2 moles each of liquids A and B are dissolve | vedclass

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(D) The mole fraction of $A$ $(X_A)$ and $B$ $(X_B)$ in the liquid phase are:$X_A = \frac{2}{2+2} = 0.5$$X_B = \frac{2}{2+2} = 0.5$The total vapour pr

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$\frac{2}{5}$

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(D) The mole fraction of $A$ $(X_A)$ and $B$ $(X_B)$ in the liquid phase are:$X_A = \frac{2}{2+2} = 0.5$$X_B = \frac{2}{2+2} = 0.5$The total vapour pressure of the solution $(P_S)$ is calculated as:$P_S = P_A^o X_A + P_B^o X_B$$P_S = 120 	imes 0.5 + 80 	imes 0.5 = 60 + 40 = 100 \ torr$According to Dalton's Law,the partial pressure of $B$ in the vapour phase is $P_B = P_B^o X_B = P_S Y_B$,where $Y_B$ is the mole fraction of $B$ in the vapour phase:$80 	imes 0.5 = 100 	imes Y_B$$40 = 100 	imes Y_B$$Y_B = \frac{40}{100} = \frac{2}{5}$

$2 \ moles$ each of liquids $A$ and $B$ are dissolved to form an ideal solution. What will be the mole fraction of $B$ in the vapour phase? Given: $P_A^o = 120 \ torr, P_B^o = 80 \ torr$

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