Two liquids A and B have vapour pressure in t | vedclass

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(A) Given the ratio of vapour pressures $P_A^o : P_B^o = 1 : 3$ and the ratio of mole fractions in the vapour phase $y_A : y_B = 4 : 3$. According to

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$\frac{1}{5}$

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(A) Given the ratio of vapour pressures $P_A^o : P_B^o = 1 : 3$ and the ratio of mole fractions in the vapour phase $y_A : y_B = 4 : 3$. According to Raoult's law and Dalton's law,the mole fraction in the vapour phase is given by $y_A = \frac{P_A}{P_{total}} = \frac{x_A P_A^o}{P_{total}}$ and $y_B = \frac{P_B}{P_{total}} = \frac{x_B P_B^o}{P_{total}}$. Therefore,$\frac{y_A}{y_B} = \frac{x_A P_A^o}{x_B P_B^o}$. Substituting the given values: $\frac{4}{3} = \frac{x_A}{x_B} 	imes \frac{1}{3}$. This simplifies to $\frac{x_A}{x_B} = 4$,which means $x_A = 4x_B$. Since the sum of mole fractions in the solution is $x_A + x_B = 1$,we substitute $x_A$: $4x_B + x_B = 1$. $5x_B = 1$,so $x_B = \frac{1}{5}$.

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^o : P_B^o = 1 : 3$ at a certain temperature. Assume $A$ and $B$ form an ideal solution and the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4 : 3$. Then the mole fraction of $B$ in the solution at the same temperature is

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