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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.
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0.65
0.75
0.85
0.95
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40 g mol−1
60 g mol−1
80 g mol−1
100 g mol−1
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120 g mol−1
180 g mol−1
200 g mol−1
240 g mol−1
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120 g mol⁻¹
160 g mol⁻¹
200 g mol⁻¹
250 g mol⁻¹
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100 g mol⁻¹
110 g mol⁻¹
120 g mol⁻¹
130 g mol⁻¹
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100 g mol⁻¹
112.5 g mol⁻¹
120 g mol⁻¹
135 g mol⁻¹
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100 g mol⁻¹
120 g mol⁻¹
140 g mol⁻¹
160 g mol⁻¹
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20%
30%
40%
50%
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40 g mol⁻¹
60 g mol⁻¹
80 g mol⁻¹
100 g mol⁻¹
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i = 0.625; association
i = 1.6; dissociation
i = 1.0; normal behaviour
i = 2.5; association
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i = 0.8; association
i = 1.25; dissociation
i = 2.0; dissociation
i = 0.5; complete dissociation
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60 g mol⁻¹
80 g mol⁻¹
100 g mol⁻¹
120 g mol⁻¹
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40 g mol⁻¹
45 g mol⁻¹
60 g mol⁻¹
75 g mol⁻¹
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25%
40%
50%
75%
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88.8 g mol⁻¹
99.9 g mol⁻¹
111 g mol⁻¹
133.2 g mol⁻¹
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24 g mol⁻¹
25 g mol⁻¹
48 g mol⁻¹
50 g mol⁻¹
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80 g mol⁻¹
96 g mol⁻¹
112 g mol⁻¹
128 g mol⁻¹
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181.8 g mol⁻¹
200 g mol⁻¹
222.2 g mol⁻¹
250 g mol⁻¹
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120 g mol⁻¹
150 g mol⁻¹
180 g mol⁻¹
240 g mol⁻¹
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50 g mol⁻¹
62.5 g mol⁻¹
75 g mol⁻¹
80 g mol⁻¹
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40 g mol⁻¹
60 g mol⁻¹
96 g mol⁻¹
120 g mol⁻¹
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100 g mol⁻¹
200 g mol⁻¹
300 g mol⁻¹
400 g mol⁻¹
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100 g mol⁻¹
125 g mol⁻¹
360 g mol⁻¹
450 g mol⁻¹
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40 g mol⁻¹
50 g mol⁻¹
60 g mol⁻¹
75 g mol⁻¹
Question 1HardLevel 2
When 1.6 g of a substance is dissolved in 200 g of water, the freezing-point depression is 0.744 K. If the normal molar mass is 80 g mol⁻¹ and Kf = 1.86 K kg mol⁻¹, what is the van’t Hoff factor i?
Correct answer: D
First calculate the theoretical molality using the normal molar mass: moles of solute = 1.6/80 = 0.020 mol, and solvent mass = 0.200 kg, so m = 0.020/0.200 = 0.100 mol kg⁻¹. The normal freezing-point depression is Kf m = 1.86 × 0.100 = 0.186 K. Thus i = observed depression/theoretical depression = 0.744/0.186 = 4. Option D is correct.
If 20% of a substance forms tetramers, what is the van’t Hoff factor?
Correct answer: C
Assume one mole of solute initially. Of this, 80% remains as individual molecules and contributes 0.80 effective particles. The remaining 20% associates into tetramers; four original molecules form one particle, so this portion contributes 0.20/4 = 0.05 particles. Therefore i = 0.80 + 0.05 = 0.85, making option C correct.
When 3 g of an AB2-type solute is dissolved in 250 g water, ΔTf = 0.558 K. If dissociation is 50% and Kf = 1.86 K kg mol−1, what is the true molar mass?
Correct answer: C
For AB2, dissociation produces one A particle and two B particles. At degree of dissociation α = 0.50, the van’t Hoff factor is i = 1 + 2α = 2. Using ΔTf = iKf m, the molality is m = 0.558/(2 × 1.86) = 0.150 mol kg−1. The solvent mass is 250 g = 0.250 kg, so moles of solute = 0.150 × 0.250 = 0.0375 mol. Hence the true molar mass is 3/0.0375 = 80 g mol−1.
