Liquids A and B form an ideal solution. The v | vedclass

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(D) Given: $P_A^0 = 50 \ mm \ Hg$, $P_B^0 = 32 \ mm \ Hg$. Mole fraction in liquid phase: $X_A = \frac{1}{1+1} = 0.5$ and $X_B = \frac{1}{1+1} = 0.5$.

Vedclass · Accepted Answer

$0.61$

Vedclass · Answer

(D) Given: $P_A^0 = 50 \ mm \ Hg$, $P_B^0 = 32 \ mm \ Hg$. Mole fraction in liquid phase: $X_A = \frac{1}{1+1} = 0.5$ and $X_B = \frac{1}{1+1} = 0.5$. Partial pressures: $P_A = P_A^0 	imes X_A = 50 	imes 0.5 = 25 \ mm \ Hg$ and $P_B = P_B^0 	imes X_B = 32 	imes 0.5 = 16 \ mm \ Hg$. Total pressure: $P_{total} = P_A + P_B = 25 + 16 = 41 \ mm \ Hg$. Mole fraction of $A$ in vapour phase $(Y_A)$: $Y_A = \frac{P_A}{P_{total}} = \frac{25}{41} \approx 0.6097 \approx 0.61$.

Liquids $A$ and $B$ form an ideal solution. The vapour pressure of pure $A$ and $B$ are $50 \ mm \ Hg$ and $32 \ mm \ Hg$ respectively at $300 \ K$. One mole of liquid $A$ is mixed with $1$ mole of liquid $B$. What is the approximate mole fraction of $A$ in the vapour phase?

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