What is the mole ratio of benzene ({P_B}^0 = | vedclass

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(A) According to Raoult's law,the total vapour pressure $P$ is given by $P = P_{B}^{0} X_{B} + P_{T}^{0} X_{T}$.Given $P = 120 \ torr$,$P_{B}^{0} = 15

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$7 : 1$

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(A) According to Raoult's law,the total vapour pressure $P$ is given by $P = P_{B}^{0} X_{B} + P_{T}^{0} X_{T}$.Given $P = 120 \ torr$,$P_{B}^{0} = 150 \ torr$,and $P_{T}^{0} = 50 \ torr$.Let $X_{B}$ be the mole fraction of benzene in the liquid phase,then $X_{T} = 1 - X_{B}$.$120 = 150(X_{B}) + 50(1 - X_{B})$$120 = 150 X_{B} + 50 - 50 X_{B}$$70 = 100 X_{B} \implies X_{B} = 0.7$.Thus,$X_{T} = 1 - 0.7 = 0.3$.The mole fraction in the vapour phase $(Y)$ is given by $Y_i = \frac{P_i}{P} = \frac{P_i^0 X_i}{P}$.$Y_{B} = \frac{150 	imes 0.7}{120} = \frac{105}{120} = 0.875$.$Y_{T} = \frac{50 	imes 0.3}{120} = \frac{15}{120} = 0.125$.The mole ratio in the vapour phase is $\frac{Y_{B}}{Y_{T}} = \frac{0.875}{0.125} = \frac{7}{1}$.

What is the mole ratio of benzene $({P_B}^0 = 150 \ torr)$ and toluene $({P_T}^0 = 50 \ torr)$ in the vapour phase if the given solution has a vapour pressure of $120 \ torr$?

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