Question

An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 g mol$^{-1}$) in 600 g of water. Calculate the freezing point of the solution. (K$_f$ for water = 1.86 K kg mol$^{-1}$)

Hint:

For non-electrolytes such as glucose, urea and ethylene glycol, always take van't Hoff factor \(i=1\).

Answer 1

Step 1: Calculate moles of ethylene glycol and molality.
Moles $= \frac{31}{62} = 0.5\,mol$. Mass of solvent $= 600\,g = 0.6\,kg$. Molality $m = \frac{0.5}{0.6} = 0.833\,mol\,kg^{-1}$.
Step 2: Determine van't Hoff factor.
Ethylene glycol is a non-electrolyte and does not ionize in solution, so $i = 1$.
Step 3: Apply the freezing point depression formula.
\[ \Delta T_f = iK_fm = 1 \times 1.86 \times 0.833 = 1.55\,K \] Freezing point of solution $= 0^\circ C - 1.55^\circ C = -1.55^\circ C$. \[ \boxed{T_f = -1.55^\circ C} \]