 342 \) g/mol. So, 342 g of sucrose contains 1 mole. Therefore, 10 g of sucrose will contain \( \frac{10}{342} \approx 0.029 \) moles.
In simple words: To find the number of moles, divide the given mass by the substance's molar mass.

🎯 Exam Tip: Always make sure to use the correct molar mass for the given substance, differentiating between atomic mass and molecular mass.

 

Question 32. What is the number of atoms present in 32 g of oxygen gas and 14 g of nitrogen gas?
Answer:
For Oxygen gas (\( O_2 \)): The molar mass of oxygen gas (\( O_2 \)) is 32 g/mol. So, 32 g of oxygen gas contains 1 mole of \( O_2 \) molecules. Since 1 mole of \( O_2 \) molecules contains \( 6.023 \times 10^{23} \) molecules, and each \( O_2 \) molecule has 2 oxygen atoms, the number of oxygen atoms in 32 g of oxygen gas = \( 1 \text{ mole } \times 6.023 \times 10^{23} \text{ molecules/mole} \times 2 \text{ atoms/molecule} = 12.046 \times 10^{23} \) atoms. For Nitrogen gas (\( N_2 \)): The molar mass of nitrogen gas (\( N_2 \)) is \( 2 \times 14 = 28 \) g/mol. In 14 g of nitrogen gas, the number of moles of \( N_2 \) is \( \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \) moles. The number of nitrogen molecules in 0.5 moles of \( N_2 \) is \( 0.5 \times 6.023 \times 10^{23} \) molecules. Since each \( N_2 \) molecule has 2 nitrogen atoms, the number of nitrogen atoms in 14 g of nitrogen gas = \( 0.5 \text{ moles} \times 6.023 \times 10^{23} \text{ molecules/mole} \times 2 \text{ atoms/molecule} = 6.023 \times 10^{23} \) atoms.
In simple words: For oxygen, 32g is one mole of \( O_2 \), so it has two times Avogadro's number of atoms. For nitrogen, 14g is half a mole of \( N_2 \), so it has one times Avogadro's number of atoms.

🎯 Exam Tip: Always distinguish between the number of molecules and the number of atoms in a molecular substance like \( O_2 \) or \( N_2 \), multiplying by the atomicity when necessary.

 

Question 33. Calculate molar mass of the following:
(i) HNO3
Answer:
(i) For \( HNO_3 \) (Nitric Acid): Molar mass of \( HNO_3 \) = Atomic mass of H + Atomic mass of N + (3 \( \times \) Atomic mass of O) Molar mass = \( 1 + 14 + (3 \times 16) = 1 + 14 + 48 = 63 \) g/mol.
(ii) For \( CO_2 \) (Carbon Dioxide): Molar mass of \( CO_2 \) = Atomic mass of C + (2 \( \times \) Atomic mass of O) Molar mass = \( 12 + (2 \times 16) = 12 + 32 = 44 \) g/mol. (iii) For \( C_2H_6 \) (Ethane): Molar mass of \( C_2H_6 \) = (2 \( \times \) Atomic mass of C) + (6 \( \times \) Atomic mass of H) Molar mass = \( (2 \times 12) + (6 \times 1) = 24 + 6 = 30 \) g/mol.
In simple words: To find the molar mass, add up the atomic weights of all atoms in the chemical formula for each compound.

🎯 Exam Tip: Practice memorizing common atomic masses or know how to quickly look them up to calculate molar masses efficiently.

