5 moles of liquid X and 10 moles of liquid Y | vedclass

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(B) $X$$Y$$n_x = 5$$n_y = 10$$P_x^0 = 63 \ torr$$P_y^0 = 78 \ torr$Mole fraction of $X$ $(x_x)$ = $\frac{5}{5+10} = \frac{1}{3}$Mole fraction of $Y$ $

Vedclass · Accepted Answer

The solution shows negative deviation.

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(B) $X$$Y$$n_x = 5$$n_y = 10$$P_x^0 = 63 \ torr$$P_y^0 = 78 \ torr$Mole fraction of $X$ $(x_x)$ = $\frac{5}{5+10} = \frac{1}{3}$Mole fraction of $Y$ $(x_y)$ = $\frac{10}{5+10} = \frac{2}{3}$According to Raoult's Law,the calculated vapour pressure $(P_{s(calc.)})$ is:$P_{s(calc.)} = P_x^0 x_x + P_y^0 x_y$$P_{s(calc.)} = (63 	imes \frac{1}{3}) + (78 	imes \frac{2}{3})$$P_{s(calc.)} = 21 + 52 = 73 \ torr$Given observed vapour pressure $(P_{s(obs.)})$ = $70 \ torr$Since $P_{s(obs.)} < P_{s(calc.)}$,the solution shows negative deviation from Raoult's Law.

$5$ moles of liquid $X$ and $10$ moles of liquid $Y$ make a solution having a vapour pressure of $70 \ torr$. The vapour pressures of pure $X$ and $Y$ are $63 \ torr$ and $78 \ torr$ respectively. Which of the following is true regarding the described solution?

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