What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass $256 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and the decrease in freezing point is 0.40 K ?

<p>To find the freezing point depression constant ($ K_f $) of the solvent, we use the formula for freezing point depression:</p> <p>$ \Delta T_f = K_f \cdot m $</p> <p>Given:</p> <p><p>The decrease in freezing point $\Delta T_f$ is 0.40 K.</p></p> <p><p>The mass of the solute is 1 g and its molar mass is 256 g/mol.</p></p> <p><p>The mass of the solvent is 50 g (or 0.050 kg).</p></p> <p>First, calculate the molality ($ m $):</p> <p><p>Molality is defined as the moles of solute per kilogram of solvent.</p></p> <p><p>Calculate moles of solute:</p></p> <p>$ \text{Moles of solute} = \frac{1 \, \text{g}}{256 \, \text{g/mol}} = \frac{1}{256} \, \text{mol} $</p> <p>Calculate molality ($ m $):</p> <p>$ m = \frac{\frac{1}{256} \, \text{mol}}{0.050 \, \text{kg}} = \frac{1}{256 \times 0.050} \, \text{mol/kg} $</p> <p>Now, substitute into the formula to find $ K_f $:</p> <p>$ 0.4 = K_f \cdot \frac{1}{256 \times 0.050} $</p> <p>Solving for $ K_f $:</p> <p>$ K_f = 0.4 \cdot 256 \times 0.050 = 5.12 \, \text{K kg/mol} $</p> <p>Thus, the freezing point depression constant of the solvent is $ 5.12 \, \text{K kg/mol} $.</p>