If vapour pressures of pure liquids A and B a | vedclass

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(B) According to Raoult's law,the theoretical vapour pressure $P_{calc}$ is given by: $P_{calc} = X_A P_A^o + X_B P_B^o$ $P_{calc} = (0.08 	imes 300)

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$\Delta H_{mix} < 0$

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(B) According to Raoult's law,the theoretical vapour pressure $P_{calc}$ is given by: $P_{calc} = X_A P_A^o + X_B P_B^o$ $P_{calc} = (0.08 	imes 300) + (0.92 	imes 800) \ torr$ $P_{calc} = 24 + 736 = 760 \ torr = 1 \ atm$. Since the observed total vapour pressure $P_{exp} = 0.95 \ atm$ is less than the calculated pressure $P_{calc} = 1 \ atm$,the solution shows negative deviation from Raoult's law. For solutions showing negative deviation,the intermolecular forces between $A-B$ are stronger than those between $A-A$ and $B-B$. Consequently,$\Delta H_{mix} < 0$ and $\Delta V_{mix} < 0$.

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