 

Question 34. Which of the following contains maximum number of molecules :
(i) 36 g water
(ii) 28 g carbon monoxide
Answer:
(i) For 36 g of water (\( H_2O \)): The molar mass of water (\( H_2O \)) is 18 g/mol. Number of moles = \( \frac{36 \text{ g}}{18 \text{ g/mol}} = 2 \) moles. 1 mole of water contains \( 6.023 \times 10^{23} \) molecules. So, 2 moles of water contain \( 2 \times 6.023 \times 10^{23} = 12.046 \times 10^{23} \) molecules. (ii) For 28 g of carbon monoxide (CO): The molar mass of carbon monoxide (CO) is Atomic mass of C + Atomic mass of O = \( 12 + 16 = 28 \) g/mol. Number of moles = \( \frac{28 \text{ g}}{28 \text{ g/mol}} = 1 \) mole. So, 1 mole of CO contains \( 6.023 \times 10^{23} \) molecules. Comparing the two, 36 g of water contains \( 12.046 \times 10^{23} \) molecules, which is more than the \( 6.023 \times 10^{23} \) molecules in 28 g of carbon monoxide. Therefore, 36 g of water contains the maximum number of molecules.
In simple words: 36 grams of water is 2 moles, so it has more molecules than 28 grams of carbon monoxide, which is only 1 mole.

🎯 Exam Tip: To compare the number of molecules, convert the given mass of each substance into moles and then use Avogadro's number. More moles mean more molecules.

 

Question 35. Which of the following contains minimum number of molecules?
(i) 46 g of ethyl alcohol
(ii) 54 g of nitrogen pentaoxide
Answer:
(i) For 46 g of ethyl alcohol (\( C_2H_5OH \)): The molar mass of ethyl alcohol (\( C_2H_5OH \)) is \( (2 \times 12) + (6 \times 1) + (1 \times 16) = 24 + 6 + 16 = 46 \) g/mol. So, 46 g of ethyl alcohol contains 1 mole, which means it has \( 6.023 \times 10^{23} \) molecules. (ii) For 54 g of nitrogen pentaoxide (\( N_2O_5 \)): The molar mass of nitrogen pentaoxide (\( N_2O_5 \)) is \( (2 \times 14) + (5 \times 16) = 28 + 80 = 108 \) g/mol. Number of moles in 54 g = \( \frac{54 \text{ g}}{108 \text{ g/mol}} = 0.5 \) moles. So, 0.5 moles of \( N_2O_5 \) will contain \( 0.5 \times 6.023 \times 10^{23} = 3.0115 \times 10^{23} \) molecules. Comparing the two, \( 3.0115 \times 10^{23} \) molecules (from 54 g of nitrogen pentaoxide) is less than \( 6.023 \times 10^{23} \) molecules (from 46 g of ethyl alcohol). Therefore, 54 g of nitrogen pentaoxide will have the minimum number of molecules.
In simple words: 46 grams of ethyl alcohol is 1 mole, but 54 grams of nitrogen pentaoxide is only half a mole. So, nitrogen pentaoxide has fewer molecules.

🎯 Exam Tip: Calculate the number of moles for each substance using its molar mass. The substance with fewer moles will have fewer molecules.

 

Question 37. What is Limiting reagent? Identify limiting reagent in reaction between 3.0 g H2 and 29 g O2.
Answer: A limiting reagent is a reactant that is completely consumed during a chemical reaction. It stops the reaction and determines the maximum amount of product that can be formed. It does not matter if other reactants are still left over. Consider the reaction of hydrogen and oxygen to form water: \( 2H_2 + O_2 \rightarrow 2H_2O \) Given: 3.0 g of \( H_2 \) and 29 g of \( O_2 \). Molar mass of \( H_2 \) = 2 g/mol. Molar mass of \( O_2 \) = 32 g/mol. From the equation, 2 moles of \( H_2 \) (which is \( 2 \times 2 = 4 \) g) react with 1 mole of \( O_2 \) (which is 32 g). So, 4 g of \( H_2 \) reacts with 32 g of \( O_2 \). Now, let's see how much \( O_2 \) is needed for 3.0 g of \( H_2 \): Mass of \( O_2 \) required = \( \frac{32 \text{ g } O_2}{4 \text{ g } H_2} \times 3.0 \text{ g } H_2 = 8 \times 3.0 = 24 \) g of \( O_2 \). We have 29 g of \( O_2 \) given, but only 24 g of \( O_2 \) is needed to react with all the 3.0 g of \( H_2 \). This means that \( O_2 \) is in excess (29 g - 24 g = 5 g of \( O_2 \) will be left over). Since \( H_2 \) will be completely used up, Hydrogen (\( H_2 \)) is the limiting reagent in this reaction.
In simple words: The limiting reagent is the chemical that gets fully used up first in a reaction. Here, 3.0 g of hydrogen needs 24 g of oxygen, but we have 29 g. So, hydrogen is the limiting reagent because it will run out first.

