CBSE Class 11 Chemistry Some Basic Concepts of Chemistry MCQs Set 03

Answer: (c) 93.3 °C

Explanation:
1. °F = \( \frac{9}{5} \)(°C) + 32
2. 200 − 32 = \( \frac{9}{5} \)(°C), so \( \frac{9}{5} \)(°C) = 168
3. °C = \( \frac{168 \times 5}{9} \) = 93.3 °C

Teacher's Note:
1. Always subtract 32 first and then multiply by 5/9; multiplying 200 by 5/9 first gives a wrong value.
2. A quick check: 212 °F is 100 °C, so 200 °F must be a little below 100 °C.

Answer: (a) 6.023 × 10²³

Explanation:
1. In scientific notation only the digits of N in N × 10ⁿ are counted: 6.023 has 4 significant figures.
2. 285 cm has 3, 0.0025 L has 2 (leading zeroes are not significant) and 0.200 g has 3 significant figures.
3. Hence only 6.023 × 10²³ has four significant figures.

Teacher's Note:
1. The power of 10 never adds significant figures; count only the digits of the coefficient.
2. Zeroes before the first non-zero digit only fix the decimal point and are never significant.

Answer: (d) 6

Explanation:
1. All non-zero digits are significant, and zeroes between non-zero digits are also significant.
2. So all six digits 1, 0, 3, 4, 0, 6 in 10.3406 are significant. Hence, there are 6 significant figures.

Teacher's Note:
1. Zeroes sandwiched between non-zero digits are always significant.
2. The decimal point does not count as a digit, so just count 1-0-3-4-0-6.

Answer: (a) \( \frac{0.02856 \times 298.15 \times 0.112}{0.5785} \)

Explanation:
1. (a) In multiplication and division the result has as many significant figures as the least precise term. The least precise term 0.112 has 3 significant figures, so the answer has 3 significant figures.
2. (b) 5 is an exact number and 5.364 has 4 significant figures, so the answer has 4 significant figures.
3. (c) In addition the answer cannot have more digits after the decimal point than either term: 0.0125 + 0.7864 + 0.0215 = 0.8204, which has 4 significant figures.
4. Hence (a) gives the result with the least number of significant figures.

Teacher's Note:
1. Exact (counted) numbers have unlimited significant figures and never limit the answer.
2. For addition look at decimal places; for multiplication and division look at significant figures.

Answer: (c) A → precision, B → accuracy

Explanation:
1. Precision refers to the closeness of various measurements for the same quantity.
2. Accuracy is the agreement of a particular value to the true value of the result.
3. Thus, A and B are precision and accuracy respectively.

Teacher's Note:
1. Precision is about agreement among the readings; accuracy is about agreement with the true value.
2. A set of readings can be precise without being accurate, for example when an instrument has a constant error.

Answer: (b) 0.01 cm

Explanation:
1. The least count of an instrument is equal to the maximum possible error in a measurement made with it.
2. Hence the most possible error of the instrument is 0.01 cm.

Teacher's Note:
1. The least count is the smallest division an instrument can read, so it is also the largest possible error.
2. An instrument cannot measure more finely than its least count.

Answer: (c) Student C

Explanation:
1. Student A reports 1.95 g and 1.93 g. The values are precise as they are close to each other, but they are not accurate.
2. Student B reports 1.94 g and 2.05 g. These observations are neither precise nor accurate.
3. Student C reports 2.01 g and 1.99 g. These values are close to each other and also close to the true value 2.00 g, so they are both precise and accurate.

Teacher's Note:
1. Check the spread between the two readings for precision and the average against the true value for accuracy.
2. Student C's average equals the true value exactly and the readings differ by only 0.02 g.

Answer: (d) All the above statements are correct.

Explanation:
1. The molecular formula shows the exact number of each type of atom in a molecule.
2. Molecular formula = n × empirical formula, where n = molar mass / empirical formula mass, so it can be found when the molar mass is known.
3. From the molecular formula, mass % of each element = (mass of element in 1 mole / molar mass) × 100.
4. Hence all the given statements are correct.

