45 g of ethylene glycol ($C_2H_6O_2$) is mixed with 600 g of water. Calculate the depression in freezing point.

($K_f$ for water = $1.86\, K\, kg\, mol^{-1}$)

2.25 K

The correct answer is Option (1) → 2.25 K

Core Concept:

Depression in freezing point:

ΔTf = i Kf m

For non-electrolyte → i = 1

Step 1: Molar mass of ethylene glycol

C₂H₆O₂ = 2(12) + 6(1) + 2(16)

= 24 + 6 + 32

= 62 g/mol

Step 2: Moles of solute

$n = \frac{m}{M}$

• $m$: Mass of the substance (grams)
• $M$: Molar mass of the substance (g/mol)

$= 45 / 62$

$= 0.726 \text{ mol}$

Step 3: Mass of solvent in kg

$600 \text{ g} = 0.6 \text{ kg}$

Step 4: Molality

$\text{Molality } (m) = \frac{\text{Moles of Solute}}{\text{Mass of Solvent in kg}}$

Alternative Formula (if mass of solvent is in grams):

$m = \frac{\text{Moles of Solute}}{\text{Mass of Solvent in g}} \times 1000$

$m = 0.726 / 0.6$

$= 1.21 \text{ mol/kg}$

Step 5: Depression in Freezing Point

$\Delta T_f = 1 \times 1.86 \times 1.21$

$= 2.25 \text{ K}$