CBSE Class 11 Chemistry Some Basic Concepts of Chemistry MCQs Set 04

Answer: (c) A is true, but R is false.

Explanation:
1. 0.200 has 3 significant figures, while 200 (written without a decimal point) has only 1 significant figure. So the Assertion is correct.
2. Zeroes at the end or right of a number are significant provided they are on the right side of the decimal point. The Reason states the opposite, so it is incorrect.

Teacher's Note:
1. Terminal zeroes count only when a decimal point is present: 0.200 has 3, 200 has 1, 200. has 3.
2. Writing 200 as 2.00 × 10² is the clear way to show three significant figures.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 138.42 = 1.3842 × 100 = 1.3842 × 10², so the Assertion is correct.
2. In scientific notation the number is written as N × 10ⁿ with N between 1 and 9.999…, which is exactly how 1.3842 × 10² is written.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Moving the decimal point two places to the left means multiplying by 10², so the exponent is +2.
2. N must have exactly one non-zero digit before the decimal point.

Answer: (d) A is false, but R is true.

Explanation:
1. All the digits 1, 2, 3 and 1 in 1.231 are non-zero, and all non-zero digits are significant.
2. So 1.231 has four significant figures, not three. The Assertion is incorrect.
3. The Reason (all non-zero digits are significant) is a correct rule.

Teacher's Note:
1. Count every non-zero digit; the decimal point does not reduce the count.
2. Only zeroes need special rules; non-zero digits are always significant.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Valency of Cu in CuO is 2: equivalent weight = \( \frac{63.6}{2} \) = 31.8
2. Valency of Cu in Cu₂O is 1: equivalent weight = \( \frac{63.6}{1} \) = 63.6
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. An element showing variable valency has a different equivalent weight in each compound.
2. Find the valency from the formula: Cu is +2 in CuO and +1 in Cu₂O.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The molecular formula of ethene is C₂H₄ (mass 28) and its empirical formula, the simplest ratio of atoms, is CH₂ (mass 14).
2. Molecular formula = 2 × empirical formula, so the empirical mass is half of the molecular mass.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Reduce the subscripts of the molecular formula by their highest common factor to get the empirical formula.
2. For benzene (C₆H₆) the empirical mass is one-sixth of the molecular mass.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The volume of a gas changes with temperature and pressure. For example, an increase in temperature increases the kinetic energy of the molecules, the gas expands and occupies more volume. So the Assertion is correct.
2. The Reason correctly defines molar volume, but the definition does not explain why it varies with temperature and pressure.
3. Both Assertion and Reason are correct, but the Reason is not the correct explanation of the Assertion.

Teacher's Note:
1. Molar volume of a gas is 22.4 L only at 273 K and 1 atm (22.7 L at 1 bar).
2. A definition rarely explains a behaviour; look for a cause in the Reason before choosing (a).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. According to the law of conservation of mass, in a chemical reaction the total mass of the products is equal to the total mass of the reactants.
2. An equation is balanced so that the number of atoms of each element, and hence the mass, is the same on both sides.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Balancing makes the atoms of each element equal on both sides, which keeps the mass equal.
2. Matter is neither created nor destroyed in a chemical reaction.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. 1 amu (u) is defined as exactly one-twelfth of the mass of one carbon-12 atom. The Assertion is correct.
2. C-12 is the most abundant isotope of carbon and is the chosen standard. Atomic masses obtained by comparing with C-12 come out close to whole numbers.
3. Both statements are correct, but the abundance of C-12 is not the reason for the definition of 1 amu. So the Reason is not the correct explanation of the Assertion.

Teacher's Note:
1. 1 u = 1.66056 × 10⁻²⁴ g; it is based on the ¹²C atom taken as exactly 12 u.
2. Earlier standards were hydrogen and then oxygen; carbon-12 replaced them in 1961.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The number of atoms present in one molecule of an element is called its atomicity. O₂ has two atoms, so its atomicity is 2.
2. 1 mole of atoms of an element contains 6.023 × 10²³ atoms, which is a correct statement about Avogadro's number.
3. Both Assertion and Reason are correct, but the Reason is not the correct explanation of the Assertion.

Teacher's Note:
1. Atomicity: noble gases are monatomic (1), O₂ is diatomic (2), O₃ is triatomic (3), P₄ is tetratomic (4).
2. 1 mole of O₂ molecules contains 2 × 6.023 × 10²³ oxygen atoms because of its atomicity.

Answer: (d) A is false, but R is true.

