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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 4View options
i = 1, normal behaviour
i = 2, dissociation
i = 0.5, association
i = 3, complete dissociation into three particles
Hard · Level 4View options
300 g mol⁻¹
400 g mol⁻¹
500 g mol⁻¹
600 g mol⁻¹
Hard · Level 4View options
120 g mol⁻¹, 25%
120 g mol⁻¹, 50%
96 g mol⁻¹, 25%
144 g mol⁻¹, 25%
Hard · Level 4View options
32 g mol⁻¹
80 g mol⁻¹
160 g mol⁻¹
200 g mol⁻¹
Hard · Level 4View options
100 g mol⁻¹
150 g mol⁻¹
200 g mol⁻¹
250 g mol⁻¹
Hard · Level 4View options
60 g mol⁻¹, 80%
60 g mol⁻¹, 60%
78 g mol⁻¹, 80%
90 g mol⁻¹, 40%
Hard · Level 4View options
200 g mol⁻¹
210 g mol⁻¹
220 g mol⁻¹
240 g mol⁻¹
Hard · Level 4View options
Include the van’t Hoff factor with i > 1
Include a van’t Hoff factor with i < 1
Set the temperature equal to zero
Remove the solvent mass from the calculation
Hard · Level 4View options
Interparticle association or dissociation effects are decreasing
The solute is disappearing
The solvent is becoming solid
Temperature has no physical meaning
Hard · Level 4View options
πV = i nRT
πV = nRT/i
π = iVnRT
πRT = inV
Hard · Level 4View options
When the solute undergoes association
When the solute completely dissociates
When temperature is measured in kelvin
When volume is measured in litres
Question 1HardLevel 4
A solute of concentration 0.02 M has an osmotic pressure of 0.984 atm at 300 K. If R = 0.082 L atm mol⁻¹ K⁻¹, what is the van’t Hoff factor (i), and what does it indicate about the solute’s behaviour?
Correct answer: B
For a dilute solution, osmotic pressure is given by π = iCRT, where i is the van’t Hoff factor. Substituting the data gives i = 0.984 ÷ (0.02 × 0.082 × 300) = 0.984 ÷ 0.492 = 2. Since i is greater than 1, the solute produces more particles in solution than its original formula units, indicating dissociation. Thus, option B is correct.
If 0.3 g of a solute in 150 mL of solution gives an osmotic pressure of 0.123 atm at 300 K, and i = 1, what is the molar mass?
Correct answer: B
Because i = 1, the osmotic-pressure equation is π = CRT. The molar concentration is C = π/(RT) = 0.123/(0.082 × 300) = 0.005 mol L⁻¹. The solution volume is 150 mL = 0.150 L, so solute moles are 0.005 × 0.150 = 0.00075 mol. Hence M = 0.3/0.00075 = 400 g mol⁻¹, option B.
An AB solute has a van’t Hoff factor i = 1.25 and an observed molar mass of 96 g mol⁻¹. What are its true molar mass and degree of dissociation?
Correct answer: A
The true molar mass is obtained from M_true = iM_observed = 1.25 × 96 = 120 g mol⁻¹. For AB dissociating into A and B, two particles form from one particle, so i = 1 + α. Therefore α = i − 1 = 0.25, or 25%. Both required values are given in option A, which is correct.
If a solute has i = 0.4 and a true molar mass of 80 g mol⁻¹, what will be the observed molar mass determined by a colligative method?
Correct answer: D
For molar mass determined from a colligative property, M_observed = M_true/i. Substitution gives M_observed = 80/0.4 = 200 g mol⁻¹. Since i is less than one, association has reduced the number of particles in solution, weakened the colligative effect, and made the experimentally observed molar mass greater than the true molar mass. Thus option D is correct.
When 0.5 g of a solute is dissolved in 250 g of water, the freezing-point depression is 0.0186 K. If i = 1 and Kf for water is 1.86 K kg mol⁻¹, what is the molar mass?
Correct answer: C
Freezing-point depression is given by ΔTf = iKf m. Since i = 1, the molality is m = 0.0186/1.86 = 0.010 mol kg⁻¹. The solvent mass is 250 g = 0.250 kg, so solute moles are 0.010 × 0.250 = 0.00250 mol. Therefore molar mass = 0.5/0.00250 = 200 g mol⁻¹, making option C correct.
An AB₂ solute has a van’t Hoff factor of 2.6, and its true molar mass is 156 g mol⁻¹. What are its observed molar mass and degree of dissociation?
Correct answer: A
The observed molar mass is related to the true molar mass by M observed = M true/i. Therefore, M observed = 156/2.6 = 60 g mol⁻¹. For AB₂ dissociating as AB₂ → A + 2B, one mole produces three particles when completely dissociated, so i = 1 + 2α. Hence 2.6 = 1 + 2α, giving α = 0.8 or 80%. Option A is correct.
A 1 L solution is prepared from 3.6 g of a solute. At 300 K, its osmotic pressure is 0.369 atm. If the solute undergoes 25% dimerization, what is its approximate true molar mass?
Correct answer: B
For dimerization, two solute molecules combine to form one particle, so the van’t Hoff factor is i = 1 − α/2 = 1 − 0.25/2 = 0.875. Using π = iCRT, C = 0.369/(0.875 × 0.0821 × 300) ≈ 0.01714 mol L⁻¹. Since the solution volume is 1 L, it contains about 0.01714 mol of solute units. Therefore, the true molar mass is M = 3.6/0.01714 ≈ 210 g mol⁻¹. Hence, option B is correct.
If the molar mass calculated from boiling-point elevation is lower than the true molar mass, which correction is most appropriate when dissociation is responsible?
Correct answer: A
For boiling-point elevation, the general relation is ΔTb = iKbm. Dissociation produces more particles than expected from the original solute units, so i becomes greater than 1 and the observed elevation is larger. If this effect is ignored, the calculation attributes the large elevation to too many solute moles and therefore gives an apparent molar mass that is too low. Including i > 1 corrects the particle-count effect and gives a value closer to the true molar mass.
If the apparent molar mass of a solute approaches its true molar mass on dilution, what does this indicate?
Correct answer: A
An apparent molar mass differs from the true value when the solute associates into larger units or dissociates into more particles, changing the van’t Hoff factor. Dilution generally reduces interactions between solute particles, so association or other non-ideal effects become weaker. As the solution approaches ideal behaviour, the van’t Hoff factor approaches unity and the apparent molar mass approaches the true molar mass.
If osmotic pressure increases due to factor (i), what is the corrected form of the osmotic-pressure relation?
Correct answer: A
For an ideal dilute solution, the osmotic-pressure equation is πV = nRT. If the solute dissociates or otherwise produces more effective particles, the van’t Hoff factor i accounts for that change. The effective number of particles becomes in, so the corrected equation is πV = i nRT. Thus i multiplies the colligative effect; it does not divide the right-hand side.
In molar-mass determination, under which condition will the value calculated from M = wRT/(πV) be higher than the true value?
Correct answer: A
Association joins two or more solute molecules into larger aggregates, reducing the number of independent particles in solution. Consequently, the observed osmotic pressure π becomes smaller than the ideal value. Since π appears in the denominator of M = wRT/(πV), a smaller π produces a calculated molar mass greater than the true molar mass. Therefore, association is the correct condition.
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