Teacher's Note:
1. The empirical formula gives only the simplest ratio, so the molar mass is needed to fix the molecular formula.
2. Percentage composition is the same for the empirical and the molecular formula.

Answer: (d) 1.4 × 10⁻²¹ g

Explanation:
1. Molecular mass of C₆₀H₁₂₂ = (60 × 12) + (122 × 1) = 720 + 122 = 842 g mol⁻¹
2. Weight of 6.022 × 10²³ molecules = 842 g
3. Weight of 1 molecule = \( \frac{842}{6.022 \times 10^{23}} \) g = 1.398 × 10⁻²¹ g ≈ 1.4 × 10⁻²¹ g

Teacher's Note:
1. Mass of one molecule = molar mass ÷ Avogadro's number.
2. A single molecule must have a tiny mass, so options with positive powers of 10 can be rejected at once.

Answer: (b) \( \frac{\text{molecular weight}}{6} \)

Explanation:
1. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
2. 6I⁻ → 3I₂ + 6e⁻; then 2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI
3. Each Cr changes from +6 to +3 (3 electrons) and there are 2 Cr atoms, so one K₂Cr₂O₇ gains 3 × 2 = 6 electrons.
4. Equivalent weight of K₂Cr₂O₇ = molecular weight / 6

Teacher's Note:
1. Equivalent weight of an oxidising agent = molar mass ÷ number of electrons gained per formula unit.
2. Remember to multiply the change per Cr atom (3) by the two Cr atoms in dichromate.

Answer: (a) 12.044 × 10²⁰ molecules

Explanation:
1. Moles of H₂SO₄ = Molarity × Volume (in L) = 0.02 × 0.1 = 2 × 10⁻³ mol
2. Number of molecules = Number of moles × N_A = 2 × 10⁻³ × 6.022 × 10²³
3. = 12.044 × 10²⁰ molecules

Teacher's Note:
1. Convert the volume to litres before multiplying by molarity (100 mL = 0.1 L).
2. Molarity × volume in mL gives millimoles, not moles.

Answer: (a) 10⁻³ N_A

Explanation:
1. 18 g of water (1 mol) contains N_A molecules.
2. 18 mg = 18 × 10⁻³ g, so number of molecules = N_A × \( \frac{18 \text{ mg}}{18 \times 10^{3} \text{ mg}} \) = 10⁻³ N_A

Teacher's Note:
1. 1 mg = 10⁻³ g, so 18 mg of water is exactly 10⁻³ mol.
2. Watch the unit prefix: confusing mg with g gives an answer 1000 times too large.

Answer: (c) 1.673 × 10²¹

Explanation:
1. Density of water = 1 g mL⁻¹, so 1 mL of water = 1 g.
2. Moles of water in 1 mL = \( \frac{1}{18} \) mol
3. Number of molecules in 1 mL = \( \frac{6.023 \times 10^{23}}{18} \) = 3.346 × 10²²
4. Number of molecules in 1 drop = \( \frac{3.346 \times 10^{22}}{20} \) = 1.673 × 10²¹ molecules

Teacher's Note:
1. Liquid water does not occupy 22.4 L per mole; that value is only for gases at STP.
2. For liquids, convert volume to mass using density and then mass to moles.

Answer: (c) 18 mL of water

Explanation:
1. Number of molecules = moles × N_A, so compare the moles.
2. (a) n = \( \frac{0.00224}{22.4} \) = 1 × 10⁻⁴ mol
3. (b) n = \( \frac{0.18}{18} \) = 0.01 mol
4. (c) 18 mL of water = 18 g (density 1 g mL⁻¹), so n = \( \frac{18}{18} \) = 1 mol
5. (d) n = 10⁻³ mol
6. Hence the maximum number of molecules (1 × N_A) is present in 18 mL of water.

Teacher's Note:
1. Bring every option to moles first; the largest number of moles has the most molecules.
2. 18 mL of liquid water is a full mole, while the same volume of vapour holds far fewer molecules.

