Q.Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Concept: Raoult's Law for Ideal Solutions

For an ideal binary solution, the partial vapor pressures follow Raoult's law: $p_i = x_i p_i^0$, where $x_i$ is the mole fraction in the liquid phase and $p_i^0$ is the pure component vapor pressure. The mole fraction in the vapor phase is given by Dalton's law.

Step 1: Calculate moles of each component.

• Moles of benzene ($\mathrm{C_6H_6}$, $M = 78$ g/mol): $n_{\text{benzene}} = \frac{80}{78} = 1.026$ mol
• Moles of toluene ($\mathrm{C_7H_8}$, $M = 92$ g/mol): $n_{\text{toluene}} = \frac{100}{92} = 1.087$ mol

Step 2: Find mole fraction of benzene in liquid phase.

Step 3: Apply Raoult's law to find partial pressures.

$p_{\text{benzene}} = 0.486 \times 50.71 = 24.65 \text{ mm Hg}$ …

When two liquids form an ideal solution, each component's vapour pressure is simply proportional to its mole fraction in the liquid phase—that's Raoult's law. The vapour above the solution is a mixture of both components, and the composition of that vapour depends on how much each liquid contributes to the total pressure. The key insight is that the more volatile component (higher pure vapour pressure) will be enriched in the vapour relative to the liquid.

We need to find what fraction of the vapour is benzene when we mix specific masses of benzene and toluene.

1. Convert masses to moles

Benzene is $\text{C}_6\text{H}_6$ with molar mass $M_{\text{benzene}} = 6(12) + 6(1) = 78 \text{ g/mol}$.

Toluene is $\text{C}_7\text{H}_8$ with molar mass $M_{\text{toluene}} = 7(12) + 8(1) = 92 \text{ g/mol}$.

2. Calculate mole fractions in the liquid phase

Total moles in solution:

Mole fraction of benzene in liquid:

Mole fraction of toluene in liquid:

3. Apply Raoult's law to find partial pressures

For an ideal solution, the partial pressure of each component is:

4. Find total vapour pressure …

Method: Raoult's Law for an Ideal Binary Solution (Vapour Composition)

This problem asks for the mole fraction of benzene in the vapour phase above an ideal benzene-toluene solution.

Steps

Step 1: Convert masses to moles

Step 2: Find liquid-phase mole fractions

Step 3: Apply Raoult's Law for each component

Step 4: Total vapour pressure (Dalton's Law)

$p_{total}=p_{benzene}+p_{toluene}=24.6+16.5=41.1\,\text{mm Hg}$ …

Common Mistakes in Raoult's Law Application (Ideal Solution Vapour Composition)

Here are the most frequent errors students make on this benzene/toluene vapour-phase mole-fraction problem, with clear explanations of why they happen and how to avoid them.

1. Forgetting to Convert Mass to Moles First

The Mistake: Plugging the given masses (80 g benzene, 100 g toluene) directly into Raoult's Law without converting to moles.

How to avoid: Always convert mass to moles first:

• Molar mass of benzene ($C_6H_6$) = $78\,\text{g/mol}$, so $n_{benzene} = 80/78 = 1.026\,\text{mol}$
• Molar mass of toluene ($C_7H_8$) = $92\,\text{g/mol}$, so $n_{toluene} = 100/92 = 1.087\,\text{mol}$

Rule: Mass -> Moles -> Mole fraction -> Raoult's Law. Never skip step 1.

2. Using Solution Mole Fraction Instead of Vapour-Phase Mole Fraction

The Mistake: Reporting the liquid-phase mole fraction of benzene ($x_{benzene} = 0.486$) as if it were the answer to "mole fraction in the vapour phase."

Why it happens: Confusion between "mole fraction of A in solution" vs "mole fraction of A in vapour."

How to avoid:

• In solution: $x_A = \dfrac{n_A}{n_A+n_B}$
• In vapour: $y_A = \dfrac{p_A}{p_{total}}$ (Dalton's Law)

3. Confusing Partial Pressure with Pure Vapour Pressure

The Mistake: Writing $p_{benzene} = 50.71\,\text{mm Hg}$ (the pure vapour pressure) instead of $p_{benzene} = x_{benzene}\times 50.71$.

How to avoid: Always write the full form: $p_A = x_A\cdot p_A^\circ$. Here $p_{benzene}=0.486\times50.71=24.6\,\text{mm Hg}$ and $p_{toluene}=0.514\times32.06=16.5\,\text{mm Hg}$.

4. Forgetting to Add Partial Pressures for Total Vapour Pressure

The Mistake: Stopping after one partial pressure instead of $p_{total}=p_{benzene}+p_{toluene}=24.6+16.5=41.1\,\text{mm Hg}$. …