Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) boils at $80.1^{\circ} \mathrm{C}\( and has a density of \)0.876 \mathrm{~g} / \mathrm{mL} .$ (a) When \(0.100 \mathrm{~mol}\) of a nondissociating solute is dissolved in $500 \mathrm{~mL}\( of \)\mathrm{C}_{6} \mathrm{H}_{6}$, the solution boils at \(79.52^{\circ} \mathrm{C}\). What is the molal boiling-point-elevation constant for \(\mathrm{C}_{6} \mathrm{H}_{6} ?\) (b) When \(10.0 \mathrm{~g}\) of a nondissociating unknown is dissolved in \(500 \mathrm{~mL}\) of $\mathrm{C}_{6} \mathrm{H}_{6}\(, the solution boils at \)79.23^{\circ} \mathrm{C}$. What is the molar mass of the unknown?

First, we have to determine the boiling-point elevation (ΔTb) that happens when the solute is added. ΔTb is calculated by subtracting the normal boiling point of the solvent from the boiling point of the solution: ΔTb = T_solution - T_solvent = \(79.52^{\circ} C - 80.1^{\circ} C = -0.58^{\circ} C\) Next, we need to calculate the molality of the solute (m) which is given by the moles of solute divided by the mass of the solvent, expressed in kilograms: m = moles of solute / mass of solvent (in kg) Since the density of benzene is 0.876 g/mL, we can find the mass of 500 mL of benzene: Mass of benzene = \(0.876 \frac{g}{mL} \times 500 \, mL = 438 \, g\) Converting grams to kilograms, we get: Mass of benzene = \(438 \, g \times \frac{1 \, kg}{1000 \, g} = 0.438 \, kg\) Now, we are given 0.100 mol of solute, so the molality (m) is: m = \(\frac{0.100 \, mol}{0.438 \, kg} = 0.228 \frac{mol}{kg}\) Finally, we can find the molal boiling-point-elevation constant (Kb) for benzene using the boiling-point elevation equation: ΔTb = Kb × m Kb = \(\frac{ΔTb}{m} = \frac{-0.58^{\circ} C}{0.228 \frac{mol}{kg}} = -2.54 \frac{^{\circ}C \cdot kg}{mol}\)

Firstly, calculate the boiling-point elevation (ΔTb) for the solution with the unknown solute using the same procedure: ΔTb = T_solution - T_solvent = \(79.23^{\circ} C - 80.1^{\circ} C = -0.87^{\circ} C\) Now, we can rearrange the boiling-point elevation equation to find the molality (m) of the unknown solute: m = ΔTb / Kb = \(\frac{-0.87^{\circ} C}{-2.54 \frac{^{\circ}C \cdot kg}{mol}} = 0.343\frac{mol}{kg}\) We are given that 10.0 g of the unknown solute was dissolved in 500 mL of benzene. To find the molar mass (M) of the unknown solute, we need to find the mass of the solute in 1 kg of benzene. Since the mass of benzene is 0.438 kg, we can find the mass of the solute in 1 kg of benzene as: Mass of unknown in 1 kg of benzene = \(\frac{10.0 \, g}{0.438 \, kg} = 22.8 \, g\) Lastly, we use the molality to find the molar mass (M) of the unknown solute: M = Mass of unknown in 1 kg of benzene / m M = \(\frac{22.8 \, g}{0.343 \frac{mol}{kg}} = 66.5 \frac{g}{mol}\) So, the molar mass of the unknown solute is 66.5 g/mol.