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(B) $1$. Calculate the moles of benzene $(A)$ and chlorobenzene $(B)$: $n_A = \frac{1560}{78} = 20 \ mol$ $n_B = \frac{1125}{112.5} = 10 \ mol$ $2$. C

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$100$

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(B) $1$. Calculate the moles of benzene $(A)$ and chlorobenzene $(B)$: $n_A = \frac{1560}{78} = 20 \ mol$ $n_B = \frac{1125}{112.5} = 10 \ mol$ $2$. Calculate the mole fractions: $X_A = \frac{20}{20+10} = \frac{2}{3}$ $X_B = \frac{10}{20+10} = \frac{1}{3}$ $3$. The boiling point is the temperature where the total vapor pressure $P_s = P_A^0 X_A + P_B^0 X_B = 1000 \ torr$. $4$. Test $100 \ ^{\circ}C$: From the graph,$P_A^0 = 1350 \ torr$ and $P_B^0 = 300 \ torr$. $P_s = (1350 	imes \frac{2}{3}) + (300 	imes \frac{1}{3}) = 900 + 100 = 1000 \ torr$. Since $P_s = 1000 \ torr$ at $100 \ ^{\circ}C$,the boiling point is $100 \ ^{\circ}C$.

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