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(B) According to Raoult's Law,$P_{total} = P_A^0 X_A + P_B^0 X_B$.For the $1:1$ molar ratio,$X_A = 1/2$ and $X_B = 1/2$:$400 = P_A^0(1/2) + P_B^0(1/2)

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$550, 250$

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(B) According to Raoult's Law,$P_{total} = P_A^0 X_A + P_B^0 X_B$.For the $1:1$ molar ratio,$X_A = 1/2$ and $X_B = 1/2$:$400 = P_A^0(1/2) + P_B^0(1/2) \implies P_A^0 + P_B^0 = 800$ (Equation $1$).For the $1:2$ molar ratio,$X_A = 1/3$ and $X_B = 2/3$:$350 = P_A^0(1/3) + P_B^0(2/3) \implies P_A^0 + 2P_B^0 = 1050$ (Equation $2$).Subtracting Equation $1$ from Equation $2$:$(P_A^0 + 2P_B^0) - (P_A^0 + P_B^0) = 1050 - 800 \implies P_B^0 = 250 \ mm \ Hg$.Substituting $P_B^0 = 250$ into Equation $1$:$P_A^0 + 250 = 800 \implies P_A^0 = 550 \ mm \ Hg$.Thus,the vapor pressures are $550 \ mm \ Hg$ and $250 \ mm \ Hg$.

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