Two liquids x and y form an ideal solution. A | vedclass

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(C) Let $P_A^0$ and $P_B^0$ be the vapor pressures of pure $x$ and $y$ respectively.For the first solution: $n_x = 1, n_y = 3$. Total moles = $4$.Mole

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$400, 600$

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(C) Let $P_A^0$ and $P_B^0$ be the vapor pressures of pure $x$ and $y$ respectively.For the first solution: $n_x = 1, n_y = 3$. Total moles = $4$.Mole fractions: $X_x = 1/4, X_y = 3/4$.$P_{total} = P_A^0 X_x + P_B^0 X_y \Rightarrow 550 = P_A^0(1/4) + P_B^0(3/4)$.$P_A^0 + 3P_B^0 = 2200$ ... $(1)$For the second solution: $n_x = 1, n_y = 4$. Total moles = $5$.$P_{total} = 550 + 10 = 560 \ mm \ Hg$.Mole fractions: $X_x = 1/5, X_y = 4/5$.$560 = P_A^0(1/5) + P_B^0(4/5)$.$P_A^0 + 4P_B^0 = 2800$ ... $(2)$Subtracting $(1)$ from $(2)$:$(P_A^0 + 4P_B^0) - (P_A^0 + 3P_B^0) = 2800 - 2200$.$P_B^0 = 600 \ mm \ Hg$.Substituting $P_B^0$ in $(1)$:$P_A^0 + 3(600) = 2200 \Rightarrow P_A^0 = 2200 - 1800 = 400 \ mm \ Hg$.Thus,the vapor pressures are $400 \ mm \ Hg$ and $600 \ mm \ Hg$.

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