Total vapour pressure of mixture of 1 mol A(pA0=150 torr) and 2 mol B(pB0=240 torr) is 200 torr. In this case

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there is positive deviation from Raoult's law

there is negative deviation from Raoult's law

there is no deviation from Raoult's law

molecular masses of A and B are also required for calculating the deviation

answer is B.

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Detailed Solution

Ps=PA0XA+PB0XB =(150×13)+(240×23) =50 +160 =210ΔPmix=Pexp-Ptheo=-ve