The vapour pressures of two liquids A and B i | vedclass

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(A) Given: $\frac{P_A^0}{P_B^0} = \frac{1}{2}$ and $\frac{n_A}{n_B} = \frac{1}{2}$.Mole fraction of $A$ in liquid phase,$x_A = \frac{n_A}{n_A + n_B} =

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$0.2$

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(A) Given: $\frac{P_A^0}{P_B^0} = \frac{1}{2}$ and $\frac{n_A}{n_B} = \frac{1}{2}$.Mole fraction of $A$ in liquid phase,$x_A = \frac{n_A}{n_A + n_B} = \frac{1}{1+2} = \frac{1}{3}$.Mole fraction of $B$ in liquid phase,$x_B = \frac{n_B}{n_A + n_B} = \frac{2}{1+2} = \frac{2}{3}$.According to Raoult's law,partial pressures are:$P_A = x_A P_A^0 = \frac{1}{3} P_A^0$$P_B = x_B P_B^0 = \frac{2}{3} P_B^0 = \frac{2}{3} (2 P_A^0) = \frac{4}{3} P_A^0$Total pressure $P_{total} = P_A + P_B = \frac{1}{3} P_A^0 + \frac{4}{3} P_A^0 = \frac{5}{3} P_A^0$.Mole fraction of $A$ in vapour phase,$y_A = \frac{P_A}{P_{total}} = \frac{\frac{1}{3} P_A^0}{\frac{5}{3} P_A^0} = \frac{1}{5} = 0.2$.

The vapour pressures of two liquids $A$ and $B$ in their pure states are in the ratio of $1:2$. $A$ binary solution of $A$ and $B$ contains $A$ and $B$ in the mole proportion of $1:2$. The mole fraction of $A$ in the vapour phase of the solution will be

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