For two volatile liquids A and B,if the vapou | vedclass

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(B) Given: $P_{A}^{0} : P_{B}^{0} = 1 : 2$,so $P_{B}^{0} = 2P_{A}^{0}$.Given: $X_{A} : X_{B} = 1 : 2$,so $X_{B} = 2X_{A}$.Since $X_{A} + X_{B} = 1$,we

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$0.20$

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(B) Given: $P_{A}^{0} : P_{B}^{0} = 1 : 2$,so $P_{B}^{0} = 2P_{A}^{0}$.Given: $X_{A} : X_{B} = 1 : 2$,so $X_{B} = 2X_{A}$.Since $X_{A} + X_{B} = 1$,we have $X_{A} + 2X_{A} = 1$,which gives $3X_{A} = 1$,so $X_{A} = 1/3$ and $X_{B} = 2/3$.According to Raoult's Law,the partial pressures are:$P_{A} = X_{A} P_{A}^{0} = (1/3) P_{A}^{0}$$P_{B} = X_{B} P_{B}^{0} = (2/3) (2P_{A}^{0}) = (4/3) P_{A}^{0}$Total pressure $P_{T} = P_{A} + P_{B} = (1/3) P_{A}^{0} + (4/3) P_{A}^{0} = (5/3) P_{A}^{0}$.The mole fraction of $A$ in the vapour state $(Y_{A})$ is given by $Y_{A} = P_{A} / P_{T}$.$Y_{A} = ((1/3) P_{A}^{0}) / ((5/3) P_{A}^{0}) = 1/5 = 0.20$.

For two volatile liquids $A$ and $B$,if the vapour pressure ratio of $P_{A}^{0} : P_{B}^{0} = 1 : 2$ and the mole fraction ratio in the liquid phase is $X_{A} : X_{B} = 1 : 2$,then find the mole fraction of component $A$ in the vapour state $(Y_{A})$?

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