If two substances A and B have P_A^0 : P_B^0 | vedclass

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(D) According to Raoult's Law,the mole fraction of a component in the vapour phase $(y_A)$ is given by the formula:$y_A = \frac{P_A}{P_{total}} = \fra

Vedclass · Accepted Answer

$0.2$

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(D) According to Raoult's Law,the mole fraction of a component in the vapour phase $(y_A)$ is given by the formula:$y_A = \frac{P_A}{P_{total}} = \frac{x_A \cdot P_A^0}{x_A \cdot P_A^0 + x_B \cdot P_B^0}$Given:$P_A^0 : P_B^0 = 1 : 2 \implies P_A^0 = 1, P_B^0 = 2$$x_A : x_B = 1 : 2 \implies x_A = 1, x_B = 2$Substituting these values into the formula:$y_A = \frac{1 	imes 1}{(1 	imes 1) + (2 	imes 2)}$$y_A = \frac{1}{1 + 4} = \frac{1}{5} = 0.2$Therefore,the mole fraction of $A$ in the vapour phase is $0.2$.

If two substances $A$ and $B$ have $P_A^0 : P_B^0 = 1 : 2$ and have mole fraction in solution $1 : 2$,then what is the mole fraction of $A$ in the vapour phase?

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