Question

Define Molal depression constant. 48 g of ethylene glycol (C₂H₆O₂) is dissolved in 600 g of water. Calculate (i) depression in freezing point and (ii) freezing point of the solution. [Given K_f(water) = 1.86 °C kg mol^-1]
OR
Define molality and explain mole fraction. Calculate the mole fraction of solute and solvent of a 20% aqueous C₂H₆O₂ solution.

Hint:

Use \( \Delta T_f = K_f \times m \) with M(C₂H₆O₂) = 62; for the OR-part take 100 g of solution (20 g solute + 80 g water) and apply \( x = n/n_{total} \).

Answer 1

Option 1: Molal depression constant and the freezing-point numerical

Step 1: Definition. The molal depression constant (cryoscopic constant, \(K_f\)) is the depression in freezing point produced when 1 mole of a non-volatile solute is dissolved in 1 kg (1000 g) of solvent. The working relation is \( \Delta T_f = K_f \times m \), where m is the molality.

Step 2: Molar mass of the solute. For ethylene glycol C₂H₆O₂: \( M = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62 \text{ g mol}^{-1}. \)

Step 3: Moles of solute. \( n = \dfrac{48}{62} = 0.774 \text{ mol}. \)

Step 4: Molality. Mass of solvent = 600 g = 0.600 kg, so \( m = \dfrac{0.774}{0.600} = 1.290 \text{ mol kg}^{-1}. \)

Step 5: (i) Depression in freezing point. \( \Delta T_f = K_f \times m = 1.86 \times 1.290 = 2.40. \)
\[\boxed{\Delta T_f = 2.4\,^\circ C}\]Step 6: (ii) Freezing point of the solution. Pure water freezes at 0 °C, so the solution freezes 2.4 °C lower: \( T_f = 0 - 2.4. \)
\[\boxed{T_f = -2.4\,^\circ C}\]
Option 2: Molality, mole fraction, and the 20% solution

Step 1: Molality. Molality (m) is the number of moles of solute dissolved per kilogram of solvent: \( m = \dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}} \). It uses mass, so it does not change with temperature.

Step 2: Mole fraction. The mole fraction of a component is the ratio of its moles to the total moles of all components: \( x_{solute} = \dfrac{n_2}{n_1+n_2} \), \( x_{solvent} = \dfrac{n_1}{n_1+n_2} \), and the two add up to 1.

Step 3: Fix a basis. A 20% aqueous solution means 20 g of solute in 100 g of solution, so solute = 20 g and water = 100 - 20 = 80 g.

Step 4: Convert to moles. \( n_{solute} = \dfrac{20}{62} = 0.3226 \text{ mol}, \quad n_{water} = \dfrac{80}{18} = 4.444 \text{ mol}. \)

Step 5: Total moles. \( n_1 + n_2 = 0.3226 + 4.444 = 4.767 \text{ mol}. \)

Step 6: Mole fractions. \( x_{solute} = \dfrac{0.3226}{4.767} = 0.0677, \quad x_{solvent} = \dfrac{4.444}{4.767} = 0.9323 \) (sum \(=1\)).
\[\boxed{x_{solute} \approx 0.068,\quad x_{solvent} \approx 0.932}\]