🎯 Exam Tip: To find the limiting reagent, calculate the amount of product formed from each reactant. The reactant that produces the smallest amount of product is the limiting reagent.

 

Question 39. Write short notes on:
(i) Dalton's Atomic Theory
(ii) Gay Lussac's Law of Combining Volumes
(iii) Avogadro's Hypothesis and its applications.
Answer:
(i) Dalton's Atomic Theory: John Dalton's Atomic Theory, proposed in the early 1800s, laid the foundation for modern chemistry. Its main ideas are:

• Matter is made up of extremely tiny particles called "atoms."
• Atoms are indivisible particles; they cannot be created or destroyed.
• All atoms of a specific element are identical in their size, shape, mass, and other properties.
• Atoms of different elements have different properties.
• Atoms combine in simple whole-number ratios to form compounds.

(ii) Gay Lussac's Law of Combining Volumes: This law, proposed by Gay Lussac in 1808, states that when gases react together, or are produced in a chemical reaction, they do so in simple whole-number ratios by volume. This condition holds true as long as all the gases are measured at the same temperature and pressure. For example:
\( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \)
1 vol. 3 vols. 2 vols.
Here, 1 volume of nitrogen reacts with 3 volumes of hydrogen to form 2 volumes of ammonia. This shows a simple ratio of 1:3:2.
\( 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \)
2 vols. 1 vol. 2 vols.
Here, 2 volumes of hydrogen combine with 1 volume of oxygen to form 2 volumes of steam. This shows a simple ratio of 2:1:2.
(iii) Avogadro's Hypothesis and its applications: In 1811, Italian physicist Amedeo Avogadro proposed a hypothesis (also known as Avogadro's law or principle) stating that for all gases, the volume of a gas is directly proportional to the number of molecules of the gas, provided they are at the same temperature and pressure. This can be written as: \( V = aN \), where V is the volume, N is the number of molecules, and 'a' is a constant. Applications of Avogadro's law have been very useful in chemistry. It has helped in:

• Explaining Gay Lussac's law of gaseous volumes.
• Determining the atomicity (number of atoms in a molecule) of gases.
• Determining the molecular formula of various gases.
• Establishing the relationship between relative molecular mass and vapour density.

In simple words: Dalton said matter is tiny, unbreakable atoms. Gay Lussac found gases combine in simple volume ratios. Avogadro said equal volumes of gases have equal molecules, which helped understand atomicity and molecular formulas.

🎯 Exam Tip: For short notes, clearly state the main principle or law, and include a brief example or key applications to show full understanding.

 