Answer: (a) A-2, B-3, C-1, D-5, E-4

Explanation:
1. A. 46 g of Na = 46/23 = 2 mol → (2)
2. B. 6.022 × 10²³ molecules of H₂O = 1 mol → (3)
3. C. 0.224 L of O₂ at STP = 0.224/22.4 = 0.01 mol → (1)
4. D. 84 g of N₂ = 84/28 = 3 mol → (5)
5. E. 1 mole of any gas contains 6.022 × 10²³ molecules → (4)

Teacher's Note:
1. Use molar mass of N₂ as 28 g, not 14 g; N₂ is diatomic.
2. Once two or three pairs are fixed, eliminate the codes rather than solving every pair.

Answer: (d) 32.37; 22.57 and 45.06

Explanation:
1. Mass per cent of an element = \( \frac{\text{Mass of that element in the compound}}{\text{Molar mass of the compound}} \) × 100
2. Molar mass of Na₂SO₄ = (2 × 22.99) + 32.06 + (4 × 16.00) = 142.04 g
3. Mass per cent of sodium = \( \frac{45.98 \times 100}{142.04} \) = 32.37
4. Mass per cent of sulphur = \( \frac{32.06 \times 100}{142.04} \) = 22.57
5. Mass per cent of oxygen = \( \frac{64 \times 100}{142.04} \) = 45.06

Teacher's Note:
1. Check that the three percentages add up to 100; option (c) adds to more than 100.
2. Two options contain the right numbers, so the order (Na, S, O) decides the answer.

Answer: (b) 27.27%

Explanation:
1. Molecular mass of CO₂ = 1 × 12 + 2 × 16 = 44 g
2. 1 mole of CO₂ (44 g) contains 12 g of carbon.
3. % of C in CO₂ = \( \frac{12}{44} \) × 100 = 27.27%

Teacher's Note:
1. 0.034% is the share of CO₂ in air, a common distractor; it is not the carbon content of CO₂.
2. Mass per cent always uses the mass of the element in one mole divided by the molar mass.

Answer: (c) C₆H₁₂O₆

Explanation:
1. Empirical formula mass of CH₂O = 12 + 2 × 1 + 16 = 30
2. n = \( \frac{\text{Molecular mass}}{\text{Empirical formula mass}} \) = \( \frac{180}{30} \) = 6
3. Molecular formula = n × empirical formula = 6 × CH₂O = C₆H₁₂O₆

Teacher's Note:
1. Multiply every subscript of the empirical formula by n.
2. Verify quickly: C₆H₁₂O₆ (glucose) has molar mass 180 g mol⁻¹.

Answer: (a) MO

Explanation:
1. Moles of M = \( \frac{68}{34} \) = 2; moles of O = \( \frac{32}{16} \) = 2
2. Simplest ratio M : O = 2 : 2 = 1 : 1
3. Hence, the empirical formula of the compound is MO.

Teacher's Note:
1. Divide each percentage by its atomic mass, then divide by the smallest result.
2. M₂O₂ is not an empirical formula because it can be simplified to MO.

Answer: (b) CHCl₃

Explanation:
1. C: \( \frac{10.06}{12} \) = 0.84; simplest ratio = 0.84/0.84 = 1
2. H: \( \frac{0.84}{1} \) = 0.84; simplest ratio = 0.84/0.84 = 1
3. Cl: \( \frac{89.10}{35.5} \) = 2.51; simplest ratio = 2.51/0.84 ≈ 3
4. Thus, the empirical formula of the substance is CHCl₃.

Teacher's Note:
1. Round the ratios only at the end; 2.98 or 3.02 rounds to 3, but 1.5 must be doubled.
2. CHCl₃ is chloroform, a familiar compound that helps you check your answer.

Answer: (a) 3 : 2

Explanation:
1. First oxide: O = 22/16 = 1.375, Fe = 78/56 = 1.392; ratio ≈ 1 : 1, so the oxide is FeO.
2. Second oxide: O = 30/16 = 1.875, Fe = 70/56 = 1.25; Fe : O = 1 : 1.5 = 2 : 3, so the oxide is Fe₂O₃.
3. Directly: iron combining with 1 g of oxygen = 78/22 = 3.545 g in the first oxide and 70/30 = 2.333 g in the second oxide.
4. Ratio = 3.545 : 2.333 = 1.52 ≈ 3 : 2 (in FeO 56 g Fe combine with 16 g O; in Fe₂O₃ 37.3 g Fe combine with 16 g O).

