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In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how the molar mass of a solute can be determined from the measurable properties of a solution. The topic connects mass, moles, concentration, and colligative properties such as relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure. Students also practise selecting suitable formulas, interpreting experimental data, and recognising how observed results can indicate association or dissociation of solute particles.
TOPIC PRACTICE
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Hard · Level 3View options
18 g mol⁻¹
20 g mol⁻¹
25 g mol⁻¹
30 g mol⁻¹
Hard · Level 3View options
40 g mol⁻¹
46.2 g mol⁻¹
60 g mol⁻¹
75 g mol⁻¹
Hard · Level 3View options
166.7 g mol⁻¹
180 g mol⁻¹
200 g mol⁻¹
225 g mol⁻¹
Hard · Level 3View options
225 g mol⁻¹
250 g mol⁻¹
300 g mol⁻¹
400 g mol⁻¹
Hard · Level 3View options
40 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
160 g mol⁻¹
Hard · Level 3View options
200 g mol⁻¹
250 g mol⁻¹
300 g mol⁻¹
400 g mol⁻¹
Hard · Level 3View options
25%
50%
75%
100%
Hard · Level 3View options
20%
30%
40%
50%
Hard · Level 3View options
100 g mol⁻¹
125 g mol⁻¹
137.5 g mol⁻¹
150 g mol⁻¹
Hard · Level 3View options
75 g mol⁻¹
90 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
Hard · Level 3View options
40 g mol⁻¹
50 g mol⁻¹
60 g mol⁻¹
80 g mol⁻¹
Hard · Level 3View options
45 g mol⁻¹
60 g mol⁻¹
75 g mol⁻¹
90 g mol⁻¹
Hard · Level 3View options
18 g mol⁻¹
19 g mol⁻¹
20 g mol⁻¹
24 g mol⁻¹
Hard · Level 3View options
200 g mol⁻¹
240 g mol⁻¹
300 g mol⁻¹
360 g mol⁻¹
Hard · Level 3View options
60 g mol⁻¹
72 g mol⁻¹
80 g mol⁻¹
90 g mol⁻¹
Hard · Level 3View options
100 g mol⁻¹
125 g mol⁻¹
150 g mol⁻¹
200 g mol⁻¹
Hard · Level 3View options
20 g mol⁻¹
25 g mol⁻¹
40 g mol⁻¹
50 g mol⁻¹
Hard · Level 3View options
i = 0.67, association
i = 1.5, dissociation
i = 2.4, dissociation
i = 1, normal
Hard · Level 3View options
112.5 g mol⁻¹
75 g mol⁻¹
100 g mol⁻¹
150 g mol⁻¹
Hard · Level 3View options
90 g mol⁻¹
120 g mol⁻¹
150 g mol⁻¹
360 g mol⁻¹
Hard · Level 3View options
50 g mol⁻¹
60 g mol⁻¹
75 g mol⁻¹
90 g mol⁻¹
Hard · Level 3View options
23 g mol⁻¹
25 g mol⁻¹
46 g mol⁻¹
50 g mol⁻¹
Hard · Level 3View options
128 g mol⁻¹
144 g mol⁻¹
160 g mol⁻¹
192 g mol⁻¹
Hard · Level 3View options
60 g mol⁻¹
75 g mol⁻¹
90 g mol⁻¹
105.9 g mol⁻¹
Hard · Level 3View options
240 g mol⁻¹
270 g mol⁻¹
300 g mol⁻¹
360 g mol⁻¹
Question 1HardLevel 3
When 5 g of a non-volatile solute is dissolved in 45 g water, the relative lowering of vapour pressure is 0.10. What is the approximate molar mass of the solute?
Correct answer: A
For a dilute solution containing a non-volatile solute, relative lowering of vapour pressure equals the mole fraction of the solute: x₂ = n₂/(n₁+n₂). Water moles are 45/18 = 2.5 mol. Thus 0.10 = n₂/(2.5+n₂), giving n₂ = 0.278 mol approximately. Therefore the solute molar mass is 5/0.278 ≈ 18 g mol⁻¹, so option A is correct.
An AB₂ solute has a true molar mass of 120 g mol⁻¹ and is 80% dissociated. What will be its observed molar mass?
Correct answer: B
For AB₂ dissociation, one formula unit produces three particles, so the van’t Hoff factor is i = 1 + 2α. With α = 0.80, i = 1 + 2(0.80) = 2.6. The observed molar mass is lower because dissociation increases the number of particles: M observed = M true/i = 120/2.6 ≈ 46.2 g mol⁻¹. Hence B is correct.