A 300 mL solution is prepared using 2.4 g solute. Its osmotic pressure at 300 K is 0.492 atm. If the solute forms dimers and i = 0.5, what is the true molar mass?
Correct answer: C
Use the osmotic-pressure equation π = iCRT, where C is the concentration of solute particles in the solution. Thus C = 0.492/(0.5 × 0.082 × 300) = 0.040 mol L−1. The solution volume is 300 mL = 0.300 L, so the number of moles represented by the colligative concentration is 0.040 × 0.300 = 0.012 mol. Therefore the true molar mass of the original solute is 2.4/0.012 = 200 g mol−1. The given i has already been included in calculating C.
A 750 mL sample of a 0.02 M solution contains 3.0 g of solute. If i = 1.25 and the given molarity is the actual solute molarity, what is the observed molar mass?
Correct answer: B
The actual moles in 750 mL are n = M V = 0.02 × 0.750 = 0.015 mol. The molar mass calculated without correcting for dissociation or association is 3.0/0.015 = 200 g mol⁻¹. For abnormal colligative behaviour, observed molar mass M_obs = M_true/i. Hence M_obs = 200/1.25 = 160 g mol⁻¹, so option B is correct.
A substance undergoes 40% dimerization. If its observed molar mass determined by a colligative method is 150 g mol⁻¹, what is its true molar mass?
Correct answer: C
For dimerization, two monomer particles become one dimer particle. If the degree of association is α, the van’t Hoff factor is i = 1 − α/2. With α = 0.40, i = 1 − 0.40/2 = 0.80. Since M_obs = M_true/i, the true molar mass is M_true = iM_obs = 0.80 × 150 = 120 g mol⁻¹. Therefore, option C is correct.
A substance forms trimers and is 30% associated. If its true molar mass is 90 g mol⁻¹, what will be the approximate observed molar mass?
Correct answer: B
For formation of trimers, three monomer particles form one trimer, so the van’t Hoff factor is i = 1 − 2α/3. With α = 0.30, i = 1 − (2 × 0.30)/3 = 0.80. The observed molar mass is related by M_obs = M_true/i. Therefore, M_obs = 90/0.80 = 112.5 g mol⁻¹, making option B correct.
An AB-type solute is 75% dissociated. If its observed molar mass is 80 g mol⁻¹, what is its true molar mass?
Correct answer: C
For AB → A + B, one solute unit produces two particles on complete dissociation. Thus, for degree of dissociation α, i = 1 + α. With α = 0.75, i = 1.75. Since M_obs = M_true/i, the true molar mass is M_true = iM_obs = 1.75 × 80 = 140 g mol⁻¹. Hence option C is correct.
The observed molar mass of an A₂B-type solute is 5/9 of its true molar mass. What is the degree of dissociation?
Correct answer: C
Because M_obs = M_true/i, the given ratio M_obs/M_true = 5/9 gives i = 9/5 = 1.8. On dissociation, A₂B produces 2A + B, or three particles in total. Therefore i = 1 + (3 − 1)α = 1 + 2α. Solving 1.8 = 1 + 2α gives α = 0.40, or 40%. Thus option C is correct.
When 2.8 g of a solute is dissolved in 200 g of water, the depression in freezing point is 0.651 K. If the solute is AB-type and 50% dissociated, what is the true molar mass?
Correct answer: B
For AB → A + B with 50% dissociation, i = 1 + α = 1.5. Using ΔT_f = iK_fm, the actual molality is m = 0.651/(1.5 × 1.86) ≈ 0.233 mol kg⁻¹. For 200 g = 0.2 kg water, moles of solute are 0.233 × 0.2 ≈ 0.0467 mol. Thus M = 2.8/0.0467 ≈ 60 g mol⁻¹, so option B is correct.
If the observed molar mass of a solute is 0.625 times its true molar mass, what are the van’t Hoff factor (i) and the type of behaviour shown by the solute?
Correct answer: B
For abnormal molar masses, the van’t Hoff factor is related by i = true molar mass / observed molar mass. Therefore, i = 1/0.625 = 1.6. Since i is greater than 1, the number of solute particles has increased. This happens when the solute dissociates into two or more particles in solution, so the behaviour is dissociation.