Teacher's Note:
1. This is an application of the law of multiple proportions: fix the mass of oxygen and compare the masses of iron.
2. Equivalent weight of Fe is 56/2 in FeO and 56/3 in Fe₂O₃, which again gives the ratio 3 : 2.

Answer: (a) 1.5 : 3

Explanation:
1. Sodium dihydrogen orthophosphate is NaH₂PO₄ and sodium hydrogen orthophosphate is Na₂HPO₄.
2. On heating they condense: NaH₂PO₄ + 2Na₂HPO₄ → Na₅P₃O₁₀ + 2H₂O
3. Check: Na = 1 + 4 = 5, P = 3, H = 2 + 2 = 4 (in 2H₂O), O = 4 + 8 = 12 = 10 + 2.
4. Hence, NaH₂PO₄ : Na₂HPO₄ = 1 : 2 = 1.5 : 3.

Teacher's Note:
1. Count sodium: 5 Na are needed and only the combination 1 NaH₂PO₄ + 2 Na₂HPO₄ supplies 5 Na with 3 P.
2. Keep the order of the ratio the same as the order in which the substances are named in the question.

Answer: (c) 3

Explanation:
1. Ca → Ca²⁺ + 2e⁻ (oxidation) …(i)
2. Al³⁺ + 3e⁻ → Al (reduction) …(ii)
3. Multiply Eq. (i) by 3 and Eq. (ii) by 2 to equalise the electrons (6e⁻).
4. Net reaction: 3Ca + 2Al³⁺ → 3Ca²⁺ + 2Al
5. Therefore, the stoichiometric coefficient of Ca is 3.

Teacher's Note:
1. Ionic equations must balance both atoms and charge: here 2 × (3+) = 3 × (2+) = 6+.
2. Use the LCM of electrons lost and gained (2 and 3) to find the multipliers.

Answer: (c) 1

Explanation:
1. 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O
2. Carbon changes from +3 in C₂O₄²⁻ to +4 in CO₂, i.e. 1 electron per carbon atom.
3. 5 oxalate ions transfer 10 electrons to produce 10 molecules of CO₂.
4. So, the number of electrons involved in producing 1 molecule of CO₂ is 1.

Teacher's Note:
1. Each oxalate ion loses 2 electrons, but it gives two CO₂ molecules, so it is 1 electron per CO₂.
2. Find the change in oxidation number of one carbon atom to answer such questions quickly.

Answer: (a) Mg, 0.16 g

Explanation:
1. Mg + \( \frac{1}{2} \)O₂ → MgO; 24 g Mg reacts with 16 g O₂.
2. 1.0 g Mg would need \( \frac{16}{24} \) × 1.0 = 0.67 g O₂, but only 0.56 g O₂ is available, so O₂ is the limiting reagent.
3. Mg that reacts with 0.56 g O₂ = \( \frac{24}{16} \) × 0.56 = 0.84 g
4. Mg left unreacted = 1.0 − 0.84 = 0.16 g. Hence, Mg is in excess and 0.16 g of Mg is left.

Teacher's Note:
1. The limiting reagent is never the one left over, so options with O₂ in excess are ruled out once O₂ is found limiting.
2. Compare the required amount with the available amount for one reactant; that settles which one is limiting.

Answer: (a) 56 g of N₂ + 10 g of H₂

Explanation:
1. From the equation, 28 g of N₂ requires 3 × 2 = 6 g of H₂.
2. (a) 56 g N₂ requires 12 g H₂, but only 10 g H₂ is present, so H₂ is the limiting reagent.
3. (b) 35 g N₂ requires \( \frac{6}{28} \) × 35 = 7.5 g H₂; 8 g is present, so H₂ is in excess (0.5 g left).
4. (c) 14 g N₂ requires 3 g H₂; 4 g is present, so H₂ is in excess.
5. (d) 28 g N₂ requires exactly 6 g H₂; the mixture is stoichiometric, so H₂ is not the limiting reagent.
6. Hence, H₂ is the limiting reagent only in option (a).

