Vidaara.orgClass 12 · Chemistry
CodeVID-C12-02-T3-01
Colligative Properties — Topic Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Which is NOT a colligative property?
- A.Osmotic pressure
- B.Elevation of boiling point
- C.Vapour pressure of pure solvent
- D.Depression of freezing point
2.
The unit of $K_f$ is:
- A.K kg mol$^{-1}$
- B.mol kg$^{-1}$
- C.K mol kg$^{-1}$
- D.dimensionless
3.
For NaCl in water the van't Hoff factor is approximately:
- A.0.5
- B.1
- C.2
- D.4
4.
Osmotic pressure of a solution is given by:
- A.$K_b m$
- B.$CRT$
- C.$p^0 x$
- D.$K_f m$
5.
For an associating solute, the van't Hoff factor is:
- A.greater than 1
- B.equal to 1
- C.less than 1
- D.equal to zero
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define osmotic pressure and state its formula.
7.
Why is molality, not molarity, used in $\Delta T_b$ and $\Delta T_f$ equations?
8.
What is the van't Hoff factor and what does $i>1$ indicate?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
6.0 g of a non-volatile solute in 100 g water raises the boiling point by 0.52 K. Find the molar mass of the solute. $K_b=0.52$.
10.
Calculate the osmotic pressure of a 0.2 M glucose solution at 300 K. $R=0.083\ \text{L bar K}^{-1}\text{mol}^{-1}$.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
A 0.5 m $\text{CaCl}_2$ solution shows a freezing-point depression of 2.42 K. Given $K_f=1.86$, find (i) the van't Hoff factor and (ii) the degree of dissociation, and explain why $i>1$.
Answer Key
Section A — Multiple Choice Questions
- (C) Vapour pressure of pure solvent
- (A) K kg mol$^{-1}$
- (C) 2
- (B) $CRT$
- (C) less than 1
Section B — Short Answer (2 marks)
- The excess pressure that must be applied to the solution side to stop the net flow of solvent through a semipermeable membrane; $\Pi=CRT$.
- Molality is mass-based and temperature-independent, whereas these measurements involve temperature changes that would alter a volume-based molarity.
- $i=\frac{\text{observed colligative property}}{\text{calculated value}}$; $i>1$ indicates dissociation of the solute into more particles.
Section C — Short Answer (3 marks)
- $m=\Delta T_b/K_b=0.52/0.52=1.0$ m. So 0.1 mol in 0.1 kg gives 1.0 mol kg$^{-1}$; moles $=0.1$, $M=6.0/0.1=60\ \text{g mol}^{-1}$.
- $\Pi=CRT=0.2\times0.083\times300=4.98$ bar.
Section D — Long Answer (5 marks)
- Calculated $\Delta T_f=1.86\times0.5=0.93$ K. $i=2.42/0.93=2.60$. For CaCl2, $n=3$; $\alpha=(i-1)/(n-1)=(1.60)/2=0.80$ (80%). $i>1$ because CaCl2 dissociates into 3 ions, increasing the number of particles.
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