At 30^circ C,liquid A and B form an ideal sol | vedclass

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(A) According to Raoult's law,the total vapor pressure $P_s = P_A^0 X_A + P_B^0 X_B$.For the first solution: $X_A = 1/3$ and $X_B = 2/3$. Thus,$250 =

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$450, 150 \ mm \ Hg$

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(A) According to Raoult's law,the total vapor pressure $P_s = P_A^0 X_A + P_B^0 X_B$.For the first solution: $X_A = 1/3$ and $X_B = 2/3$. Thus,$250 = P_A^0(1/3) + P_B^0(2/3)$,which simplifies to $P_A^0 + 2P_B^0 = 750$ (Equation $1$).For the second solution: $1 \ mol$ of $A$ is added,so $n_A = 2 \ mol$ and $n_B = 2 \ mol$. Thus,$X_A = 2/4 = 0.5$ and $X_B = 2/4 = 0.5$. The total pressure is $300 = P_A^0(0.5) + P_B^0(0.5)$,which simplifies to $P_A^0 + P_B^0 = 600$ (Equation $2$).Subtracting Equation $2$ from Equation $1$: $(P_A^0 + 2P_B^0) - (P_A^0 + P_B^0) = 750 - 600$,giving $P_B^0 = 150 \ mm \ Hg$.Substituting $P_B^0$ into Equation $2$: $P_A^0 + 150 = 600$,so $P_A^0 = 450 \ mm \ Hg$.

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