At 25,^oC,the vapour pressure of pure liquid | vedclass

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(A) Moles of $A = \frac{20}{40} = 0.5 \, mol$Moles of $B = \frac{20}{80} = 0.25 \, mol$Total moles $= 0.5 + 0.25 = 0.75 \, mol$Mole fraction of $A$ $(

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$80$

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(A) Moles of $A = \frac{20}{40} = 0.5 \, mol$Moles of $B = \frac{20}{80} = 0.25 \, mol$Total moles $= 0.5 + 0.25 = 0.75 \, mol$Mole fraction of $A$ $(X_A)$ $= \frac{0.5}{0.75} = \frac{2}{3} \approx 0.667$Mole fraction of $B$ $(X_B)$ $= \frac{0.25}{0.75} = \frac{1}{3} \approx 0.333$According to Raoult's Law,$P_{total} = P_A^{\circ} X_A + P_B^{\circ} X_B$$P_{total} = (100 	imes \frac{2}{3}) + (40 	imes \frac{1}{3}) = \frac{200}{3} + \frac{40}{3} = \frac{240}{3} = 80 \, torr$

At $25\,^oC$,the vapour pressure of pure liquid $A$ (mol. wt. $= 40$) is $100\,torr$,while that of pure liquid $B$ is $40\,torr$ (mol. wt. $= 80$). The vapour pressure at $25\,^oC$ of a solution containing $20\,g$ of each $A$ and $B$ is ........... $torr$.

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