Question 40. Briefly explain Law of chemical combinations.
Answer: The Laws of Chemical Combinations are fundamental principles that govern how elements combine to form compounds. There are five main laws:
1. Law of Conservation of Mass: This law states that during any chemical change or physical change, matter can neither be created nor destroyed. In simpler terms, the total mass of the reactants before a reaction must equal the total mass of the products formed after the reaction. The overall mass remains constant. 2. Law of Definite Proportions: Joseph Proust, a French chemist, stated this law. It says that in a pure chemical compound, the elements are always present in the same fixed proportion by mass, regardless of how the compound was made or its source. For example, pure water (\( H_2O \)) always consists of hydrogen and oxygen combined in a 1:8 mass ratio. 3. Law of Multiple Proportions: Proposed by Dalton in 1803, this law states that if two elements can combine to form more than one compound, then the masses of one element that combine with a fixed mass of the other element are in ratios of small whole numbers. For instance, carbon and oxygen can form CO and \( CO_2 \). In CO, 12g of C combines with 16g of O. In \( CO_2 \), 12g of C combines with 32g of O. The masses of oxygen (16g and 32g) that combine with a fixed mass of carbon (12g) are in a simple ratio of 16:32, or 1:2. 4. Law of Reciprocal Proportion: This law states that if two different elements combine separately with a third element, then the ratio of their combining masses will be either the same or a simple multiple of the ratio in which they combine with each other. This shows how elements behave in relation to a common third element. 5. Gay Lussac's Law of Gaseous Volumes: As discussed in Question 39, this law states that when gases react or are produced in a chemical reaction, their volumes bear a simple whole-number ratio to one another, provided all measurements are done at the same temperature and pressure. For example, 1 volume of nitrogen combines with 3 volumes of hydrogen to form 2 volumes of ammonia (\( N_2 + 3H_2 \rightarrow 2NH_3 \)), showing a 1:3:2 ratio by volume.
In simple words: These laws explain how chemicals join together: mass is always conserved, specific compounds always have the same element ratios, if elements make more than one compound their ratios are simple, elements reacting with a third have related ratios, and gases combine in simple volume ratios.

🎯 Exam Tip: When explaining the Laws of Chemical Combinations, state each law clearly and provide a concise example for each to illustrate its meaning.

 

Question 40. Briefly explain Law of chemical combinations.
Answer: There are five main laws of chemical combinations that explain how elements combine to form compounds:

1. Law of Conservation of Mass: This law states that matter cannot be created or destroyed. It means the total mass of all the things you start with (reactants) will always be the same as the total mass of all the new things you end up with (products).
2. Law of Definite Proportions: Joseph Proust, a French chemist, said that in any pure compound, the elements are always present in the same fixed ratio by weight. For example, water will always have hydrogen and oxygen combined in a 1:8 mass ratio, no matter where the water comes from.
3. Law of Multiple Proportions: This law, given by Dalton in 1803, states that if two elements can combine to form more than one compound, then the different masses of one element that combine with a fixed mass of the other element will be in a simple whole number ratio.
4. Law of Reciprocal Proportion: This law says that if two elements react separately with a third element, and they also react with each other, then their combining ratios will be related in a simple way.
5. Gay-Lussac's Law of Gaseous Volumes: In 1808, Gay-Lussac found that when gases react together, or when products are gases, they do so in simple whole number ratios by volume, as long as the temperature and pressure are kept the same. For instance, 1 volume of nitrogen and 3 volumes of hydrogen combine to make 2 volumes of ammonia.

In simple words: These are the basic rules that explain how different chemicals come together and react to form new substances. They help us understand why chemical reactions happen in certain ways and how much of each substance is involved.

🎯 Exam Tip: Remember the name of the scientist associated with each law (e.g., Dalton for Multiple Proportions, Proust for Definite Proportions) and a simple example to illustrate each law.

RBSE Class 12 Chemistry Chapter 1 Numerical Problems

 

Question 41. In an organic compound, 40% carbon, 6.66% hydrogen and rest oxygen is present. Its vapour destiny is 30. Calculate the empirical formula and molecular formula.
Answer:

First, calculate the percentage of oxygen:

Percentage of oxygen \( = 100 - (40 + 6.66) = 100 - 46.66 = 53.4\% \)

Now, we find the empirical formula:

Thus, the empirical formula is \( \text{CH}_2\text{O} \).

Now, calculate the empirical formula mass:

Empirical formula mass \( = (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30 \) g/mol

Next, find the molecular mass using the vapour density:

Molecular mass \( = 2 \times \text{Vapour density} = 2 \times 30 = 60 \) g/mol

To find the molecular formula, calculate 'n':

\( n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2 \)

Finally, calculate the molecular formula:

Molecular formula \( = (\text{Empirical formula})_n = (\text{CH}_2\text{O})_2 = \text{C}_2\text{H}_4\text{O}_2 \)

In simple words: We first found out how much oxygen is in the compound. Then, by looking at the percentages and atomic weights of carbon, hydrogen, and oxygen, we figured out the simplest ratio of atoms, which gave us the empirical formula. After that, we used the vapour density to find the actual molecular mass and then calculated the full molecular formula.

