Consider a binary mixture of volatile liquids. If at XA = 0.4 the Vapour pressure of solution is 580 torr then the mixture could be (PA0 = 300 torr, PB0 = 800 torr).

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Consider a binary mixture of volatile liquids. If at X_A = 0.4 the Vapour pressure of solution is 580 torr then the mixture could be ( $P_{A}^{0}$ = 300 torr, $P_{B}^{0}$ = 800 torr).

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C₆H₆ & C₆H₅CH₃

nC₆H₁₄ & nC₇H₁₆

C₆H₅Cl & C₆H₅Br

CHCl₃ & CH₃COCH₃

answer is A.

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Detailed Solution

$\begin{array}{l} P_{Cal} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B} \\ = 300 \times 0 .4 + 800 \times 0 .6 = 600 mm \end{array}$
So, $P_{obs} = 580 ∴ P_{obs} < P_{Cal}$
Sol. shows (-ve) deviation from Raoult’s law mixture of CHCl₃ & CH₃COCH₃ shows (-ve) deviation