Explanation:
1. Number of gram molecules = \( \frac{\text{Mass of substance}}{\text{Gram molecular mass (GMM)}} \)
2. GMM of SO₂Cl₂ = 32 + 2 × 16 + 2 × 35.5 = 135 g
3. Number of g-molecules = \( \frac{13.5}{135} \) = 0.1, not 0.2. So the Assertion is incorrect.
4. A gram-molecule is the molecular mass expressed in grams (1 mole of molecules), so the Reason is correct.

Teacher's Note:
1. Gram-molecule is an older name for one mole of molecules.
2. Do not forget the two chlorine atoms (71 g) when finding the molar mass of SO₂Cl₂.

Answer: (c) A is true, but R is false.

Explanation:
1. One mole of any substance contains 6.023 × 10²³ entities irrespective of its mass, so the Assertion is correct.
2. Molecular weight of SO₂ = 32 + 2 × 16 = 64 g mol⁻¹ and of O₂ = 2 × 16 = 32 g mol⁻¹.
3. So the molecular weight of SO₂ is double (not half) that of O₂. The Reason is incorrect.

Teacher's Note:
1. Equal moles always mean equal numbers of molecules, whatever the molar masses.
2. A mole is a count, like a dozen; it does not depend on the mass of each particle.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Molecular weight = 2 × vapour density = 2 × 22 = 44. The Assertion is correct.
2. This relation is the basis of the vapour density method for determining molecular mass.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Vapour density is the density of a gas relative to hydrogen, whose molar mass is 2; hence the factor 2.
2. CO₂ (M = 44) is an example of a gas with vapour density 22.

Answer: (d) A is false, but R is true.

Explanation:
1. CH₄ + 2O₂ → CO₂ + 2H₂O
2. 1 mol CH₄ (16 g) gives 2 mol H₂O = 2 × 18 = 36 g of water, not 18 g. So the Assertion is incorrect.
3. Water is indeed one of the products of combustion of methane, so the Reason is correct.

Teacher's Note:
1. Always use the stoichiometric coefficient: 2 moles of water are formed per mole of methane.
2. Each CH₄ has four H atoms, which need two H₂O molecules.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The reactant that is completely consumed in a reaction is called the limiting reactant (reagent).
2. The amount of product formed is decided by the amount of the limiting reagent, because the reaction stops once it is used up.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Once the limiting reagent is used up, no more product can form even if other reactants remain.
2. All product calculations in a stoichiometry problem must start from the limiting reagent.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Concentration means the amount of solute present in a given volume of solution.
2. The amount of solute can be measured in moles, so molarity (moles of solute per litre of solution) is a measure of concentration.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Molarity is the most common laboratory unit of concentration, expressed in mol L⁻¹.
2. Because volume changes with temperature, molarity also changes slightly with temperature.

Answer: (b) 3.816 g

Explanation:
1. CuSO₄·5H₂O contains 1 mol of Cu and 4 + 5 = 9 mol of O atoms per mole.
2. So 63.6 g of Cu is present with 9 × 16 = 144 g of oxygen.
3. Mass of Cu with 8.64 g of oxygen = \( \frac{63.6}{9 \times 16} \) × 8.64 = 3.816 g (≈ 3.81 g with Cu = 63.5)

Teacher's Note:
1. Count oxygen in the water of crystallisation too: 4 in SO₄ and 5 in 5H₂O.
2. Missing the 5 water oxygens (using only 4 O) gives a wrong answer of 8.6 g.

Answer: (d) 80.03%

Explanation:
1. Equal volumes of gases contain equal numbers of moles, so the molar ratio He : CH₄ = 1 : 1.
2. Ratio by weight = 4 : 16 = 1 : 4 (using He = 4.003 and CH₄ = 16.04)
3. CH₄ by weight = \( \frac{16.04}{4.003 + 16.04} \) × 100 = 80.03%

Teacher's Note:
1. Volume per cent equals mole per cent for gases, but not mass per cent.
2. The heavier gas (CH₄) has the larger mass share for the same number of moles.

Answer: (d) half of the gram atomic mass of oxygen

Explanation:
1. 22.4 L of O₂ gas at STP = 1 gram molecule (1 mol) = 32 g
2. 5.6 L of O₂ at STP = \( \frac{5.6}{22.4} \) = \( \frac{1}{4} \) gram molecule = 8 g
3. \( \frac{1}{4} \) gram molecule = \( \frac{1}{4} \) × 2 gram atoms = \( \frac{1}{2} \) gram atom of oxygen
4. 8 g is half of the gram atomic mass of oxygen (16 g).

