Liquids A and B form an ideal solution. At 30 | vedclass

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(C) Let the vapour pressure of pure $A$ be $P_A^o$ and pure $B$ be $P_B^o$.In the first solution,the mole fraction of $A$ is $x_A = \frac{1}{1+2} = \f

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$450, 150\,mm\,Hg$

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(C) Let the vapour pressure of pure $A$ be $P_A^o$ and pure $B$ be $P_B^o$.In the first solution,the mole fraction of $A$ is $x_A = \frac{1}{1+2} = \frac{1}{3}$ and the mole fraction of $B$ is $x_B = \frac{2}{1+2} = \frac{2}{3}$.According to Raoult's law,the total vapour pressure is $P_{total} = P_A^o x_A + P_B^o x_B$.$250 = \frac{1}{3} P_A^o + \frac{2}{3} P_B^o \implies P_A^o + 2P_B^o = 750$ . . . $(i)$In the second solution,$1\,mol$ of $A$ is added,so total moles of $A = 2$ and $B = 2$.The mole fraction of $A$ is $x_A = \frac{2}{2+2} = \frac{1}{2}$ and the mole fraction of $B$ is $x_B = \frac{2}{2+2} = \frac{1}{2}$.$300 = \frac{1}{2} P_A^o + \frac{1}{2} P_B^o \implies P_A^o + P_B^o = 600$ . . . $(ii)$Subtracting equation $(ii)$ from equation $(i)$:$(P_A^o + 2P_B^o) - (P_A^o + P_B^o) = 750 - 600$$P_B^o = 150\,mm\,Hg$Substituting $P_B^o = 150$ into equation $(ii)$:$P_A^o + 150 = 600 \implies P_A^o = 450\,mm\,Hg$.Thus,the vapour pressures of pure $A$ and $B$ are $450\,mm\,Hg$ and $150\,mm\,Hg$ respectively.

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