A solute forms trimers to the extent of 25%. If its true molar mass is 150 g mol⁻¹, what will be the approximate observed molar mass?
Correct answer: B
For association into trimers, three original particles become one particle. If α is the fraction associated, the van’t Hoff factor is i = 1 − 2α/3. With α = 0.25, i = 1 − (2 × 0.25)/3 = 0.8333 approximately. Therefore M observed = M true/i = 150/0.8333 ≈ 180 g mol⁻¹. Association lowers i and raises the observed molar mass, so B is correct.
A 1 L solution is prepared from 2.0 g of a solute. At 300 K, its osmotic pressure is 0.164 atm. If the solute forms dimers to the extent of 50%, what is the true molar mass?
Correct answer: A
For 50% dimerisation, the van’t Hoff factor is i = 1 − α/2 = 1 − 0.50/2 = 0.75, because two solute molecules combine to form one particle. Using π = iCRT, C = 0.164/(0.75 × 0.082 × 300) = 0.00889 mol L⁻¹. In 1 L, the amount is 0.00889 mol. Therefore, the true molar mass is 2.0/0.00889 ≈ 225 g mol⁻¹, so option A is correct.
When 4.0 g of an AB₂-type solute is dissolved in 200 g of water, the freezing-point depression is 0.744 K. If dissociation is 50% and Kf = 1.86 K kg mol⁻¹, what is the true molar mass?
Correct answer: B
For AB₂, dissociation produces three ions, so at degree of dissociation α = 0.50, i = 1 + 2α = 2. The freezing-point relation is ΔTf = iKf m; hence m = 0.744/(2 × 1.86) = 0.200 mol kg⁻¹. Since the solvent mass is 0.200 kg, moles of solute = 0.200 × 0.200 = 0.0400 mol. Therefore, molar mass = 4.0/0.0400 = 100 g mol⁻¹, so B is correct.
A 250 mL solution is prepared from 3.0 g of a solute. At 300 K, its osmotic pressure is 0.738 atm. If the solute has i = 0.75, what is the true molar mass?
Correct answer: C
Use the osmotic-pressure equation π = iCRT. The molar concentration is C = 0.738/(0.75 × 0.082 × 300) = 0.0400 mol L⁻¹. The volume is 250 mL = 0.250 L, so the amount of solute is n = 0.0400 × 0.250 = 0.0100 mol. Thus, the true molar mass is M = 3.0/0.0100 = 300 g mol⁻¹. Therefore, option C is correct.
The true molar mass of MgCl₂ is 95 g mol⁻¹. If its observed molar mass from a colligative-property method is 47.5 g mol⁻¹, what is the degree of dissociation?
Correct answer: B
The van’t Hoff factor is the ratio of true to observed molar mass: i = Mtrue/Mobserved = 95/47.5 = 2. For MgCl₂, dissociation into Mg²⁺ and 2Cl⁻ gives three particles at complete dissociation, so i = 1 + 2α. Substituting i = 2 gives 2 = 1 + 2α, hence α = 0.50 or 50%. Therefore, option B is correct.
A solute forms dimers. Its true molar mass is 120 g mol⁻¹ and its observed molar mass is 150 g mol⁻¹. What is the degree of association?
Correct answer: C
For abnormal molar mass, i = Mtrue/Mobserved = 120/150 = 0.80. When monomer molecules associate to form dimers, the number of particles decreases and i = 1 − α/2, where α is the fraction associated. Thus, 0.80 = 1 − α/2, so α/2 = 0.20 and α = 0.40. The degree of association is therefore 40%, making option C correct.
An AB₃-type solute is 40% dissociated. If its observed molar mass is 62.5 g mol⁻¹, what is its true molar mass?
Correct answer: C
For AB₃, dissociation produces four particles from one formula unit, so i = 1 + 3α. With α = 0.40, i = 1 + 3(0.40) = 2.20. The relation between true and observed molar masses is Mtrue = i × Mobserved. Therefore, Mtrue = 2.20 × 62.5 = 137.5 g mol⁻¹. Thus, option C is correct.
A 0.04 M solution has an osmotic pressure of 1.476 atm at 300 K. If the true molar mass is 150 g mol⁻¹, what will be the observed molar mass?
