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: One-shot formula method for the freezing-point drop

Step 1: Meaning of \(K_f\). Think of \(K_f\) as the freezing-point depression that a 1 molal ideal solution would show; it is the proportionality constant in \( \Delta T_f = K_f\,m \). For water \(K_f = 1.86\,^\circ C\,\text{kg mol}^{-1}\).

Step 2: Combine everything into a single formula. Writing molality in terms of masses gives \( \Delta T_f = \dfrac{1000 \times K_f \times w_2}{M_2 \times w_1} \), where \(w_2\) = mass of solute (g), \(w_1\) = mass of solvent (g) and \(M_2\) = molar mass of solute.

Step 3: Insert the data. Here \(w_2 = 48\), \(w_1 = 600\), \(M_2 = 62\), \(K_f = 1.86\):
\( \Delta T_f = \dfrac{1000 \times 1.86 \times 48}{62 \times 600}. \)

Step 4: Arithmetic. Numerator \(= 1000 \times 1.86 \times 48 = 89280\); denominator \(= 62 \times 600 = 37200\); so \( \Delta T_f = \dfrac{89280}{37200} = 2.4\,^\circ C. \)
\[\boxed{\Delta T_f = 2.4\,^\circ C}\]Step 5: Freezing point. The solution freezes 2.4 °C below pure water, hence \[\boxed{T_f = 0 - 2.4 = -2.4\,^\circ C}\]
Option 2: 100 g basis for the mole fractions

Step 1: Molality restated. Molality is moles of solute per kilogram of solvent (mol/kg). Because it depends on mass and not on volume, it stays constant when the temperature changes.

Step 2: Mole fraction restated. Mole fraction is the fraction of the total number of moles contributed by one component; it is dimensionless and the mole fractions of all components sum to 1, which is why it links neatly to partial pressures in Raoult's law.

Step 3: Take 100 g of the solution. Since it is 20% by mass, the solute (ethylene glycol) mass is 20 g and the solvent (water) mass is 80 g.

Step 4: Moles of each. \( n_{glycol} = \dfrac{20}{62} = 0.3226 \text{ mol}, \quad n_{water} = \dfrac{80}{18} = 4.4444 \text{ mol}. \)

Step 5: Solvent mole fraction first. \( x_{water} = \dfrac{4.4444}{4.4444 + 0.3226} = \dfrac{4.4444}{4.7670} = 0.9323. \)

Step 6: Solute mole fraction by difference. \( x_{glycol} = 1 - 0.9323 = 0.0677. \)
\[\boxed{x_{solute} \approx 0.068,\quad x_{solvent} \approx 0.932}\]