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

We need to find the freezing point depression constant ($$K_f$$) of a solvent.

Mass of solvent = 50 g = 0.050 kg

Mass of non-volatile solute = 1 g

Molar mass of solute = 256 g/mol

Decrease in freezing point $$\Delta T_f = 0.40$$ K

$$\Delta T_f = K_f \times m$$

where $$m$$ is the molality of the solution.

Moles of solute = $$\frac{1}{256}$$ mol = 0.003906 mol

Molality $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1/256}{0.050} = \frac{1}{12.8} = 0.078125$$ mol/kg

$$K_f = \frac{\Delta T_f}{m} = \frac{0.40}{0.078125} = 5.12$$ K kg mol$$^{-1}$$

The correct answer is Option 4: $$5.12$$ K kg mol$$^{-1}$$.