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An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to second mixture, the vapour pressure comes down to 1atm. Vapour pressure of C, $P_{C}^{0} = 0.8$ atm. Calculate the vapour pressure of pure A and B.

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$P_{A}^{0} = 1.4 a t m, P_{B}^{0} = 0.7 a t m$

$P_{A}^{0} = 1.2 a t m, P_{B}^{0} = 0.6 a t m$

$P_{A}^{0} = 1.4 a t m, P_{B}^{0} = 0.6 a t m$

$P_{A}^{0} = 0.6 a t m, P_{B}^{0} = 4.4 a t m$

answer is D.

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Detailed Solution

$1 = \frac{1}{2} P_{A}^{0} + \frac{1}{2} P_{B}^{0} 1 < \frac{1}{4} P_{A}^{0} + \frac{3}{4} P_{B}^{0} 1 = \frac{1}{8} P_{A}^{0} + \frac{3}{8} P_{B}^{0} + \frac{4}{8} \times 0.8 \Rightarrow P_{A}^{0} + 3 P_{B}^{0} = 4.8$
Only (d) option satisfies.

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