Q18 / 180 · COMEDK 2026 (9 May Shift 2) · Chemistry
When 1 mole of benzene is mixed with 1 mole of toluene, the vapours will contain
- AEqual number of moles of benzene and toluene since it is an ideal solution
- Correct answerBHigher percentage of benzene
- CEqual mass of benzene and toluene since it is an ideal solution
- DHigher percentage of toluene
Solution
Benzene is more volatile than toluene, which means the vapour pressure of pure benzene is greater than that of pure toluene ($P^{\circ}_B > P^{\circ}_T$).
Given that $1$ mole of benzene is mixed with $1$ mole of toluene, the mole fractions in the liquid phase are $x_B = 0.5$ and $x_T = 0.5$.
According to Raoult's law, the partial pressures in the vapour phase are $P_B = x_B P^{\circ}_B$ and $P_T = x_T P^{\circ}_T$.
Since $P^{\circ}_B > P^{\circ}_T$ and $x_B = x_T$, it follows that $P_B > P_T$.
By Dalton's law of partial pressures, the mole fraction in the vapour phase is $y = \dfrac{P}{P_{\text{total}}}$.
Therefore, $y_B > y_T$, meaning the vapour phase will contain a higher percentage of benzene.
Answer: Higher percentage of benzene
Given that $1$ mole of benzene is mixed with $1$ mole of toluene, the mole fractions in the liquid phase are $x_B = 0.5$ and $x_T = 0.5$.
According to Raoult's law, the partial pressures in the vapour phase are $P_B = x_B P^{\circ}_B$ and $P_T = x_T P^{\circ}_T$.
Since $P^{\circ}_B > P^{\circ}_T$ and $x_B = x_T$, it follows that $P_B > P_T$.
By Dalton's law of partial pressures, the mole fraction in the vapour phase is $y = \dfrac{P}{P_{\text{total}}}$.
Therefore, $y_B > y_T$, meaning the vapour phase will contain a higher percentage of benzene.
Answer: Higher percentage of benzene
Hence, the correct answer is option: 'B'.