The water-soluble nonelectrolyte \(\mathrm{X}\) has a molar mass of \(410 \mathrm{~g} / \mathrm{mol}\). A \(0.100\) -g mixture containing this substance and sugar \((\mathrm{MM}=342 \mathrm{~g} / \mathrm{mol})\) is added to \(1.00 \mathrm{~g}\) of water to give a solution whose freezing point is \(-0.500^{\circ} \mathrm{C}\). Estimate the mass percent of \(\mathrm{X}\) in the mixture.

Knowing that the freezing point of pure water is \(0^{\circ} \mathrm{C}\), and the freezing point of the solution is \(-0.500^{\circ} C\), we can calculate the freezing-point depression \(\Delta T_f\) as: \(\Delta T_f = T_f^\text{pure} - T_f^\text{solution} = 0 - (-0.500) = 0.500^{\circ} \mathrm{C}\) For water, the freezing-point depression constant (\(K_f\)) is \(1.86 \, \mathrm{K \, kg/mol}\). Now, we can find the molality (m) of solute using the formula: \(\Delta T_f = K_f \cdot m\) Rearranging to solve for molality (m): \(m = \frac{\Delta T_f}{K_f} = \frac{0.500^{\circ} \mathrm{C}}{1.86 \, \mathrm{K \, kg/mol}} = 0.2691 \, \mathrm{mol/kg}\)

Since total mass of the mixture (\(0.100 \, \mathrm{g}\)) was added to \(1.00 \, \mathrm{g}\) of water, the total mass of the solution is \(1.10 \, \mathrm{g}\). Molality (m) is the number of moles of solute per kilogram solute. We can use molality, total mass of the solution, and given molar masses to calculate the moles of substance X and sugar in the solution. Let \(w_X\) be the mass of substance X in the mixture. Then, the mass of sugar in the mixture is \((0.100 - w_X)\, \mathrm{g}\). Substance X: \(n_X = \frac{w_X}{\mathrm{MM_X}}\) Sugar: \(n_S = \frac{0.100 - w_X}{\mathrm{MM_S}}\) Since the molality of the solution is the sum of the molalities of substance X and sugar: \(m = 0.2691 \, \mathrm{mol/kg} = \frac{n_X}{w_w} + \frac{n_S}{w_w}\) Replacing moles and masses in terms of \(w_X\): \(0.2691 \, \mathrm{mol/kg} = \frac{\frac{w_X}{\mathrm{MM_X}}}{w_w} + \frac{\frac{0.100 - w_X}{\mathrm{MM_S}}}{w_w}\)

We can solve for \(w_X\) to find the mass of substance X in the mixture: \(0.2691 \, \mathrm{mol/kg} = w_X\,(\frac{1}{410\, \mathrm{g/mol}}-\frac{1}{342\, \mathrm{g/mol}})\frac{1}{1.00\, \mathrm{g}}\) Now, isolate \(w_X\): \(w_X = \frac{0.2691 \, \mathrm{mol/kg}}{(\frac{1}{410\, \mathrm{g/mol}}-\frac{1}{342\, \mathrm{g/mol}})\frac{1}{1.00\, \mathrm{g}}} \approx 0.0432\, \mathrm{g}\) Finally, we can calculate the mass percent of substance X in the mixture: Mass Percent \(= \frac{0.0432\, \mathrm{g}}{0.100\, \mathrm{g}} \times 100\% \approx 43.2\%\) Thus, the mass percent of substance X in the mixture is approximately 43.2%.