A solution that contains 22.0 g of a nonelectrolyte in 100 g of water freezes at - 2.33 ° C. What is the molar mass of the solute given k_f= 1.86 ° C/m for water?

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The freezing point depression formula is $\Delta T_f = i \cdot K_f \cdot m$, where $i$ is the van't Hoff factor (1 for nonelectrolytes), $K_f$ is the cryoscopic constant for water (1.86 °C/m), and $m$ is the molality of the solution. Given $\Delta T_f = 2.33 °C$, we can solve for $m$, the molality, which is moles of solute per kilogram of solvent. First, find $m = \frac{\Delta T_f}{K_f} = \frac{2.33}{1.86} = 1.25 \, \text{mol/kg}$. Since there are 22.0 g of solute in 0.1 kg of water, we can find the molar mass ($M$) of the solute by dividing the mass of the solute by the number of moles: $M = \frac{22.0 \, \text{g}}{1.25 \, \text{mol}} = 176 \, \text{g/mol}$.

A Solution That Contains 22