How many grams of methyl alcohol should be added to a 10 litre tank of water to prevent its freezing at 268 K? ($$K_f$$ for water is 1.86 K kg mol$$^{-1}$$)

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

$$\Delta T_f = K_f \times \frac{W_{\text{solute}} \text{ (in g)}}{M_{\text{solute}} \text{ (in g/mol)} \times W_{\text{solvent}} \text{ (in kg)}}$$

$$\Delta T_f$$ = $$273.15 - 268 = 5.15\text{ K}$$

Volume of water = $$10\text{ L} \implies$$ Mass of solvent ($$W_{\text{solvent}}$$) = $$10\text{ kg}$$ (assuming density = $$1\text{ g/mL}$$)

$$5.05 = 1.86 \times \frac{W_{\text{solute}}}{32 \times 10}$$

$$W_{\text{solute}} = \frac{5.05 \times 320}{1.86} = \frac{1616}{1.86} \approx 868.81\text{ g}$$