Teacher's Note:
1. Use the mass ratio N₂ : H₂ = 28 : 6 from the balanced equation, not a 1 : 3 mass ratio.
2. If H₂ present is less than the H₂ required, H₂ is the limiting reagent.

Answer: (d) 0.10

Explanation:
1. 3BaCl₂ + 2Na₃PO₄ → Ba₃(PO₄)₂ + 6NaCl
2. 0.5 mol BaCl₂ would require \( \frac{2}{3} \) × 0.5 = 0.33 mol Na₃PO₄, but only 0.20 mol is present. So Na₃PO₄ is the limiting reagent.
3. 2 mol Na₃PO₄ give 1 mol Ba₃(PO₄)₂.
4. 0.20 mol Na₃PO₄ gives \( \frac{1}{2} \) × 0.20 = 0.10 mol Ba₃(PO₄)₂.

Teacher's Note:
1. The amount of product is always calculated from the limiting reagent.
2. Dividing moles by the stoichiometric coefficient (0.5/3 = 0.167, 0.20/2 = 0.10) shows the limiting reagent at a glance.

Answer: (c) 5 L

Explanation:
1. C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
2. At the same temperature and pressure, volumes of gases are in the ratio of their moles (Gay-Lussac's law).
3. 1 volume of C₃H₈ needs 5 volumes of O₂, so 1 L of propane needs 5 L of O₂.

Teacher's Note:
1. For gases under the same conditions, the mole ratio in the equation is also the volume ratio.
2. You do not need 22.4 L at all when both volumes are measured under the same conditions.

Answer: (d) 0.029

Explanation:
1. PbO + 2HCl → PbCl₂ + H₂O
2. Moles of PbO = \( \frac{6.5}{223.2} \) = 0.029 mol; moles of HCl = \( \frac{3.2}{36.5} \) = 0.087 mol
3. 0.029 mol PbO needs 2 × 0.029 = 0.058 mol HCl, which is available. So PbO is the limiting reagent.
4. 1 mol of PbO gives 1 mol of PbCl₂, so 0.029 mol of PbO gives 0.029 mol of PbCl₂.

Teacher's Note:
1. Molar mass of PbO = 207.2 + 16 = 223.2 g mol⁻¹.
2. HCl appears to be smaller in mass but is present in excess in terms of moles; always compare moles, not grams.

Answer: (b) 84.3

Explanation:
1. M₂CO₃ + 2HCl → 2MCl + H₂O + CO₂
2. Moles of M₂CO₃ reacted = moles of CO₂ evolved
3. \( \frac{1}{M} \) = 0.01186, where M = molar mass of M₂CO₃
4. M = \( \frac{1}{0.01186} \) = 84.3 g mol⁻¹

Teacher's Note:
1. One mole of any carbonate gives one mole of CO₂ with excess acid.
2. Moles = mass/molar mass, so molar mass = mass/moles = 1/0.01186.

Answer: (a) 45.0 g conc. HNO₃

Explanation:
1. Mass of HNO₃ = Molarity × Molar mass × Volume (in L) = 2 × 63 × \( \frac{250}{1000} \) = 31.5 g
2. This is the mass of 100% HNO₃. The given acid is 70% HNO₃.
3. Mass of concentrated acid = 31.5 × \( \frac{100}{70} \) = 45.0 g

Teacher's Note:
1. The concentrated acid is only 70% HNO₃, so you need more than 31.5 g of it.
2. Molar mass of HNO₃ = 1 + 14 + 48 = 63 g mol⁻¹.

Answer: (c) 22.05 g

Explanation:
1. Mohr's salt is FeSO₄·(NH₄)₂SO₄·6H₂O; only Fe²⁺ is oxidised.
2. 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O
3. Moles of Fe²⁺ = 0.6 × 0.750 = 0.450 mol
4. 6 mol Fe²⁺ ≡ 1 mol Cr₂O₇²⁻, so moles of K₂Cr₂O₇ = \( \frac{0.450}{6} \) = 0.075 mol
5. Mass of K₂Cr₂O₇ = 0.075 × 294 = 22.05 g

Teacher's Note:
1. One dichromate ion oxidises six Fe²⁺ ions because it accepts 6 electrons.
2. The molar mass of Mohr's salt is not needed when the molarity of its solution is given.

