At $\mathrm{T}(\mathrm{K})$ two liquids A and B form an ideal solution. The vapour pressures of pure liquids…

At $\mathrm{T}(\mathrm{K})$ two liquids A and B form an ideal solution. The vapour pressures of pure liquids A and B at that temperature are 400 and 600 mm Hg respectively. If the mole fraction of liquid $B$ is 0.3 in the mixture, the mole fractions of $A$ and $B$ in vapour phase respectively are
  1. 0.391,0.609
  2. 0.509,0.491
  3. 0.609,0.391
  4. 0.491,0.509

Solution

$\begin{aligned} & \text { } \mathrm{P}_{\mathrm{A}}^0=400 \\ & \mathrm{P}_{\mathrm{B}}^0=600 \\ & \mathrm{x}_{\mathrm{A}}=0.7 \\ & \mathrm{x}_{\mathrm{B}}=0.3 \quad \left[\mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}=1\right]\\\ & \mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^0 \mathrm{x}_{\mathrm{A}} \\ & \mathrm{P}_{\mathrm{A}}=0.7 \times 400=280 \\ & \mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{B}}^0 \mathrm{x}_{\mathrm{B}} \\ & =0.3 \times 600=180 \\ & \mathrm{P}_{\mathrm{T}}=\mathrm{P}_{\mathrm{A}}+\mathrm{P}_{\mathrm{B}}=460 \\ & \left.\therefore \quad \mathrm{In} \mathrm{vapour} \mathrm{phase}^2+\mathrm{x}_{\mathrm{B}}=1\right] \\ & \mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{T}} \mathrm{X}_1 \mathrm{~A} \\ & \mathrm{P}_{\mathrm{B}}=\mathrm{P}_{\mathrm{T}} \mathrm{X}_1 \mathrm{~B} \\ & \therefore \quad \mathrm{X}_1 \mathrm{~A}=\frac{280}{460}=0.61 \\ & \therefore \quad \mathrm{X}_1 \mathrm{~B}=\frac{180}{460}=0.39\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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