Vidaara.orgClass 12 · Chemistry
CodeVID-C12-02-T2-01
Vapour Pressure & Raoult's Law — 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.
Raoult's law for a volatile component states $p_A=$
- A.$p_A^0/x_A$
- B.$p_A^0 x_A$
- C.$x_A/p_A^0$
- D.$p_A^0+x_A$
2.
An ideal solution has $\Delta_{mix}H$ equal to:
- A.positive
- B.negative
- C.zero
- D.infinite
3.
Benzene + toluene is an example of:
- A.positive deviation
- B.negative deviation
- C.an ideal solution
- D.a maximum-boiling azeotrope
4.
Acetone + chloroform shows:
- A.positive deviation
- B.negative deviation
- C.ideal behaviour
- D.no mixing
5.
Azeotropes are mixtures that:
- A.always separate easily
- B.boil at a constant composition
- C.have zero vapour pressure
- D.are always solids
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Raoult's law for a solution of two volatile liquids.
7.
Distinguish positive and negative deviations in terms of intermolecular forces.
8.
Why can an azeotropic mixture not be separated by fractional distillation?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
At 300 K, pure A and B have vapour pressures 100 and 60 mm Hg. For $x_A=0.5$, find $p_{total}$ and the vapour composition $y_A$.
10.
45 g of a non-volatile solute (M = 90) is dissolved in 360 g water. If $p^0_{water}=24$ mm Hg, find the solution's vapour pressure.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Explain ideal and non-ideal solutions with one example each, the sign of $\Delta_{mix}H$, and how each relates to azeotrope formation.
Answer Key
Section A — Multiple Choice Questions
- (B) $p_A^0 x_A$
- (C) zero
- (C) an ideal solution
- (B) negative deviation
- (B) boil at a constant composition
Section B — Short Answer (2 marks)
- The partial vapour pressure of each component equals its pure vapour pressure times its mole fraction: $p_A=p_A^0 x_A$ and $p_B=p_B^0 x_B$.
- Positive deviation: A–B forces weaker than A–A/B–B, vapour pressure higher than ideal. Negative deviation: A–B forces stronger, vapour pressure lower than ideal.
- At the azeotropic composition the liquid and vapour have identical compositions, so distillation produces vapour of the same composition and no further separation occurs.
Section C — Short Answer (3 marks)
- $p_A=100\times0.5=50$; $p_B=60\times0.5=30$; $p_{total}=80$ mm Hg. $y_A=50/80=0.625$.
- $n_{solute}=0.5$; $n_{water}=20$; $x_{water}=20/20.5=0.9756$; $p=24\times0.9756=23.4$ mm Hg.
Section D — Long Answer (5 marks)
- Ideal: obeys Raoult's law throughout, $\Delta_{mix}H=0$ (benzene + toluene), no azeotrope. Positive deviation: $\Delta_{mix}H>0$ (ethanol + water), forms minimum-boiling azeotrope. Negative deviation: $\Delta_{mix}H<0$ (acetone + chloroform / HNO3 + water), forms maximum-boiling azeotrope.
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