Teacher's Note:
1. One gram molecule of O₂ contains two gram atoms of oxygen.
2. Option (b) is the trap: it is one fourth of the gram molecular mass, not of the gram atomic mass.

Answer: (b) 1.993 × 10⁻¹⁹ mg

Explanation:
1. Yellow (white) phosphorus exists as P₄ molecules.
2. Molecular mass of P₄ = 4 × 30 = 120 g mol⁻¹; 6.022 × 10²³ molecules weigh 120 g.
3. Mass of one molecule = \( \frac{120}{6.022 \times 10^{23}} \) = 1.993 × 10⁻²² g = 1.993 × 10⁻¹⁹ mg

Teacher's Note:
1. Yellow phosphorus is P₄, so multiply the atomic mass by four.
2. Watch the unit: 1 g = 1000 mg, so the power of 10 rises by 3 when converting g to mg.

Answer: (d) 0.0093 mol

Explanation:
1. Volume of O₂ in 1 L (1000 cm³) of air = 21% of 1000 cm³ = 210 cm³
2. 22400 cm³ of a gas at STP = 1 mol
3. Moles of O₂ = \( \frac{210}{22400} \) = 0.0093 mol

Teacher's Note:
1. First find the volume of oxygen alone, then divide by the molar volume.
2. 0.21 mol is a trap: 0.21 is a volume in litres, not a number of moles.

Answer: (b) 0.5 L

Explanation:
1. NH₃ and HCl react in the volume ratio 1 : 1.
2. 1 L NH₃ reacts with 1 L HCl to form solid NH₄Cl, which has negligible volume.
3. HCl left = 1.5 − 1 = 0.5 L. Hence, the final mixture contains only 0.5 L of HCl gas.

Teacher's Note:
1. Solids formed from gases contribute almost no volume to the gaseous mixture.
2. NH₃ is the limiting reagent here, so all of it is used up.

Answer: (a) 1.12 L, 1.12 L

Explanation:
1. H₂(g) + Cl₂(g) → 2HCl(g); 22.4 L H₂ and 22.4 L Cl₂ give 2 × 36.5 = 73 g HCl.
2. For 3.65 g HCl, volume of H₂ (or Cl₂) required = \( \frac{22.4}{2 \times 36.5} \) × 3.65 = 1.12 L
3. Hence 1.12 L of H₂ and 1.12 L of Cl₂ are required.

Teacher's Note:
1. 3.65 g HCl is 0.1 mol, which needs 0.05 mol each of H₂ and Cl₂.
2. Since H₂ and Cl₂ react 1 : 1, their volumes must be equal, which removes options (b) and (c).

Answer: (d) 80.03%

Explanation:
1. Equal volumes of gases contain equal numbers of moles, so the molar ratio He : CH₄ = 1 : 1.
2. Ratio by weight = 4 : 16 = 1 : 4
3. CH₄ present by weight = \( \frac{16.04}{4.003 + 16.04} \) × 100 = 80.03%

Teacher's Note:
1. 19.97% is the mass per cent of helium, so read the question carefully.
2. By Avogadro's law, volume ratio of gases equals their mole ratio.

Answer: (b) 17.0 g

Explanation:
1. AgNO₃ + NaCl → AgCl + NaNO₃
2. By the law of conservation of mass, x + 5.85 = 14.35 + 8.5
3. x = 22.85 − 5.85 = 17.0 g

Teacher's Note:
1. Mass of reactants = mass of products; no molar masses are needed here.
2. Check: 17.0 g AgNO₃ is 0.1 mol, which matches 5.85 g NaCl (0.1 mol).

Answer: (a) 46%

Explanation:
1. Urea is NH₂CONH₂; its molar mass = 60 g mol⁻¹.
2. There are two nitrogen atoms in urea, so mass of nitrogen = 28 g.
3. Percentage of N = \( \frac{28}{60} \) × 100 = 46.7% ≈ 46%

Teacher's Note:
1. Urea has two N atoms, so use 28 g and not 14 g.
2. Urea's high nitrogen content (about 46%) is why it is a popular fertiliser.

Answer: (d) A is false, but R is true.

Explanation:
1. 22.4 L of N₂ at NTP = 1 mol = 6.022 × 10²³ molecules.
2. 5.6 L of O₂ at NTP = \( \frac{5.6}{22.4} \) = 0.25 mol = 1.506 × 10²³ molecules.
3. The numbers of molecules are not equal, so the Assertion is incorrect.
4. The Reason is a correct statement of Avogadro's law: equal volumes of all gases under similar conditions of temperature and pressure contain equal numbers of molecules.