Correct answer: C
From π = iCRT, the van’t Hoff factor is i = 1.476/(0.04 × 0.082 × 300) = 1.50. Since i is greater than 1, dissociation has increased the number of particles, causing the observed molar mass to be smaller than the true value. The relation is Mobserved = Mtrue/i = 150/1.50 = 100 g mol⁻¹. Therefore, option C is correct.
When 2.4 g of a solute is dissolved in 200 g of solvent, the elevation in boiling point is 0.156 K. If Kb = 0.52 K kg mol⁻¹ and i = 1.5, what is the true molar mass?
Correct answer: C
Use the colligative-property equation ΔTb = iKb m. Thus, the molality is m = 0.156/(1.5 × 0.52) = 0.20 mol kg⁻¹. The solvent mass is 200 g = 0.200 kg, so moles of solute = 0.20 × 0.200 = 0.040 mol. Therefore, molar mass = 2.4/0.040 = 60 g mol⁻¹, so option C is correct.
When 1.8 g of a solute is dissolved in 150 g of water, the depression in freezing point is 0.279 K. If i = 0.75, what is the true molar mass?
Correct answer: B
For freezing-point depression, ΔTf = iKf m. Taking Kf for water as 1.86 K kg mol⁻¹, the effective molality is 0.279/1.86 = 0.15 mol kg⁻¹. Since i = 0.75, the actual molality is 0.15/0.75 = 0.20 mol kg⁻¹. In 0.150 kg water, moles are 0.030; hence M = 1.8/0.030 = 60 g mol⁻¹. Option B is correct.
In the vapour-pressure method, p⁰ = 120 mmHg and p = 114 mmHg. If 3 g of solute is dissolved in 54 g of water, what is the approximate molar mass of the solute?
Correct answer: B
The relative lowering of vapour pressure is (p⁰ − p)/p⁰ = (120 − 114)/120 = 0.05. For a dilute solution, this equals the solute mole fraction: x₂ = n₂/(n₁ + n₂). Water moles are 54/18 = 3. Hence 0.05 = n₂/(3 + n₂), giving n₂ = 0.1579 mol. The molar mass is 3/0.1579 ≈ 19 g mol⁻¹, so option B is correct.
A solution contains 2.4 g of solute in 400 mL of solution. Its osmotic pressure at 300 K is 0.492 atm. If i = 0.8, what is the true molar mass?
Correct answer: B
For osmotic pressure, π = iCRT, where C is the molar concentration. Therefore, C = π/(iRT) = 0.492/(0.8 × 0.082 × 300) = 0.025 mol L⁻¹. The solution volume is 0.400 L, so moles of solute = 0.025 × 0.400 = 0.010 mol. Thus, molar mass = 2.4/0.010 = 240 g mol⁻¹. Option B is correct.
When 3 g of a solute is dissolved in 250 g of water, the depression in freezing point is 0.465 K. If the van’t Hoff factor is 1.5, what is the true molar mass?
Correct answer: B
For water, Kf = 1.86 K kg mol⁻¹. The observed depression gives effective molality = ΔTf/Kf = 0.465/1.86 = 0.25 mol kg⁻¹. Since ΔTf = iKf m, the actual molality is 0.25/1.5 = 0.1667 mol kg⁻¹. With 0.250 kg solvent, moles of solute = 0.1667 × 0.250 = 0.04167 mol. Therefore, molar mass = 3/0.04167 ≈ 72 g mol⁻¹, so option B is correct.
When 5 g of a solute is dissolved in 500 g of solvent, the elevation in boiling point is 0.078 K. If Kb = 0.52 K kg mol⁻¹ and i = 1.5, what is the true molar mass?
Correct answer: A
Use the elevation formula ΔTb = iKb m. Thus, the actual molality is m = 0.078/(1.5 × 0.52) = 0.100 mol kg⁻¹. The solvent mass is 500 g = 0.500 kg, so the amount of solute is 0.100 × 0.500 = 0.050 mol. Hence, molar mass = mass/moles = 5/0.050 = 100 g mol⁻¹. Therefore, option A is the only correct answer.
When 2 g of a solute is dissolved in 0.4 kg of solvent, the effective molality is 0.10 m. If i = 0.5, what is the true molar mass?
Correct answer: B
Effective molality is i times the true molality, so mtrue = meffective/i = 0.10/0.5 = 0.20 mol kg⁻¹. For 0.4 kg of solvent, moles of solute = 0.20 × 0.4 = 0.08 mol. The molar mass is therefore mass divided by amount: M = 2 g/0.08 mol = 25 g mol⁻¹. Thus, option B is correct.
