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 = $1.86\, K\, kg\, mol^{-1}$)
Answer & explanation
Correct answer: option 1
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
$\Delta T_f = 1 \times 1.86 \times 1.21$
$= 2.25 \text{ K}$