Two liquids A and B have P_A^o and P_B^o in t | vedclass

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(A) Given: $\frac{P_A^o}{P_B^o} = \frac{1}{3}$ and $\frac{n_A}{n_B} = \frac{1}{3}$.Calculate the mole fraction of $A$ in the liquid phase $(x_A)$:$x_A

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$0.1$

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(A) Given: $\frac{P_A^o}{P_B^o} = \frac{1}{3}$ and $\frac{n_A}{n_B} = \frac{1}{3}$.Calculate the mole fraction of $A$ in the liquid phase $(x_A)$:$x_A = \frac{n_A}{n_A + n_B} = \frac{1}{1 + 3} = \frac{1}{4} = 0.25$.Consequently,$x_B = 1 - x_A = 1 - 0.25 = 0.75$.Using Raoult's Law,the partial pressures are:$P_A = P_A^o \cdot x_A$$P_B = P_B^o \cdot x_B = (3 P_A^o) \cdot (0.75) = 2.25 P_A^o$.The total pressure $(P_{total})$ is:$P_{total} = P_A + P_B = P_A^o(0.25) + 2.25 P_A^o = 2.5 P_A^o$.The mole fraction of $A$ in the vapour phase $(y_A)$ is given by:$y_A = \frac{P_A}{P_{total}} = \frac{P_A^o \cdot 0.25}{2.5 P_A^o} = \frac{0.25}{2.5} = 0.1$.

Two liquids $A$ and $B$ have $P_A^o$ and $P_B^o$ in the ratio of $1 : 3$ and the ratio of number of moles of $A$ and $B$ in the liquid phase is $1 : 3$. The mole fraction of $A$ in the vapour phase in equilibrium with the solution is equal to:

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