The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} /\) mol. Usually \(t\) -butanol absorbs water on exposure to air. If the freezing point of a 10.0 -g sample of \(t\) -butanol is \(24.59^{\circ} \mathrm{C},\) how many grams of water are present in the sample?

We are given: - Freezing point of pure \(t\)-butanol: \(T_{f_{pure}} = 25.50^{\circ} \mathrm{C}\) - Freezing point depression constant: \(K_f = 9.1^{\circ} \mathrm{C} \cdot \mathrm{kg}/\) mol - Freezing point of the mixture: \(T_{f_{mixture}} = 24.59^{\circ} \mathrm{C}\) - Sample mass of \(t\)-butanol: \(m_{t-butanol} = 10.0\, \mathrm{g}\) The freezing point depression formula is: \(\Delta T_f = K_f \cdot m_w\), where - \(\Delta T_f = T_{f_{pure}} - T_{f_{mixture}}\) is the freezing point depression - \(m_w\) is the molality of water in the mixture (mol/kg)

Using the freezing point depression formula, we first find the depression of the freezing point: \(\Delta T_f = T_{f_{pure}} - T_{f_{mixture}} = 25.50^{\circ} \mathrm{C} - 24.59^{\circ} \mathrm{C} = 0.91^{\circ} \mathrm{C}\)

Using the freezing point depression formula and the value of \(K_f\): \(m_w = \frac{\Delta T_f}{K_f} = \frac{0.91^{\circ} \mathrm{C}}{9.1^{\circ} \mathrm{C} \cdot \mathrm{kg}/ \mathrm{mol}} = 0.1 \, \mathrm{mol/kg}\)

We know that the molality of water in the mixture is \(0.1 \, \mathrm{mol/kg}\). Since molality is defined as moles of solute (water) per kilogram of solvent (\(t\)-butanol), we can write: \(m_w = \frac{n_{\mathrm{water}}}{m_{t-butanol} / 1000}\), where \(n_{\mathrm{water}}\) is moles of water and we divided by 1000 to convert grams to kilograms. Now we can solve for the moles of water: \(n_{\mathrm{water}} = m_w \cdot \frac{m_{t-butanol}}{1000} = 0.1 \, \mathrm{mol/kg} \cdot \frac{10.0 \, \mathrm{g}}{1000} = 0.001 \, \mathrm{mol}\) Since the molar mass of water is 18.015 g/mol, we can now find the mass of water present in the sample: \(m_{\mathrm{water}} = n_{\mathrm{water}} \cdot M_{\mathrm{water}} = 0.001 \, \mathrm{mol} \cdot 18.015 \, \mathrm{g}/\mathrm{mol} = 0.018 \, \mathrm{g}\)

There are 0.018 grams of water present in the 10.0-gram sample of \(t\)-butanol.