A solution has a 1:4 mole ratio of pentane to | vedclass

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(D) Given mole ratio of pentane $(n_p)$ to hexane $(n_h)$ is $1:4$. Mole fraction of pentane $(x_p)$ = $\frac{1}{1+4} = 0.2$. Mole fraction of hexane

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$0.478$

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(D) Given mole ratio of pentane $(n_p)$ to hexane $(n_h)$ is $1:4$. Mole fraction of pentane $(x_p)$ = $\frac{1}{1+4} = 0.2$. Mole fraction of hexane $(x_h)$ = $\frac{4}{1+4} = 0.8$. Total vapour pressure $(P_T)$ = $P_p^0 x_p + P_h^0 x_h$. $P_T = (440 	imes 0.2) + (120 	imes 0.8) = 88 + 96 = 184 \ mm \ Hg$. According to Dalton's law,the mole fraction of pentane in the vapour phase $(y_p)$ is given by $y_p = \frac{P_p^0 x_p}{P_T}$. $y_p = \frac{88}{184} \approx 0.478$.

$A$ solution has a $1:4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $20 \ ^oC$ are $440 \ mm \ Hg$ for pentane and $120 \ mm \ Hg$ for hexane. The mole fraction of pentane in the vapour phase would be

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