Answer: (c) 1.7 N

Explanation:
1. 10% (w/V) means 10 g of acetic acid in 100 mL of solution.
2. Acetic acid (CH₃COOH) is monobasic, so its equivalent weight = molar mass = 60.
3. N = \( \frac{w \times 1000}{E \times V \text{(mL)}} \) = \( \frac{10 \times 1000}{60 \times 100} \) = 1.66 N ≈ 1.7 N

Teacher's Note:
1. Acetic acid has only one replaceable hydrogen, so for it normality equals molarity.
2. w/V per cent always means grams of solute per 100 mL of solution.

Answer: (b) 1.66 M

Explanation:
1. M₁V₁ = M₂V₂
2. 5 × 500 = M₂ × 1500
3. M₂ = \( \frac{5 \times 500}{1500} \) = 1.66 M

Teacher's Note:
1. Dilution changes the volume, not the number of moles of solute.
2. Here the final volume is 1500 mL in total; do not add 1500 mL to 500 mL.

Answer: (c) 6.023 × 10²⁰

Explanation:
1. Normality = Molarity × valence factor; for H₂SO₄ the valence factor is 2.
2. M = \( \frac{N}{2} \) = \( \frac{0.02}{2} \) = 0.01 M
3. Moles of H₂SO₄ = \( \frac{0.01 \times 100}{1000} \) = 1 × 10⁻³ mol
4. Number of molecules = 1 × 10⁻³ × 6.023 × 10²³ = 6.023 × 10²⁰

Teacher's Note:
1. H₂SO₄ is dibasic, so 0.02 N is only 0.01 M; forgetting this doubles the answer.
2. Always convert normality to molarity before finding moles or molecules.

Answer: (b) 0.2 M

Explanation:
1. Molar mass of CaCl₂ = 40 + 2 × 35.5 = 111 g mol⁻¹
2. M = \( \frac{\text{Mass of solute}}{\text{Molar mass} \times \text{Volume (L)}} \) = \( \frac{22.2}{111 \times 1} \) = 0.2 mol L⁻¹ = 0.2 M

Teacher's Note:
1. Molarity is moles per litre of solution, so with 1 L the molarity simply equals the moles.
2. Take Cl = 35.5, not 35 or 71, when finding molar mass of CaCl₂ (two Cl atoms give 71).

Answer: (c) 2.05 M

Explanation:
1. Moles of urea = 120/60 = 2
2. Mass of solution = mass of solvent + mass of solute = 1000 + 120 = 1120 g
3. Volume of solution = \( \frac{1120 \text{ g}}{1.15 \text{ g/mL}} \) = 973.9 mL = 0.974 L
4. Molarity = \( \frac{2}{0.974} \) = 2.05 M

Teacher's Note:
1. Molarity needs the volume of the solution, found from its total mass and density.
2. Using 1 L of water as the solution volume gives 2.00 M, which is the common trap.

Answer: (b) 0.0292

Explanation:
1. Molar mass of C₁₂H₂₂O₁₁ = (12 × 12.01) + (22 × 1.008) + (11 × 16.00) = 342.3 g mol⁻¹ ≈ 342 g mol⁻¹
2. Molarity = \( \frac{w}{m \times V \text{(in L)}} \) = \( \frac{20}{342 \times 2} \) = 0.0292 mol L⁻¹

Teacher's Note:
1. The molar mass of sucrose (342 g mol⁻¹) is worth remembering; it appears often in numericals.
2. Divide by the final volume of solution (2 L), not by the volume of water added.

Answer: (d) 1 m

Explanation:
1. Molality (m) = \( \frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} \)
2. Molar mass of HCl = 1 + 35.5 = 36.5 g mol⁻¹; moles of HCl = \( \frac{18.25}{36.5} \) = 0.5
3. Mass of water = 500 g = 0.5 kg
4. m = \( \frac{0.5}{0.5} \) = 1 m

Teacher's Note:
1. Note the unit: 'm' is molality and 'M' is molarity, so 1 M is a different quantity.
2. Molality uses the mass of solvent in kilograms, not the volume of solution.