🎯 Exam Tip: Always remember that the empirical formula shows the simplest whole-number ratio of atoms, while the molecular formula shows the actual number of atoms in a molecule. Use vapour density to find molecular mass directly.

 

Question 42. In an organic compound, the ratio of masses of C, H and N are 9:1:3.5 and molecular mass is 108. What will be the empirical formula and molecular formula of the compound?
Answer:

First, let's find the empirical formula based on the mass ratio:

So, the empirical formula is \( \text{C}_3\text{H}_4\text{N} \).

Next, calculate the empirical formula mass:

Empirical formula mass \( = (3 \times 12) + (4 \times 1) + (1 \times 14) = 36 + 4 + 14 = 54 \) g/mol

The given molecular mass is 108 g/mol.

Now, find 'n':

\( n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{108}{54} = 2 \)

Finally, calculate the molecular formula:

Molecular formula \( = (\text{Empirical formula})_n = (\text{C}_3\text{H}_4\text{N})_2 = \text{C}_6\text{H}_8\text{N}_2 \)

In simple words: We used the given mass ratios and atomic masses to find the simplest atom ratio, which gave us the empirical formula. Then, we compared the mass of this empirical formula with the total molecular mass to find the actual molecular formula of the compound.

🎯 Exam Tip: When given mass ratios, treat them like percentages for calculating empirical formula. Remember to find the "n" factor to move from empirical to molecular formula.

 

Question 43. In chemical analysis, it is found that in 10 g iron chloride, 3.438 g iron and 6.650 g chlorine are present. Calculate the empirical formula of iron chloride. (Fe = 55.8, CI= 35.5)
Answer:

First, calculate the percentage of each element:

Percentage of iron \( = \frac{3.438 \text{ g}}{10 \text{ g}} \times 100 = 34.38\% \)

Percentage of chlorine \( = \frac{6.650 \text{ g}}{10 \text{ g}} \times 100 = 66.50\% \)

Now, we find the empirical formula:

Thus, the empirical formula of iron chloride is \( \text{FeCl}_3 \).

In simple words: We first found out the percentage of iron and chlorine in the compound. Then, we used their atomic masses to find the ratio of their atoms in the simplest form, which gave us the empirical formula.

🎯 Exam Tip: Always calculate the percentages of each element first if not given directly, then use atomic masses to find mole ratios for the empirical formula.

 

Question 44. Calculate number of oxygen atoms in 88 g of CO2. Also calculate mass of CO having same number of oxygen atoms.
Answer:

First, calculate the molar mass of \( \text{CO}_2 \):

Molar mass of \( \text{CO}_2 = 12 + (2 \times 16) = 12 + 32 = 44 \) g/mol

Next, find the number of moles of \( \text{CO}_2 \) in 88 g:

Moles of \( \text{CO}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{88 \text{ g}}{44 \text{ g/mol}} = 2 \) mol

Now, calculate the number of oxygen atoms in 2 moles of \( \text{CO}_2 \):

One molecule of \( \text{CO}_2 \) contains 2 oxygen atoms.

One mole of \( \text{CO}_2 \) contains 2 moles of oxygen atoms.

So, 2 moles of \( \text{CO}_2 \) contain \( 2 \times 2 = 4 \) moles of oxygen atoms.

Number of oxygen atoms \( = 4 \text{ moles} \times 6.023 \times 10^{23} \text{ atoms/mol} = 2.4092 \times 10^{24} \) oxygen atoms.

Now, calculate the mass of CO that has the same number of oxygen atoms (i.e., 4 moles of oxygen atoms):

One molecule of CO contains 1 oxygen atom.

So, to have 4 moles of oxygen atoms, we need 4 moles of CO.

