Raoult's law — calculate vapor pressure of ideal binary solution

Question

A solution contains 2 mol of benzene and 3 mol of toluene. The vapor pressures of pure benzene and toluene are 75 mmHg and 22 mmHg respectively. Assuming ideal behaviour, calculate the total vapor pressure of the solution using Raoult’s law.

(NCERT Class 12, Chapter 2)

Solution — Step by Step

For an ideal solution, the partial vapor pressure of each component equals the product of its mole fraction and its pure vapor pressure:

p_A = x_A \cdot p_A^0 \quad \text{and} \quad p_B = x_B \cdot p_B^0

Total vapor pressure: $p_{total} = p_A + p_B$

Moles of benzene ( $A$ ) = 2, moles of toluene ( $B$ ) = 3. Total moles = 5.

x_A = \frac{2}{5} = 0.4, \quad x_B = \frac{3}{5} = 0.6

p_A = 0.4 \times 75 = 30 \text{ mmHg}

p_B = 0.6 \times 22 = 13.2 \text{ mmHg}

p_{total} = 30 + 13.2 = \mathbf{43.2 \text{ mmHg}}

The composition of vapor can also be found: $y_A = p_A / p_{total} = 30/43.2 = 0.694$ . Benzene is enriched in the vapor phase because it is more volatile.

Why This Works

Raoult’s law works for ideal solutions where intermolecular forces between A-A, B-B, and A-B are nearly equal. Benzene-toluene is a classic example of an ideal solution because both are structurally similar non-polar hydrocarbons.

The more volatile component (higher $p^0$ ) is always enriched in the vapor phase compared to the liquid phase. This principle is the basis of fractional distillation.

Alternative Method — Graphical Approach

Plot $p_{total}$ vs $x_A$ . For an ideal solution, this is a straight line from $p_B^0$ (at $x_A = 0$ ) to $p_A^0$ (at $x_A = 1$ ). At $x_A = 0.4$ : read off the line to get 43.2 mmHg.

JEE/NEET often asks about positive and negative deviations from Raoult’s law. Positive deviation (e.g., ethanol + acetone): A-B interactions weaker than A-A and B-B → actual vapor pressure is higher than Raoult’s prediction. Negative deviation (e.g., chloroform + acetone): A-B interactions stronger → actual vapor pressure is lower.

Common Mistake

Students sometimes apply Raoult’s law using mass fractions instead of mole fractions. Raoult’s law strictly requires mole fractions. If the problem gives masses, always convert to moles first using molar mass before calculating $x_A$ and $x_B$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Graphical Approach

Common Mistake

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