Liquids A and B form an ideal solution for al | vedclass

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(C) For an ideal solution,the total vapor pressure is given by Raoult's Law: $P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}$.Since $X_{B} = 1 - X_

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(C) For an ideal solution,the total vapor pressure is given by Raoult's Law: $P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}$.Since $X_{B} = 1 - X_{A}$,we have $P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} (1 - X_{A})$.For the first solution: $0.3 = P_{A}^{\circ} (0.25) + P_{B}^{\circ} (0.75) \implies 0.3 = 0.25 P_{A}^{\circ} + 0.75 P_{B}^{\circ} \implies 1.2 = P_{A}^{\circ} + 3 P_{B}^{\circ} \quad (I)$.For the second solution: $0.4 = P_{A}^{\circ} (0.50) + P_{B}^{\circ} (0.50) \implies 0.4 = 0.5 P_{A}^{\circ} + 0.5 P_{B}^{\circ} \implies 0.8 = P_{A}^{\circ} + P_{B}^{\circ} \quad (II)$.Subtracting equation $(II)$ from equation $(I)$: $(P_{A}^{\circ} + 3 P_{B}^{\circ}) - (P_{A}^{\circ} + P_{B}^{\circ}) = 1.2 - 0.8 \implies 2 P_{B}^{\circ} = 0.4 \implies P_{B}^{\circ} = 0.2 \ bar$.

Liquids $A$ and $B$ form an ideal solution for all compositions of $A$ and $B$ at $25^{\circ} C$. Two such solutions with $0.25$ and $0.50$ mole fractions of $A$ have total vapor pressures of $0.3 \ bar$ and $0.4 \ bar$,respectively. What is the vapor pressure of pure liquid $B$ in $bar$?

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