A solution contains a 1:4 molar ratio of pent | vedclass

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(B) Let the moles of pentane be $n_A = 1$ and moles of hexane be $n_B = 4$. Total moles $= 1 + 4 = 5$. Mole fraction of pentane in liquid phase,$X_A =

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$0.478$

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(B) Let the moles of pentane be $n_A = 1$ and moles of hexane be $n_B = 4$. Total moles $= 1 + 4 = 5$. Mole fraction of pentane in liquid phase,$X_A = \frac{1}{5} = 0.2$. Mole fraction of hexane in liquid phase,$X_B = \frac{4}{5} = 0.8$. Using Raoult's Law,partial pressure of pentane $P_A = P_A^0 	imes X_A = 440 	imes 0.2 = 88 \ mm \ Hg$. Partial pressure of hexane $P_B = P_B^0 	imes X_B = 120 	imes 0.8 = 96 \ mm \ Hg$. Total pressure $P_{total} = P_A + P_B = 88 + 96 = 184 \ mm \ Hg$. Mole fraction of pentane in vapor phase $(Y_A)$ is given by $Y_A = \frac{P_A}{P_{total}} = \frac{88}{184} \approx 0.478$.

$A$ solution contains a $1:4$ molar ratio of pentane and hexane. At $20^{\circ}C$,the vapor pressure of pure pentane is $440 \ mm \ Hg$ and that of hexane is $120 \ mm \ Hg$. What is the mole fraction of pentane in the vapor phase?

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