Calculate the molar mass of CO:

Molar mass of CO \( = 12 + 16 = 28 \) g/mol

Mass of 4 moles of CO \( = 4 \text{ moles} \times 28 \text{ g/mol} = 112 \) g

In simple words: We found out how many moles of carbon dioxide are in 88 grams and then counted the oxygen atoms within them. Then, we figured out how much carbon monoxide you would need to have the same number of oxygen atoms.

🎯 Exam Tip: Pay close attention to the subscripts in chemical formulas (like \( \text{O}_2 \) vs. single O) as they determine the number of atoms per molecule, which is crucial for mole calculations.

 

Question 45. Define atomic mass, molecular mass and equivalent weight. In 500 mL solution, 20.7 g potassium carbonate is dissolved. Calculate its molarity. (Molecular mass of K2CO3=138)
Answer:

• Atomic Mass: This is a number that tells us how many times heavier an atom of an element is compared to \( \frac{1}{12} \) the mass of a carbon-12 atom. It is often measured in atomic mass units (amu) or unified mass (u).
• Molecular Mass: This is the sum of the atomic masses of all the atoms present in one molecule of a substance. It gives us the relative mass of a molecule compared to \( \frac{1}{12} \) the mass of a carbon-12 atom, also expressed in amu or u.
• Equivalent Weight: This refers to the mass of a substance that will react with or replace a fixed quantity of another substance. For acids, it's the molecular mass divided by its basicity; for bases, it's molecular mass divided by its acidity; and for salts, it's the molecular mass divided by the total positive charge.

Now, let's calculate the molarity of the potassium carbonate solution:

Given molar mass of \( \text{K}_2\text{CO}_3 = 138 \) g/mol

Given volume of solution \( = 500 \text{ mL} = 0.5 \text{ L} \)

Given mass of \( \text{K}_2\text{CO}_3 = 20.7 \) g

First, calculate the number of moles of \( \text{K}_2\text{CO}_3 \):

Number of moles \( = \frac{\text{Mass}}{\text{Molar mass}} = \frac{20.7 \text{ g}}{138 \text{ g/mol}} = 0.15 \) mol

Now, calculate the molarity:

Molarity \( = \frac{\text{Number of moles of solute}}{\text{Volume of solution in L}} = \frac{0.15 \text{ mol}}{0.5 \text{ L}} = 0.3 \text{ M} \)

In simple words: We defined what atomic mass, molecular mass, and equivalent weight mean. Then, we used the given amount of potassium carbonate and the volume of the solution to calculate its concentration, called molarity.

🎯 Exam Tip: Clearly state the definitions for each term. For molarity calculations, ensure the volume is converted to liters and use the correct molar mass for the solute.

 

Question 46. In industrial preparation of nitric acid, how many moles of NO2 are required to form 7.33 moles of HNO3 if the reaction is. \( 3\text{NO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{HNO}_3 (\text{l}) + \text{NO}(\text{g}) \)
Answer:

From the given balanced chemical equation:

\( 3\text{NO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{HNO}_3 (\text{l}) + \text{NO}(\text{g}) \)

This equation shows that 3 moles of \( \text{NO}_2 \) are needed to produce 2 moles of \( \text{HNO}_3 \).

We want to find out how many moles of \( \text{NO}_2 \) are needed to form 7.33 moles of \( \text{HNO}_3 \).

Let 'x' be the moles of \( \text{NO}_2 \) required.

Using the mole ratio from the balanced equation:

\( \frac{\text{Moles of NO}_2}{\text{Moles of HNO}_3} = \frac{3}{2} \)

\( \implies \frac{x}{7.33 \text{ mol}} = \frac{3}{2} \)

\( \implies x = \frac{3}{2} \times 7.33 \text{ mol} \)

\( \implies x = 1.5 \times 7.33 \text{ mol} \)

\( \implies x = 10.995 \text{ mol} \)

So, approximately 11.00 moles of \( \text{NO}_2 \) are required.

