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

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(C) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{T} = P^{\circ}_{A} X_{A} + P^{\circ}_{B} X_{B}$.First,calculat

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{T} = P^{\circ}_{A} X_{A} + P^{\circ}_{B} X_{B}$.First,calculate the number of moles of $A$ and $B$:$n_{A} = \frac{20\, g}{40\, g/mol} = 0.5\, mol$$n_{B} = \frac{20\, g}{80\, g/mol} = 0.25\, mol$Next,calculate the mole fractions:$X_{A} = \frac{n_{A}}{n_{A} + n_{B}} = \frac{0.5}{0.5 + 0.25} = \frac{0.5}{0.75} = \frac{2}{3}$$X_{B} = \frac{n_{B}}{n_{A} + n_{B}} = \frac{0.25}{0.75} = \frac{1}{3}$Finally,calculate the total vapour pressure:$P_{T} = (100\, torr 	imes \frac{2}{3}) + (40\, torr 	imes \frac{1}{3})$$P_{T} = \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 $B$ (mol. $wt. = 80$) is $40\, torr$. The vapour pressure at $25\,^{\circ}C$ of a solution containing $20\, g$ of each $A$ and $B$ is .......... $torr$.

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