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(C) According to Raoult's law,the total pressure $P$ of a binary mixture is given by $P = P_A^0 X_A + P_B^0 X_B$.Since $X_A + X_B = 1$,we have $X_B =

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$3 : 1$

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(C) According to Raoult's law,the total pressure $P$ of a binary mixture is given by $P = P_A^0 X_A + P_B^0 X_B$.Since $X_A + X_B = 1$,we have $X_B = 1 - X_A$.Substituting this into the equation: $P = P_A^0 X_A + P_B^0 (1 - X_A) = (P_A^0 - P_B^0) X_A + P_B^0$.Comparing this with the given equation $P = 180 X_A + 90$,we get:$P_B^0 = 90 \, mm \, Hg$ (when $X_A = 0$)$P_A^0 - P_B^0 = 180 \implies P_A^0 = 180 + 90 = 270 \, mm \, Hg$.The ratio of vapour pressures of pure liquids $A$ and $B$ is $P_A^0 : P_B^0 = 270 : 90 = 3 : 1$.

If the total vapour pressure of the liquid mixture $A$ and $B$ is given by the equation: $P = 180X_A + 90 \, mm \, Hg$,then the ratio of the vapour pressure of the pure liquids $A$ and $B$ is given by:

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