Nitrobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2},\) and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) are completely miscible in each other. Other properties of the two liquids are nitrobenzene: \(\mathrm{fp}=5.7^{\circ} \mathrm{C}, K_{\mathrm{f}}=\) \(8.1^{\circ} \mathrm{C} m^{-1} ;\) benzene: \(\mathrm{fp}=5.5^{\circ} \mathrm{C}, K_{\mathrm{f}}=5.12^{\circ} \mathrm{C} m^{-1} . \mathrm{It}\) is possible to prepare two different solutions with these two liquids having a freezing point of \(0.0^{\circ} \mathrm{C}\) What are the compositions of these two solutions, expressed as mass percent nitrobenzene?

Since the freezing point of both solutions is $0.0^{\circ}C$, the depression of freezing point \(\Delta T_{f}\) for nitrobenzene is \(5.7^{\circ}C - 0.0^{\circ}C = 5.7^{\circ}C\) and for benzene is \(5.5^{\circ}C - 0.0^{\circ}C = 5.5^{\circ}C.\)

Using the freezing point depression formula: \(\Delta T_{f} = K_{f} * m\), where \(m\) is the molality, \(K_{f}\) is the freezing point depression constant and \(\Delta T_{f}\) is the change in freezing point. Solving for \(m\): For nitrobenzene: \(m = \Delta T_{f} / K_{f} = 5.7 ^{\circ}C / 8.1^{\circ}C * m^{-1} = 0.7037 m\); For benzene: \(m = \Delta T_{f} / K_{f} = 5.5 ^{\circ}C / 5.12^{\circ}C * m^{-1} = 1.0742 m\)

First calculate the moles of nitrobenzene in each solution. Use the molecular weight of nitrobenzene, 123.11 g/mol: For nitrobenzene solution: \(\text{Moles of nitrobenzene} = \text{Molality} * \text{Mass of benzene} = 0.7037 m * 100 g = 70.37 \text{mol}\); For benzene solution: \(\text{Moles of nitrobenzene} = \text{Molality} * \text{Mass of benzene} = 1.0742 m * 100 g = 107.42 \text{mol}\). Now calculate the mass of nitrobenzene in each solution: For nitrobenzene solution: \(\text{Mass of nitrobenzene} = \text{Moles of nitrobenzene} * \text{Molar mass of nitrobenzene}\) ; For benzene solution: \(\text{Mass of nitrobenzene} = \text{Moles of nitrobenzene} * \text{Molar mass of nitrobenzene}\). Finally, calculate the mass percent of nitrobenzene in each solution: Mass percent = (\(\text{Mass of nitrobenzene} / (\text{Mass of nitrobenzene} + \text{Mass of benzene})\) )* 100