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The molar volume of $X (l) (d = 0.9 g / m L)$ increases by a factor of 3000 as it vaporizes at $27^{o} C$ and the $Y (l) (d = 0.88 g / m L)$ increases by a factor of 8000 at $27^{o} C .$ A miscible liquid solution of X and Y at $27^{o} C$ has a vapour pressure of 50 torr. The mole fraction of Y in solution is : (Given: R = 0.082 atm/L/mol/K; Molar mass of X = 75; Molar mass of Y = 88)

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0.48

0.52

0.62

0.247

answer is A.

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Detailed Solution

$V_{m} (X, g) = \frac{75}{0.90} \times \frac{3000}{1000} = 250 l i t r e$

By $P V = n R T; P_{X}^{o} = 74.78 t o r r a n d P_{Y}^{o} = 23.37 t o r r$

$V_{m} (Y, g) = \frac{88}{0.88} \times \frac{8000}{1000} = 800 l i t r e$

$P = P_{X}^{o} X_{X} + P_{Y}^{o} P_{Y}$

$P = P_{X}^{o} + (P_{X}^{o} - P_{Y}^{o}) X_{Y} ≃ 0.48$

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