At a certain temperature, the vapor pressure of pure benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 0.930 atm. A solution was prepared by dissolving \(10.0 \mathrm{g}\) of a nondissociating, nonvolatile solute in \(78.11 \mathrm{g}\) of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 atm. Assuming the solution behaves ideally, determine the molar mass of the solute.

To use Raoult's Law, we need to determine the mole fractions of the solute and solvent in the solution. Since we have the mass of the solvent (benzene) and its molar mass, we can calculate its moles. We'll denote the moles of benzene as \(n_1\) and the moles of the solute as \(n_2\), and the molar mass of the solute as \(M_2\). \[n_1 =\frac{mass_{solvent}}{molar\_mass_{solvent}}=\frac{78.11 \,\mathrm{g}}{78.11\, \mathrm{g/mol}} =1\, \mathrm{mol} \] To find the mole fraction of solute and solvent, we need to use the equation: \[ \chi_1 = \frac{n_1}{n_1+n_2} \] and \[ \chi_2 = \frac{n_2}{n_1+n_2} \] Since the sum of mole fractions is always equal to 1, we also have: \[ 1 = \chi_1 + \chi_2 \]

Raoult's law states that the partial pressure of a component in an ideal solution is equal to its mole fraction multiplied by its vapor pressure in the pure state. For benzene, we can write: \[ P_1 = \chi_1 \times P^*_1 \] Where \(P_1\) is the partial pressure of benzene in the solution, \(\chi_1\) is its mole fraction, and \(P^*_1\) is its vapor pressure as a pure liquid (0.930 atm). For the solute, since it is nonvolatile, its partial pressure will be zero, so the total pressure of the solution is solely due to benzene. Therefore, we can write: \[ P_\text{solution} = P_1 = \chi_1 \times P^*_1 \] We know \(P_\text{solution}\) is 0.900 atm, so we can write: \[ 0.900\, \mathrm{atm} = \chi_1 \times 0.930\, \mathrm{atm} \]

Now, we will solve for \(\chi_1\): \[ \chi_1 = \frac{0.900\, \mathrm{atm}}{0.930\, \mathrm{atm}} = 0.968 \] Since we know that \(\chi_1 + \chi_2 = 1\), we can calculate the mole fraction of the solute (\(\chi_2\)): \[ \chi_2 = 1 - \chi_1 = 1 - 0.968 = 0.032 \]

Now, we can find the moles of the solute: \[ n_2 = \frac{\chi_2}{\chi_1} \times n_1 = \frac{0.032}{0.968} \times 1\, \mathrm{mol} = 0.0330\, \mathrm{mol} \] Now that we have the moles of the solute, we can calculate the molar mass as follows: \[ M_2 = \frac{mass_{solute}}{n_2} = \frac{10.0\, \mathrm{g}}{0.0330\, \mathrm{mol}} = 302\, \mathrm{g/mol} \] So, the molar mass of the solute is estimated to be 302 g/mol.