At 25,^{circ}C,the vapour pressure of pure li | vedclass

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(A) Step $1$: Calculate the number of moles of $A$ and $B$.$n_A = \frac{20 \, g}{40 \, g/mol} = 0.5 \, mol$$n_B = \frac{20 \, g}{80 \, g/mol} = 0.25 \

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$80$

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(A) Step $1$: Calculate the number of moles of $A$ and $B$.$n_A = \frac{20 \, g}{40 \, g/mol} = 0.5 \, mol$$n_B = \frac{20 \, g}{80 \, g/mol} = 0.25 \, mol$Step $2$: Calculate the mole fractions of $A$ and $B$.$X_A = \frac{n_A}{n_A + n_B} = \frac{0.5}{0.5 + 0.25} = \frac{0.5}{0.75} = \frac{2}{3} \approx 0.667$$X_B = \frac{n_B}{n_A + n_B} = \frac{0.25}{0.5 + 0.25} = \frac{0.25}{0.75} = \frac{1}{3} \approx 0.333$Step $3$: Calculate the total vapour pressure using Raoult's Law.$P_{	ext{total}} = P_A^{\circ} X_A + P_B^{\circ} X_B$$P_{	ext{total}} = (100 \, torr 	imes \frac{2}{3}) + (40 \, torr 	imes \frac{1}{3})$$P_{	ext{total}} = \frac{200}{3} + \frac{40}{3} = \frac{240}{3} = 80 \, torr$

At $25\,^{\circ}C$,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 $25\,^{\circ}C$ of a solution containing $20\, g$ of each $A$ and $B$ is .......... $torr$.

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