Answer the following. (a) What type of deviation from Roult's Law is expected when phenol and aniline are mixed with each other? What change in the net volume of the mixture is expected? Graphically represent the deviation. (b) The vapour pressure of pure water at a certain temperature is 23.80 mm Hg. If 1 mole of a non-volatile non-electrolytic solute is dissolved in 100g water, Calculate the resultant vapour pressure of the solution.

ChemistryCBSE Class 12 · Sample paper, session 2023-24 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2023-24: question 32(b)

From the official CBSE sample question paper for 2023-24, section E, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerSolutions

Answer the following.

(a)
What type of deviation from Roult's Law is expected when phenol and aniline are mixed with each other? What change in the net volume of the mixture is expected? Graphically represent the deviation.
0 marks
(b)
The vapour pressure of pure water at a certain temperature is 23.80 mm Hg. If 1 mole of a non-volatile non-electrolytic solute is dissolved in 100g water, Calculate the resultant vapour pressure of the solution.
0 marks

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.

Hints

Open one hint at a time and try again before the next.

Hint 1

Compare the attraction between phenol and aniline molecules with that in the pure liquids.

Hint 2

Stronger attractions on mixing mean molecules pack closer — what happens to volume?

Hint 3

For (b), find moles of water from 100 g, then the mole fraction of solute.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(a) Phenol and aniline show negative deviation from Raoult's law. Strong hydrogen bonding between phenol's –OH and aniline's –NH₂ makes the A–B attractions stronger than the A–A and B–B attractions, so escaping tendency decreases.
1 mark
The net volume decreases ( $Δ V_{mi x} < 0$ , and also $Δ H_{mi x} < 0$ ). Graph: plot vapour pressure against mole fraction (x₁ = 1 to x₂ = 1). The total vapour pressure curve lies below the straight dashed line of Raoult's law and passes through a minimum. The partial pressure curves of each component also lie below their ideal lines.
1 mark
(b) Relative lowering of vapour pressure: $\frac{p ^{\circ} - p}{p ^{\circ}} = x_{2} = \frac{n _{2}}{n _{1} + n _{2}}$
½ mark
$n_{1} = \frac{100}{18} = 5.556$ mol, $n_{2} = 1$ mol, so $x_{2} = \frac{1}{5.556 + 1} = 0.1525$
½ mark
$\frac{23.80 - p}{23.80} = 0.1525$
½ mark
$23.80 - p = 3.63$
½ mark
$p = 23.80 - 3.63 = 20.17$ mm Hg
1 mark

Answer: (a) Negative deviation; volume decreases (ΔV < 0); total vapour pressure curve below the Raoult line with a minimum; (b) p ≈ 20.17 mm Hg

Where the marks are

(a) Negative deviation with the stronger phenol–aniline H-bonding reason (1 mark)
(a) ΔV < 0 and the vapour pressure curve below the ideal line (1 mark)
(b) Relative lowering formula (½ mark)
(b) Mole fraction of solute (½ mark)
(b) Substitution (½ mark) and finding p° – p (½ mark)
(b) Final vapour pressure (1 mark)

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CBSE Class 12 Chemistry sample paper 2023-24: question 32(b)

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