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 (integer)
314
Solution
Let $V . P$ of pure $A$ be $P_A^0$<br/><br/>
Let $V . P$ of pure $B$ be $P_B^0$<br/><br/>
When $\mathrm{X}_{\mathrm{A}}=0.7 \& \mathrm{X}_{\mathrm{B}}=0.3$<br/><br/>
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{s}}=350 \\\\
& \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.7+\mathrm{P}_{\mathrm{B}}^0 \times 0.3=350
\end{aligned}
$$<br/><br/>
When $\mathrm{X}_{\mathrm{A}}=0.2 \& \mathrm{X}_{\mathrm{B}}=0.8$<br/><br/>
$$
\begin{aligned}
& P_{\mathrm{s}}=410 \\\\
& \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.2+\mathrm{P}_{\mathrm{B}}^0 \times 0.8=410
\end{aligned}
$$<br/><br/>
Solving (i) and (ii)<br/><br/>
$$
\begin{aligned}
& \mathrm{P}_{\mathrm{A}}^0=314 \mathrm{~mm} \mathrm{Hg} \\\\
& \mathrm{P}_{\mathrm{B}}^0=434 \mathrm{~mm} \mathrm{Hg} \\\\
& \therefore \text{Answer} = 314
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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