Answer: (c) 1.67 m

Explanation:
1. Take 1 L (1000 mL) of solution: it contains 2 mol NaOH = 2 × 40 = 80 g.
2. Mass of solution = density × volume = 1.28 g cm⁻³ × 1000 mL = 1280 g
3. Mass of solvent = 1280 − 80 = 1200 g = 1.2 kg
4. Molality = \( \frac{2}{1.2} \) = 1.67 m
5. Alternatively, m = \( \frac{M \times 1000}{1000d - M \times M_{\text{solute}}} \) = \( \frac{2 \times 1000}{1280 - 80} \) = 1.67 m

Teacher's Note:
1. Subtract the mass of solute from the mass of solution to get the mass of solvent.
2. In the shortcut formula, the term M × molar mass (2 × 40 = 80) is subtracted, not the molar mass alone.

Answer: (b) 1.41 g/cc

Explanation:
1. Take 100 g of solution: it contains 69 g HNO₃ (molar mass 63) and its volume = \( \frac{100}{d} \) mL.
2. Molarity = \( \frac{69}{63} \times \frac{1000}{100/d} \)
3. 15.44 = \( \frac{69}{63} \times \frac{1000 d}{100} \)
4. d = \( \frac{63 \times 15.44}{69 \times 10} \) = 1.409 g/cc ≈ 1.41 g/cc

Teacher's Note:
1. Shortcut: M = (mass % × density × 10)/molar mass, so d = M × molar mass/(mass % × 10).
2. Choosing 100 g of solution makes the mass per cent directly equal to the grams of solute.

Answer: (a) 0.0177

Explanation:
1. For a dilute aqueous solution, a 1.00 molar solution is taken to contain about 1.0 mol of solute in 1000 g of water.
2. n_solute = 1; n_solvent = \( \frac{1000}{18} \) = 55.56
3. χ_solute = \( \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \) = \( \frac{1}{1 + 55.56} \) = 0.0177

Teacher's Note:
1. One kilogram of water always contains 55.56 mol; this number is very useful in mole fraction problems.
2. A mole fraction can never exceed 1, so 1.7700 is ruled out at once.

Answer: (d) 0.167, 11.11

Explanation:
1. n_NaOH = \( \frac{8}{40} \) = 0.2 mol; n(H₂O) = \( \frac{18}{18} \) = 1 mol
2. χ_NaOH = \( \frac{0.2}{0.2 + 1} \) = \( \frac{0.2}{1.2} \) = 0.167
3. Molality = \( \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} \) = \( \frac{0.2}{0.018} \) = 11.11 mol kg⁻¹

Teacher's Note:
1. In mole fraction the denominator is the total moles (0.2 + 1), not just the moles of water.
2. Convert 18 g of water to 0.018 kg before finding molality.

Answer: (b) A-4, B-5, C-3, D-1, E-2

Explanation:
1. A. Mole fraction is the ratio of moles of a component to the total moles of the solution → (4)
2. B. Molarity is the number of moles of solute in 1 L of solution → (5)
3. C. Molality is the number of moles of solute in 1 kg of solvent → (3)
4. D. M₁ × V₁ = M₂ × V₂ (dilution formula) → (1)
5. E. A stock solution is a solution of higher concentration that is diluted when needed → (2)

Teacher's Note:
1. Molarity is per litre of solution, while molality is per kilogram of solvent.
2. The dilution formula M₁V₁ = M₂V₂ is used to prepare a dilute solution from a stock solution.

Answer: (a) A-3, B-2, C-1, D-4

Explanation:
1. A. Molarity: moles per litre of solution → mol L⁻¹ (3)
2. B. Mole fraction is a ratio of moles → unitless (2)
3. C. Mole → mol (1)
4. D. Molality: moles per kilogram of solvent → mol kg⁻¹ (4)

Teacher's Note:
1. The unit of a concentration term follows directly from its definition.
2. Mole fraction and mass per cent are ratios of like quantities, so they have no units.