Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol} )\) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density $=0.879 \mathrm{g} / \mathrm{cm}^{3} )$ The elevated boiling point of the solution was 355.4 \(\mathrm{K}\) . What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 \(\mathrm{kJ} / \mathrm{mol}\) , the entropy of vaporization is 95.95 \(\mathrm{J} / \mathrm{K} \cdot\) mol, and the boiling- point elevation constant is 2.5 \(\mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}\) .

The boiling-point elevation can be calculated using the formula: \(\Delta T_b = K_b \cdot m\) where \(\Delta T_b\) is the boiling-point elevation, \(K_b\) is the boiling-point elevation constant, and \(m\) is the molality of the solute in the solution. We are given the boiling-point elevation constant, \(K_b = 2.5\ \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}\), and the boiling point of the solution, \(T_{solution} = 355.4\ \mathrm{K}\). First, we find the normal boiling point of the solvent, \(T_{solvent}\), using the given values of enthalpy and entropy of vaporization: \(T_{solvent}=\frac{\Delta H_v}{\Delta S_v}\) where \(\Delta H_v = 33.90\ \mathrm{kJ} / \mathrm{mol} = 33900\ \mathrm{J} / \mathrm{mol}\) and \(\Delta S_v = 95.95\ \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\). \(T_{solvent}=\frac{33900\ \mathrm{J} / \mathrm{mol}}{95.95\ \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}} = 353.16\ \mathrm{K}\) Now, we can find the boiling-point elevation: \(\Delta T_b = T_{solution} - T_{solvent} = 355.4\ \mathrm{K} - 353.16\ \mathrm{K} = 2.24\ \mathrm{K}\)

Now that we have the boiling-point elevation, we can calculate the molality of the solute using the formula: \(m=\frac{\Delta T_b}{K_b}\) \(m=\frac{2.24\ \mathrm{K}}{2.5\ \mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}}=0.896 \ \mathrm{mol} / \mathrm{kg}\)

To calculate the mass of the solvent, we use the given volume and density of the solvent. Mass of solvent = Volume × Density where Volume = \(150\ \mathrm{mL} = 0.150\ \mathrm{L}\) and Density = \(0.879\ \mathrm{g} / \mathrm{cm}^{3}\). To convert volume from liters to cubic centimeters, we can multiply by 1000. Volume in cubic centimeters = \(0.150\ \mathrm{L} \times 1000 = 150\ \mathrm{cm}^{3}\) Now, we can find the mass of the solvent: Mass of solvent = \(150\ \mathrm{cm}^{3} \times 0.879\ \mathrm{g} / \mathrm{cm}^{3} = 131.85\ \mathrm{g}\)

Now that we know the molality of the solute and the mass of the solvent, we can find the moles of solute using the formula: moles of solute = \(\frac{m \times \text{Mass of solvent}}{\text{Molar mass of solvent}}\) Moles of solute = \(\frac{0.896\ \mathrm{mol} / \mathrm{kg} \times 131.85\ \mathrm{g}}{1000\ \mathrm{g}/\mathrm{kg}} = 0.1182\ \mathrm{mol}\) Finally, we find the mass of the solute using the given molar mass of the solute, \(M = 142\ \mathrm{g} / \mathrm{mol}\). Mass of solute = moles of solute × molar mass of solute Mass of solute = \(0.1182\ \mathrm{mol} \times 142\ \mathrm{g} / \mathrm{mol} = 16.79\ \mathrm{g}\) So, the mass of solute dissolved in the solvent is 16.79 g.