Question

45 g ethylene glycol is mixed with 600 g H₂O. Calculate the depression of freezing point of this solution. \( (K_f = 1.86\ \text{K kg mol}^{-1}) \)

Hint:

Use \( \Delta T_f = K_f \, m \); molar mass of ethylene glycol is 62 g/mol and solvent mass is 0.6 kg.

Answer 1

Step 1: Formula. The depression in freezing point is \( \Delta T_f = K_f \times m \), where \( m \) is the molality of the solution and \( K_f \) is the cryoscopic (molal freezing point depression) constant.
Step 2: Moles of solute. Ethylene glycol is \( C_2H_6O_2 \), molar mass \( = 2(12) + 6(1) + 2(16) = 62\ \text{g mol}^{-1} \).
\[ n = \frac{45}{62} = 0.7258\ \text{mol} \]
Step 3: Molality. Mass of solvent (water) = 600 g = 0.600 kg.
\[ m = \frac{0.7258}{0.600} = 1.2097\ \text{mol kg}^{-1} \]
Step 4: Substitute and compute.
\[ \Delta T_f = 1.86 \times 1.2097 = 2.25\ \text{K} \]
\[ \boxed{\Delta T_f \approx 2.25\ \text{K}} \]
So the freezing point of the solution is lowered by about 2.25 K (2.25 °C) below that of pure water.