Question

Define molal depression constant. Calculate the freezing point of the aqueous solution containing 25 g of ethylene glycol (C\(_2\)H\(_6\)O\(_2\)) in 300 g of water. The value of molal depression constant for water is 1.86 K kg mol\(^{-1}\) and freezing point of water is 273.15 K.

Hint:

Molal depression constant (\(K_f\)) is a colligative property, depending only on the number of solute particles.

Answer 1

Step 1: Formula for Freezing Point Depression: \[ \Delta T_f = i \cdot K_f \cdot m \] Step 2: Calculate Moles of Ethylene Glycol: \[ \text{Moles} = \frac{25}{62} = 0.403 \text{ mol} \] Step 3: Convert Solvent Mass to kg: \[ 300 \text{ g} = 0.300 \text{ kg} \] Step 4: Calculate Molality: \[ m = \frac{0.403}{0.300} = 1.343 \text{ mol/kg} \] Step 5: Calculate Depression in Freezing Point: \[ \Delta T_f = 1.86 \times 1.343 = 2.50 \text{ K} \] Step 6: Calculate Freezing Point of Solution: \[ T_f = 273.15 - 2.50 = 270.65 \text{ K} \] Thus, the freezing point of the solution is 270.65 K.