Two liquids A and B have vapour pressures in | vedclass

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(A) Let $X_A$ and $X_B$ be the mole fractions of $A$ and $B$ in the liquid solution,and $Y_A$ and $Y_B$ be the mole fractions in the vapour phase.Acco

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(A) Let $X_A$ and $X_B$ be the mole fractions of $A$ and $B$ in the liquid solution,and $Y_A$ and $Y_B$ be the mole fractions in the vapour phase.According to Raoult's Law,the partial pressures are $P_A = X_A P_A^0$ and $P_B = X_B P_B^0$.In the vapour phase,$Y_A = \frac{P_A}{P_A + P_B}$ and $Y_B = \frac{P_B}{P_A + P_B}$.Therefore,$\frac{Y_A}{Y_B} = \frac{P_A}{P_B} = \frac{X_A P_A^0}{X_B P_B^0}$.Given $\frac{Y_A}{Y_B} = \frac{4}{3}$ and $\frac{P_A^0}{P_B^0} = \frac{1}{3}$,we have $\frac{4}{3} = \frac{X_A}{X_B} 	imes \frac{1}{3}$.Since $X_A = 1 - X_B$,we substitute: $\frac{4}{3} = \frac{1 - X_B}{X_B} 	imes \frac{1}{3}$.$4 = \frac{1 - X_B}{X_B} \implies 4 X_B = 1 - X_B \implies 5 X_B = 1 \implies X_B = \frac{1}{5}$.

Column $I$	Column $II$
$(a)$ $XX-XO$ method of sex determination	$(i)$ Turner's syndrome
$(b)$ $XX-XY$ method of sex determination	$(ii)$ Female heterogametic
$(c)$ Karyotype-$45$	$(iii)$ Grasshopper
$(d)$ $ZW-ZZ$ method of sex determination	$(iv)$ Female homogametic

Two liquids $A$ and $B$ have vapour pressures in the ratio $P_A^0 : P_B^0 = 1 : 3$ at a certain temperature. If 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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