P = P°ₐxₐ + P°ᵦxᵦ. Calculate VP for benzene-toluene mixture.

Raoult's Law — Vapour Pressure of Ideal Solution

Question

A solution is prepared by mixing 2 moles of benzene (P° = 150 mmHg) and 3 moles of toluene (P° = 60 mmHg) at 25°C. Assuming ideal behaviour, calculate:

The mole fractions of benzene and toluene
The vapour pressure of each component above the solution
The total vapour pressure of the solution

(JEE Main 2024 — similar pattern)

Solution — Step by Step

Total moles = 2 + 3 = 5 mol.

x_A (\text{benzene}) = \frac{2}{5} = 0.4 \qquad x_B (\text{toluene}) = \frac{3}{5} = 0.6

Quick check: $x_A + x_B = 1$ ✓

Raoult’s Law says each component contributes partial pressure proportional to its mole fraction:

P_A = P°_A \cdot x_A = 150 \times 0.4 = 60 \text{ mmHg}

P_B = P°_B \cdot x_B = 60 \times 0.6 = 36 \text{ mmHg}

The more volatile component (benzene, higher P°) still dominates even though it has fewer moles — this is the key insight.

For an ideal solution, total vapour pressure is simply the sum of partial pressures:

P_{total} = P_A + P_B = 60 + 36 = \mathbf{96 \text{ mmHg}}

We can also write this directly as:

P_{total} = P°_A x_A + P°_B x_B = (150)(0.4) + (60)(0.6) = 96 \text{ mmHg}

Why This Works

Raoult’s Law rests on one physical idea: in an ideal solution, solute and solvent interactions are identical to pure-component interactions. No energy is released or absorbed on mixing. This means each molecule’s escaping tendency (fugacity) is simply scaled down by how “crowded out” it is — i.e., by its mole fraction.

P_{total} = \sum_i P°_i \cdot x_i

Valid only for ideal solutions (similar molecular size, similar intermolecular forces). Benzene–toluene is the textbook ideal pair because both are non-polar aromatic hydrocarbons.

The total vapour pressure always lies between the two pure vapour pressures ( $P°_B = 60$ and $P°_A = 150$ ). If your answer falls outside this range, something went wrong — use this as a sanity check in the exam.

Alternative Method — Composition of Vapour Phase

JEE sometimes asks for the mole fraction in the vapour phase (not the liquid). This requires one extra step using Dalton’s Law:

y_A = \frac{P_A}{P_{total}} = \frac{60}{96} = 0.625

y_B = \frac{P_B}{P_{total}} = \frac{36}{96} = 0.375

Notice $y_A = 0.625 > x_A = 0.4$ . The vapour is richer in the more volatile component — this is the entire basis of fractional distillation. If a JEE problem asks why distillation works, this inequality is your one-line answer.

When the question says “ideal solution” and gives two vapour pressures — it’s a Raoult’s Law problem. When it says “non-ideal” or mentions “azeotrope” — Raoult’s Law breaks down, and you’ll need Henry’s Law or deviation analysis instead.

Common Mistake

Using mole % as mole fraction. Students calculate $x_A = 40\%$ and then plug $40$ into the formula instead of $0.40$ . This gives $P_A = 150 \times 40 = 6000$ mmHg — a physically impossible answer. Mole fraction is always a number between 0 and 1. If it’s greater than 1, you’ve made this error.

A second trap: mixing up P° (pure component VP) with P (partial VP after mixing). Raoult’s Law calculates P from P°. Both values appear in problems, and swapping them will flip your logic entirely. In this question, P°(benzene) = 150 is given; P(benzene) = 60 is what we calculated.

Question

Solution — Step by Step

Why This Works

Alternative Method — Composition of Vapour Phase

Common Mistake

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