In simple words: The chemical recipe tells us that 3 parts of \( \text{NO}_2 \) make 2 parts of \( \text{HNO}_3 \). So, to make 7.33 parts of \( \text{HNO}_3 \), we need 1.5 times that amount of \( \text{NO}_2 \).

🎯 Exam Tip: Always use the stoichiometric coefficients from the balanced chemical equation to set up mole ratios for calculations. This is fundamental for solving stoichiometry problems.

 

Question 47. 1.68 g iron contains how many moles of iron? Calculate, the number of atoms in same amount of iron. (At. mass of Fe = 56).
Answer:

Given mass of iron \( = 1.68 \) g

Given atomic mass of iron \( (\text{Fe}) = 56 \)

First, calculate the number of moles of iron:

Number of moles \( = \frac{\text{Mass of substance}}{\text{Atomic mass}} = \frac{1.68 \text{ g}}{56 \text{ g/mol}} = 0.03 \) mol

Next, calculate the number of atoms in this amount of iron:

We know that 1 mole of any substance contains Avogadro's number of particles (\( 6.023 \times 10^{23} \)).

Number of atoms \( = \text{Number of moles} \times \text{Avogadro's number} \)

Number of atoms \( = 0.03 \text{ mol} \times 6.023 \times 10^{23} \text{ atoms/mol} \)

Number of atoms \( = 0.18069 \times 10^{23} \text{ atoms} \)

Number of atoms \( = 1.8069 \times 10^{22} \text{ atoms} \)

In simple words: We first found out how many groups of atoms (moles) are in 1.68 grams of iron by using its atomic weight. Then, we used Avogadro's number to count the total number of iron atoms in that many moles.

🎯 Exam Tip: Remember the relationship: mass \( \leftrightarrow \) moles \( \leftrightarrow \) number of particles (atoms/molecules). Use atomic mass for elements and molecular mass for compounds to convert to moles, and Avogadro's number for converting moles to particles.

 

Question 48. Calculate the mass of \( \text{CaCO}_3 \) that reacts completely with 25 mL of 0.75 M HCl, according to the reaction: \( \text{CaCO}_3 (\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2 (\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \).
Answer:

First, calculate the number of moles of HCl in 25 mL of 0.75 M solution:

Molarity \( = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} \)

Moles of HCl \( = \text{Molarity} \times \text{Volume in L} \)

Volume in L \( = 25 \text{ mL} = \frac{25}{1000} \text{ L} = 0.025 \text{ L} \)

Moles of HCl \( = 0.75 \text{ mol/L} \times 0.025 \text{ L} = 0.01875 \) mol

From the balanced chemical equation:

\( \text{CaCO}_3 (\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2 (\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \)

We see that 1 mole of \( \text{CaCO}_3 \) reacts with 2 moles of HCl.

So, the mole ratio of \( \text{CaCO}_3 \) to HCl is 1:2.

Moles of \( \text{CaCO}_3 \) that react \( = \frac{1}{2} \times \text{Moles of HCl} \)

Moles of \( \text{CaCO}_3 = \frac{1}{2} \times 0.01875 \text{ mol} = 0.009375 \) mol

Now, calculate the molar mass of \( \text{CaCO}_3 \):

Molar mass of \( \text{CaCO}_3 = 40 (\text{Ca}) + 12 (\text{C}) + (3 \times 16) (\text{O}) = 40 + 12 + 48 = 100 \) g/mol

Finally, calculate the mass of \( \text{CaCO}_3 \):

Mass of \( \text{CaCO}_3 = \text{Moles} \times \text{Molar mass} \)

Mass of \( \text{CaCO}_3 = 0.009375 \text{ mol} \times 100 \text{ g/mol} = 0.9375 \) g

In simple words: We first found out how many moles of HCl are in the given solution. Then, using the chemical equation, we determined how many moles of \( \text{CaCO}_3 \) would react with that amount of HCl. Finally, we converted those moles of \( \text{CaCO}_3 \) into its mass.

