Q22 / 160 · TS EAMCET 2024 (11 May Shift 1) · Chemistry

At 300 K , the vapour pressure of toluene and benzene are 3.63 kPa and 9.7 kPa respectively. What is the composition of vapour in equilibrium with the solution containing 0.4 mole fraction of toluene? (Assume the solution is ideal)

  • A
    $0.40$
  • B
    $0.60$
  • C
    $0.80$
  • Correct answer
    D
    $0.20$

Solution

$\begin{aligned} & \text { } \mathrm{P}_{\mathrm{T}}^0=3.63 \mathrm{kPa} \quad \mathrm{X}_{\mathrm{A}}=0.4 \quad\left[\mathrm{X}_{\mathrm{A}}+\mathrm{X}_{\mathrm{B}}=1\right] \\ & \mathrm{P}_{\mathrm{B}}=9.7 \mathrm{kPa} \quad \mathrm{X}_{\mathrm{A}}=0.6 \\ & \therefore \mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}^0+\mathrm{P}_{\mathrm{B}}^0 \\ & =\mathrm{P}_{\mathrm{T}}^0 \mathrm{X}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}^0 \mathrm{X}_{\mathrm{B}} \Rightarrow 3.63 \times 0.4+9.7 \times 0.6 \end{aligned}$ $\begin{aligned} & =1.452+5.82 \\ & \mathrm{P}_{\mathrm{T}}=7.272 \end{aligned}$ $\therefore$ Composition of toluene $\left(\mathrm{X}_{\mathrm{T}}\right)=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{T}}}$ $\begin{aligned} \mathrm{X}_{\mathrm{T}} & =\frac{1.452}{7.272} \\ \mathrm{X}_{\mathrm{T}} & =0.1996 \\ \mathrm{X}_{\mathrm{T}} & \approx 0.20 \end{aligned}$

Hence, the correct answer is option: 'D'.

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