If the observed molar mass of a solute is 1.25 times its true molar mass, what are the van’t Hoff factor (i) and the type of behaviour shown by the solute?
Correct answer: A
The van’t Hoff factor is calculated from i = true molar mass / observed molar mass. Here the observed value is 1.25 times the true value, so i = 1/1.25 = 0.80. A factor below 1 indicates fewer solute particles than expected because individual molecules combine. Thus, the solute shows association, such as dimerization.
When 1.2 g of a solute is dissolved in 150 g of solvent, the elevation in boiling point is 0.104 K. If Kb = 0.52 K kg mol⁻¹ and the solute has a van’t Hoff factor i = 2, what is its true molar mass?
Correct answer: B
For boiling-point elevation, ΔTb = iKb m, where m is the actual molality of the solute. Thus, the effective molality is 0.104/0.52 = 0.20 mol kg⁻¹. Since i = 2, the actual molality is 0.20/2 = 0.10 mol kg⁻¹. The solvent mass is 150 g = 0.150 kg, so moles of solute = 0.10 × 0.150 = 0.015 mol. Therefore, molar mass = 1.2/0.015 = 80 g mol⁻¹.
When 0.9 g of a solute is dissolved in 100 g of water, the depression in freezing point is 0.279 K. If Kf = 1.86 K kg mol⁻¹ and the solute has i = 0.75 because of dimer formation, what is its true molar mass?
Correct answer: B
For freezing-point depression, ΔTf = iKf m. Hence the true molality is m = 0.279/(0.75 × 1.86) = 0.20 mol kg⁻¹. Since 100 g water equals 0.100 kg, the amount of solute is 0.20 × 0.100 = 0.020 mol. Its true molar mass is therefore 0.9/0.020 = 45 g mol⁻¹.
An A₂B₃-type solute has a van’t Hoff factor i = 3. If complete dissociation produces five particles, what is the degree of dissociation?
Correct answer: C
For one formula unit of A₂B₃, complete dissociation produces 2 + 3 = 5 ions or particles. For a compound producing ν particles, the relation is i = 1 + α(ν − 1), where α is the degree of dissociation. Substituting i = 3 and ν = 5 gives 3 = 1 + 4α. Hence α = 2/4 = 0.5, or 50%. Therefore, option C is correct.
CaCl₂ is 75% dissociated in solution. If its observed molar mass is 44.4 g mol⁻¹, what is its true molar mass?
Correct answer: C
CaCl₂ dissociates as CaCl₂ → Ca²⁺ + 2Cl⁻, producing three particles when completely dissociated. Therefore, i = 1 + α(3 − 1) = 1 + 2α. With α = 0.75, i = 1 + 1.5 = 2.5. The observed molar mass is related to the true molar mass by Mobserved = Mtrue/i. Hence Mtrue = i × Mobserved = 2.5 × 44.4 = 111 g mol⁻¹.
In the vapour-pressure method, p⁰ = 100 mm Hg and p = 96 mm Hg. If 2 g of a non-volatile solute is dissolved in 36 g of water, what is the approximate molar mass of the solute?
Correct answer: A
For a dilute solution, relative lowering of vapour pressure is approximately the mole fraction of solute: (p⁰ − p)/p⁰ = 4/100 = 0.04. Water contributes 36/18 = 2 mol. Using x₂ = n₂/(2 + n₂), 0.04 = n₂/(2 + n₂), giving n₂ ≈ 0.0833 mol. Thus the molar mass is 2/0.0833 ≈ 24 g mol⁻¹.
A solute has an observed molar mass of 64 g mol⁻¹. If it is an AB₂-type electrolyte and is 25% dissociated, what is its true molar mass?
Correct answer: B
An AB₂ electrolyte produces three particles on complete dissociation: AB₂ → A²⁺ + 2B⁻. Thus, for degree of dissociation α, i = 1 + α(3 − 1) = 1 + 2α. With α = 0.25, i = 1 + 0.50 = 1.50. Since dissociation makes the observed molar mass smaller, Mobserved = Mtrue/i. Therefore, Mtrue = 1.50 × 64 = 96 g mol⁻¹.
