(A) Let the vapour pressure of pure $A$ be $P_{A}^0$ and pure $B$ be $P_{B}^0$.According to Raoult's law,$P_{total} = X_{A}P_{A}^0 + X_{B}P_{B}^0$.For

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(A) Let the vapour pressure of pure $A$ be $P_{A}^0$ and pure $B$ be $P_{B}^0$.According to Raoult's law,$P_{total} = X_{A}P_{A}^0 + X_{B}P_{B}^0$.For the first case: $0.7P_{A}^0 + 0.3P_{B}^0 = 350$ $(i)$For the second case: $0.2P_{A}^0 + 0.8P_{B}^0 = 410$ $(ii)$Multiply equation $(i)$ by $8$ and equation $(ii)$ by $3$:$5.6P_{A}^0 + 2.4P_{B}^0 = 2800$ $(iii)$$0.6P_{A}^0 + 2.4P_{B}^0 = 1230$ $(iv)$Subtracting $(iv)$ from $(iii)$:$5.0P_{A}^0 = 1570$$P_{A}^0 = 314 \ mm \ Hg$.

Class 12 Chemistry Solutions Ideal and Non-ideal solution

Medium

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.

JEE MAIN 2023 All JEE MAIN

A
$314$
B
$312$
C
$311$
D
$310$

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