The freezing point of an aqueous solution containing 23 g of \( C_2H_5OH \) in 1000 g of water, is
(Given \( K_f \) for water = \( 1.86\, K\, kg\, mol^{-1} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → -0.93 °C
Adding a non-volatile solute lowers the freezing point of the solvent. The depression of freezing point is a colligative property, proportional to the molality of solute particles:
\( \displaystyle \Delta T_f = i\,K_f\,m \)
Ethanol is a non-electrolyte; it does not dissociate in water, so i = 1. Its molar mass is 46 g mol−1.
\( \displaystyle n = \frac{23\,\mathrm{g}}{46\,\mathrm{g\,mol^{-1}}} = 0.5\,\mathrm{mol} \)
\( \displaystyle m = \frac{0.5\,\mathrm{mol}}{1\,\mathrm{kg}} = 0.5\,\mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_f = 1 \times 1.86 \times 0.5 = 0.93\,\mathrm{K} \)
Water freezes at 0 °C, and the solution freezes 0.93 degrees lower:
\( \displaystyle T_f = 0 - 0.93 = -0.93\,^\circ\mathrm{C} \)
So the freezing point is −0.93 °C.
Shortcut: A solution always freezes below the pure solvent, so the answer must be negative: Options (1) or (2). Half a mole in a kilogram gives half of \( K_f \), 0.93 K, which is Option (2).