50 g of ethylene glycol (\( C_2H_6O_2 \)) is mixed with 900 g of water. The freezing point of the solution will be (\( K_f \) of water = \( 1.86\, K\, kg\, mol^{-1} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 271.49 K
Adding a non-volatile solute lowers the freezing point of the solvent. For a dilute solution the depression of freezing point is proportional to the molality:
\( \displaystyle \Delta T_f = i\,K_f\,m \)
Ethylene glycol is a non-electrolyte; it does not dissociate, so 1 particle per molecule and i = 1.
\( \displaystyle M(\ce{C2H6O2}) = 2(12) + 6(1) + 2(16) = 62\ \mathrm{g\,mol^{-1}} \)
\( \displaystyle n = \frac{50}{62} = 0.806\ \mathrm{mol} \)
\( \displaystyle m = \frac{0.806\ \mathrm{mol}}{0.900\ \mathrm{kg}} = 0.896\ \mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_f = 1.86 \times 0.896 = 1.67\ \mathrm{K} \)
The freezing point of water is 273.15 K, so
\( \displaystyle T_f = 273.15 - 1.67 = 271.48\ \mathrm{K} \approx 271.49\ \mathrm{K} \)
So the solution freezes at about 271.49 K.
Shortcut: The depression must be small and positive (a dilute solution), so the answer lies a little below 273.15 K. Option (4) is pure water with no depression, and Options (2) and (3) would need 3–3.6 K, about twice the 0.9 molal value; only 271.49 K fits.
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Depression of Freezing Point – …