🎯 Exam Tip: Always balance the chemical equation first. Then, convert all given quantities to moles, use mole ratios from the balanced equation, and convert back to the desired quantity (mass, volume, etc.).

 

Question 49. In laboratory preparation of chlorine, manganese dioxide (MnO2) reacts with aqueous hydrochloric acid according to the reaction: \( 4\text{HCl (aq)} + \text{MnO}_2 (\text{s}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{MnCl}_2 (\text{aq}) + \text{Cl}_2 (\text{g}) \) How many grams of HCl react with 5.0 g of manganese dioxide?
Answer:

First, calculate the molar mass of \( \text{MnO}_2 \):

Molar mass of \( \text{MnO}_2 = 55 (\text{Mn}) + (2 \times 16) (\text{O}) = 55 + 32 = 87 \) g/mol

Next, find the number of moles of \( \text{MnO}_2 \) in 5.0 g:

Moles of \( \text{MnO}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{5.0 \text{ g}}{87 \text{ g/mol}} \approx 0.05747 \) mol

From the balanced chemical equation:

\( 4\text{HCl (aq)} + \text{MnO}_2 (\text{s}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{MnCl}_2 (\text{aq}) + \text{Cl}_2 (\text{g}) \)

We see that 4 moles of HCl react with 1 mole of \( \text{MnO}_2 \).

So, the mole ratio of HCl to \( \text{MnO}_2 \) is 4:1.

Moles of HCl that react \( = 4 \times \text{Moles of MnO}_2 \)

Moles of HCl \( = 4 \times 0.05747 \text{ mol} = 0.22988 \) mol

Now, calculate the molar mass of HCl:

Molar mass of HCl \( = 1 (\text{H}) + 35.5 (\text{Cl}) = 36.5 \) g/mol

Finally, calculate the mass of HCl required:

Mass of HCl \( = \text{Moles} \times \text{Molar mass} \)

Mass of HCl \( = 0.22988 \text{ mol} \times 36.5 \text{ g/mol} \approx 8.39 \) g

In simple words: We first found out how many moles of manganese dioxide are present. Then, using the chemical reaction, we figured out how many moles of hydrochloric acid are needed to react with it. Finally, we converted those moles of hydrochloric acid into its mass in grams.

🎯 Exam Tip: Always write down the balanced chemical equation. If any reactant amount is given in grams, convert it to moles first before using stoichiometric ratios.

 

Question 50. Hydrochloric acid is 38% by mass. If the density of solution is 1.19 gm cm\(^{-3}\), then what will be the molarity of solution?
Answer:

Given: The HCl solution is 38% by mass. This means that 38 g of HCl is present in 100 g of the solution.

Given density of solution \( = 1.19 \) g/cm\(^3\)

First, calculate the volume of 100 g of the solution:

Density \( = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \)

Volume of solution \( = \frac{100 \text{ g}}{1.19 \text{ g/cm}^3} \approx 84.03 \) cm\(^3\)

Convert the volume to liters:

Volume of solution \( = 84.03 \text{ cm}^3 = 0.08403 \) L

Now, calculate the molar mass of HCl:

Molar mass of HCl \( = 1 (\text{H}) + 35.5 (\text{Cl}) = 36.5 \) g/mol

Next, calculate the number of moles of HCl in 38 g:

Moles of HCl \( = \frac{\text{Mass}}{\text{Molar mass}} = \frac{38 \text{ g}}{36.5 \text{ g/mol}} \approx 1.041 \) mol

Finally, calculate the molarity of the solution:

Molarity \( = \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{1.041 \text{ mol}}{0.08403 \text{ L}} \approx 12.39 \) M

In simple words: We know that 38% of the solution is HCl by weight. Using the solution's density, we found the volume of the solution that contains 38 grams of HCl. Then, we divided the moles of HCl by this volume to find its molarity, which is how concentrated it is.

🎯 Exam Tip: When given percentage by mass and density, assume a convenient total mass of solution (like 100 g) to easily find the mass of solute and then calculate the volume of that solution using density.