JEE MAIN - Chemistry (2023 - 24th January Evening Shift - No. 21)
The total pressure observed by mixing two liquids A and B is 350 mm Hg when their mole fractions are 0.7 and 0.3 respectively.
The total pressure becomes 410 mm Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is __________ mm Hg. (Nearest integer)
Consider the liquids and solutions behave ideally.
Answer
314
Explanation
Let $V . P$ of pure $A$ be $P_A^0$
Let $V . P$ of pure $B$ be $P_B^0$
When $\mathrm{X}_{\mathrm{A}}=0.7 \& \mathrm{X}_{\mathrm{B}}=0.3$
$$ \begin{aligned} & \mathrm{P}_{\mathrm{s}}=350 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.7+\mathrm{P}_{\mathrm{B}}^0 \times 0.3=350 \end{aligned} $$
When $\mathrm{X}_{\mathrm{A}}=0.2 \& \mathrm{X}_{\mathrm{B}}=0.8$
$$ \begin{aligned} & P_{\mathrm{s}}=410 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.2+\mathrm{P}_{\mathrm{B}}^0 \times 0.8=410 \end{aligned} $$
Solving (i) and (ii)
$$ \begin{aligned} & \mathrm{P}_{\mathrm{A}}^0=314 \mathrm{~mm} \mathrm{Hg} \\\\ & \mathrm{P}_{\mathrm{B}}^0=434 \mathrm{~mm} \mathrm{Hg} \\\\ & \therefore \text{Answer} = 314 \end{aligned} $$
Let $V . P$ of pure $B$ be $P_B^0$
When $\mathrm{X}_{\mathrm{A}}=0.7 \& \mathrm{X}_{\mathrm{B}}=0.3$
$$ \begin{aligned} & \mathrm{P}_{\mathrm{s}}=350 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.7+\mathrm{P}_{\mathrm{B}}^0 \times 0.3=350 \end{aligned} $$
When $\mathrm{X}_{\mathrm{A}}=0.2 \& \mathrm{X}_{\mathrm{B}}=0.8$
$$ \begin{aligned} & P_{\mathrm{s}}=410 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.2+\mathrm{P}_{\mathrm{B}}^0 \times 0.8=410 \end{aligned} $$
Solving (i) and (ii)
$$ \begin{aligned} & \mathrm{P}_{\mathrm{A}}^0=314 \mathrm{~mm} \mathrm{Hg} \\\\ & \mathrm{P}_{\mathrm{B}}^0=434 \mathrm{~mm} \mathrm{Hg} \\\\ & \therefore \text{Answer} = 314 \end{aligned} $$
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