Teacher's Note:
1. Avogadro's law applies only to equal volumes; 22.4 L and 5.6 L are different volumes.
2. Read the numbers in an Assertion carefully before applying a familiar law.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The reactant that is completely used up is the limiting reagent.
2. Because it is used up first, it decides the maximum amount of product that can form.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. The other reactants are called excess reagents and some of them remain after the reaction.
2. To find the limiting reagent, divide the moles of each reactant by its coefficient; the smallest value wins.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 44 g of CO₂ = 1 mol, which contains 1 g atom of C.
2. 16 g of CH₄ = 1 mol, which also contains 1 g atom of C.
3. 1 g atom of C = 12 g of C = 6.023 × 10²³ carbon atoms.
4. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Equal moles of compounds with one C atom per molecule contain equal numbers of C atoms.
2. One gram atom means one mole of atoms of the element.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The concentration of a solute is commonly expressed in moles, e.g. molarity (moles of solute per litre of solution). The Assertion is correct.
2. The number of reacting molecules taking part in an elementary reaction is called its molecularity, e.g. PCl₅ → PCl₃ + Cl₂ is unimolecular and 2HI → H₂ + I₂ is bimolecular. The Reason is also correct.
3. Molarity and molecularity are used in different senses, so the Reason is not the correct explanation of the Assertion.

Teacher's Note:
1. Do not confuse molarity (a concentration) with molecularity (a property of an elementary reaction).
2. When the Reason talks about a different idea from the Assertion, the answer cannot be (a).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. According to the mole concept, 1 mole of a compound contains 6.023 × 10²³ formula units (molecules). The Assertion is correct.
2. Molar mass of NaCl = 23 + 35.5 = 58.5 g, so 58.5 g of NaCl is 1 mole and also contains 6.023 × 10²³ formula units. The Reason is correct.
3. Both are correct, but the Reason only restates the same fact in terms of mass; it is not the explanation of the Assertion.

Teacher's Note:
1. NaCl is ionic, so strictly we speak of formula units rather than molecules.
2. One mole of any substance contains N_A particles and weighs its molar mass in grams.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Percentage composition of a compound is the mass percentage of each element in it. The Assertion is correct.
2. Mass % of an element = \( \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \) × 100, so it does depend on the molar mass. The Reason is correct.
3. However, the Reason does not explain why mass percentage is used for percentage composition, so it is not the correct explanation.

Teacher's Note:
1. Percentage composition is the first step in finding the empirical formula of a compound.
2. The mass percentages of all elements in a compound add up to 100.

Answer: (c) A is true, but R is false.

Explanation:
1. χ_A = \( \frac{n_A}{n_A + n_B} \) and χ_B = \( \frac{n_B}{n_A + n_B} \)
2. χ_A + χ_B = \( \frac{n_A + n_B}{n_A + n_B} \) = 1, so the Assertion is correct.
3. Moles depend only on mass, and mass does not change with temperature, so mole fraction is independent of temperature. The Reason is incorrect.

Teacher's Note:
1. Concentration terms based only on mass (mole fraction, molality, mass %) do not change with temperature.
2. Terms based on volume (molarity, normality, volume %) change with temperature.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Concentration means the amount of solute present in a given volume of solution.
2. As the amount can be measured in moles, molarity (moles of solute per litre of solution) expresses the concentration of the solution.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Molarity is written with a capital M and has the unit mol L⁻¹.
2. Molarity × volume in litres gives the moles of solute present.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Molality, m = \( \frac{\text{Moles of solute}}{\text{Weight of solvent (in g)}} \) × 1000
2. It depends on the mass of the solvent, and mass is unaffected by temperature, so molality does not change with temperature.
3. Both Assertion and Reason are correct and the Reason is the correct explanation of the Assertion.

Teacher's Note:
1. Molality is preferred over molarity in colligative property experiments because it does not change with temperature.
2. Mass does not change on heating, but volume does.

Answer: (d) A is false, but R is true.

Explanation:
1. Molarity, M = \( \frac{\text{Weight} \times 1000}{\text{Molecular weight} \times \text{Volume (mL)}} \)
2. Volume of a solution changes with temperature, so molarity changes with temperature. The Assertion is incorrect.
3. The Reason is a correct statement.

Teacher's Note:
1. Liquids expand on heating, so the same moles occupy a larger volume and molarity falls.
2. Any concentration term with volume in its formula is temperature dependent.