If the true molar mass of a solute is 240 g mol⁻¹ and the colligative method gives an observed molar mass of 160 g mol⁻¹, what are i and the probable behaviour of the solute?
Correct answer: B
For abnormal molar mass, the relation is Mobserved = Mtrue/i, or i = Mtrue/Mobserved. Therefore, i = 240/160 = 1.5. A value of i greater than one means that the number of solute particles has increased, which generally occurs through dissociation into ions or smaller particles. Hence, option B is correct.
When 1.5 g of a solute is dissolved in 100 g of water, the depression in freezing point is 0.186 K. If i = 0.75, what is the true molar mass?
Correct answer: A
For water, Kf = 1.86 K kg mol⁻¹. The effective molality is 0.186/1.86 = 0.100 mol kg⁻¹. Since i = 0.75, the true molality is 0.100/0.75 = 0.1333 mol kg⁻¹. In 0.100 kg water, moles of solute = 0.1333 × 0.100 = 0.01333 mol. Hence M = 1.5/0.01333 ≈ 112.5 g mol⁻¹, so option A is correct.
A 300 mL solution is prepared using 1.8 g of solute. At 300 K, its osmotic pressure is 0.492 atm. If i = 1.2, what is the true molar mass? Use R = 0.082 L atm K⁻¹ mol⁻¹.
Correct answer: D
For osmotic pressure, π = iCRT, so C = π/(iRT) = 0.492/(1.2 × 0.082 × 300) = 0.01667 mol L⁻¹. The solution volume is 300 mL = 0.300 L, giving moles = 0.01667 × 0.300 = 0.00500 mol. Therefore, the true molar mass is 1.8/0.00500 = 360 g mol⁻¹. Thus, option D is correct.
When 4.5 g of a solute is dissolved in 300 g of water, the depression in freezing point is 0.837 K. If i = 1.5, what is the true molar mass?
Correct answer: A
For water, Kf = 1.86 K kg mol⁻¹. The effective molality is ΔTf/Kf = 0.837/1.86 = 0.450 mol kg⁻¹. Correcting for i, the true molality is 0.450/1.5 = 0.300 mol kg⁻¹. Since 300 g water = 0.300 kg, moles of solute = 0.300 × 0.300 = 0.090 mol. Thus M = 4.5/0.090 = 50 g mol⁻¹, so option A is correct.
When 4 g of a non-volatile solute is dissolved in 36 g of water, the relative lowering of vapour pressure is 0.08. What is the approximate molar mass of the solute?
Correct answer: A
For a dilute solution containing a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute: x₂ = n₂/(n₁+n₂). Water moles are 36/18 = 2 mol. Thus 0.08 = n₂/(2+n₂), giving n₂ = 0.16/0.92 ≈ 0.174 mol. Therefore, molar mass = 4/0.174 ≈ 23 g mol⁻¹. Hence option A is correct.
A solute has a van’t Hoff factor i = 2.25 and an observed molar mass of 64 g mol⁻¹. What is its true molar mass?
Correct answer: B
The molar mass obtained from a colligative property is affected by the van’t Hoff factor. The relation is M_observed = M_true/i, because dissociation increases the number of particles and makes the observed molar mass appear smaller. Rearranging gives M_true = i × M_observed = 2.25 × 64 = 144 g mol⁻¹. Therefore, option B is correct.
An AB₂ solute has a true molar mass of 180 g mol⁻¹ and is 70% dissociated. What is its approximate observed molar mass?
Correct answer: B
For AB₂, one formula unit produces three ions on complete dissociation. At degree of dissociation α, the van’t Hoff factor is i = 1 + (3 − 1)α = 1 + 2α. With α = 0.70, i = 1 + 1.40 = 2.40. The observed molar mass is M_true/i = 180/2.40 = 75 g mol⁻¹. Hence option B is correct.
A solute forms trimers to the extent of 60%. If its true molar mass is 180 g mol⁻¹, what will be its observed molar mass?
Correct answer: C
For association into trimers, three original solute molecules combine to form one particle. If α is the fraction associated, the particle factor is i = 1 − α(1 − 1/3) = 1 − 2α/3. With α = 0.60, i = 1 − 0.40 = 0.60. Therefore M_observed = M_true/i = 180/0.60 = 300 g mol⁻¹, so option C is correct.
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