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The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600$ mmHg, respectively at $298 K.$ On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquids $B$ is $0.5$ in the mixture. The vapour pressure of the final solution, the mole fractions of components $A$ and $B$ in vapour phase, respectively are :

Option 1)
$450\; mmHg,0.5,0.5$
Option 2)
$450\; mmHg,0.4,0.6$
Option 3)
$500\; mmHg,0.4,0.6$

Option 4)
$500\; mmHg,0.5,0.5$

Answers (1)

best_answer

$P_{A}^{o}=400\; mmHg\; \; \; \; \; \; \; P_{B}^{o}=600\; mmHg$

$X_{B}=0.5\; \; \; \; \; X_{A}=0.5$

$\rightarrow P_{s}=X_{A}P_{A}^{o}+X_{B}P_{B}^{o}=0.5\times400+0.5\times600=500\; mmHg$

$P_{A}=x_{A}P_{A}^{o}=Y_{A}P_{s}^{o}(for \; vapour\; phase)$

$\downarrow$

mole fraction of A in vapour phase

$Y_{A}=\frac{P_{A}}{P_{S}}=\frac{x_{A}P_{A}^{o}}{P_{s}}=\frac{0.5\times400}{500}=0.4$

$Y_{B}=1-0.4=0.6$

$P_{s}=500\; mmHg\; \; \; Y_{A}=0.4\; \; \; Y_{B}=0.6$

Option 1)

$450\; mmHg,0.5,0.5$

Option 2)

$450\; mmHg,0.4,0.6$

Option 3)
$500\; mmHg,0.4,0.6$

Option 4)

$500\; mmHg,0.5,0.5$

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