If the true molar mass of a solute is 200 g mol⁻¹ and 20% dimerization occurs, what will be its approximate observed molar mass?
Correct answer: C
For dimerization, two original solute molecules combine to form one particle. Therefore, when the fraction dimerized is α, i = 1 − α/2. With α = 0.20, i = 1 − 0.10 = 0.90. Because i = true molar mass/observed molar mass, the observed molar mass is 200/0.90 ≈ 222.2 g mol⁻¹.
A solution contains 3.0 g of solute in 600 mL of solution. Its osmotic pressure at 300 K is 0.615 atm. If the van’t Hoff factor is i = 0.75, what is the true molar mass of the solute?
Correct answer: B
Osmotic pressure is given by π = iCRT, so the molar concentration is C = π/(iRT). Using R = 0.082 L atm mol⁻¹ K⁻¹, C = 0.615/(0.75 × 0.082 × 300) ≈ 0.0333 mol L⁻¹. The solution volume is 600 mL = 0.600 L, so moles of solute = 0.0333 × 0.600 ≈ 0.0200 mol. Therefore, true molar mass = 3.0/0.0200 = 150 g mol⁻¹.
When 2.5 g of a solute is dissolved in 200 g of water, the depression in freezing point is 0.465 K. If the van’t Hoff factor is 1.25, what is the true molar mass?
Correct answer: B
For freezing-point depression, ΔTf = iKf m. Thus the actual molality is m = 0.465 ÷ (1.25 × 1.86) = 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 = 2.5 g ÷ 0.040 mol = 62.5 g mol⁻¹. Hence option B is correct.
When 3.6 g of a solute is dissolved in 400 g of solvent, the elevation in boiling point is 0.117 K. If Kb = 0.52 K kg mol⁻¹ and i = 1.5, what is the true molar mass?
Correct answer: B
The elevation in boiling point is given by ΔTb = iKb m. Hence the true molality is m = 0.117 ÷ (1.5 × 0.52) = 0.150 mol kg⁻¹. The solvent mass is 400 g = 0.400 kg, so the amount of solute is 0.150 × 0.400 = 0.060 mol. Therefore, molar mass = 3.6 ÷ 0.060 = 60 g mol⁻¹, making option B correct.
When 2.0 g of a solute is dissolved in 100 g of water, the depression in freezing point is 0.186 K. If the van’t Hoff factor is 0.5, what is the true molar mass?
Correct answer: A
Using ΔTf = iKf m, the true molality is m = 0.186 ÷ (0.5 × 1.86) = 0.20 mol kg⁻¹. The solvent mass is 100 g = 0.100 kg, so the moles of solute are 0.20 × 0.100 = 0.020 mol. Hence the molar mass is 2.0 ÷ 0.020 = 100 g mol⁻¹. Therefore, option A is correct.
A 500 mL solution is prepared from 4.5 g of solute. At 300 K, its osmotic pressure is 0.738 atm. If the van’t Hoff factor is 1.2, what is the true molar mass? Use R = 0.082 L atm K⁻¹ mol⁻¹.
Correct answer: C
For osmotic pressure, π = iCRT. Thus the molar concentration is C = 0.738 ÷ (1.2 × 0.082 × 300) ≈ 0.025 mol L⁻¹. The solution volume is 0.500 L, so moles of solute are 0.025 × 0.500 = 0.0125 mol. Therefore, molar mass = 4.5 ÷ 0.0125 = 360 g mol⁻¹. Option C is correct.
When 3.0 g of a solute is dissolved in 200 g of water, the depression in freezing point is 0.558 K. If the van’t Hoff factor is 1.2, what is the true molar mass?
Correct answer: C
Using ΔTf = iKf m, the true molality is m = 0.558 ÷ (1.2 × 1.86) = 0.25 mol kg⁻¹. The solvent mass is 200 g = 0.200 kg, so moles of solute are 0.25 × 0.200 = 0.050 mol. Therefore, the molar mass is 3.0 g ÷ 0.050 mol = 60 g mol⁻¹. Hence option C is correct.
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