50 g of ethylene glycol ($C_2H_6O_2$) is mixed with 900 g of water. The freezing point of the solution will be ($K_f$ of water = $1.86\, K\, kg\, mol^{-1}$)

271.49 K

The correct answer is Option (1) → 271.49 K

Core Concept — Freezing Point Depression

$\Delta T_f = i K_f m$

Where,

• $\Delta T_f$ = Freezing point depression
• $i$ = Van’t hoff factor
• $K_f$ = Cryoscopic constant
• $m$ = molality

Ethylene glycol is non-electrolyte, so

$i = 1$

Step 1: Moles of solute

Molar mass of ethylene glycol

$\text{C}_2\text{H}_6\text{O}_2 = 2(12) + 6(1) + 2(16) = 62 \text{ g/mol}$

$\text{Moles} = \frac{50}{62} = 0.806 \text{ mol}$

Step 2: Molality ($m$)

$\text{Mass of solvent} = 900\text{ g} = 0.9\text{ kg}$

$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \text{}$

$m = \frac{0.806}{0.9} = 0.895\text{ mol/kg}$

Step 3: Depression in Freezing Point ($\Delta T_f$)

$\Delta T_f = i K_f m \text{}$

$\Delta T_f = 1 \times 1.86 \times 0.895 = 1.665\text{ K}$

Step 4: New Freezing Point ($T_f$)

$\text{Normal freezing point of water} = 273.15\text{ K}$

$T_f = 273.15 - 1.665 = 271.49\text{ K}$