What is the mole ratio of benzene (P_B^o = 15 | vedclass

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(A) Given $P_B^o = 150\, torr$,$P_T^o = 50\, torr$,and $P_{total} = 120\, torr$. Using Raoult's law: $P_{total} = P_B^o X_B + P_T^o X_T$. $120 = 150 X

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$7 : 1$

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(A) Given $P_B^o = 150\, torr$,$P_T^o = 50\, torr$,and $P_{total} = 120\, torr$. Using Raoult's law: $P_{total} = P_B^o X_B + P_T^o X_T$. $120 = 150 X_B + 50(1 - X_B)$. $120 = 150 X_B + 50 - 50 X_B$. $70 = 100 X_B \Rightarrow X_B = 0.7$ and $X_T = 0.3$. Mole fraction in vapour phase: $Y_B = \frac{P_B^o X_B}{P_{total}} = \frac{150 	imes 0.7}{120} = \frac{105}{120} = \frac{7}{8}$ and $Y_T = \frac{P_T^o X_T}{P_{total}} = \frac{50 	imes 0.3}{120} = \frac{15}{120} = \frac{1}{8}$. Mole ratio in vapour phase $Y_B : Y_T = 7 : 1$.

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

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