45 g of ethylene glycol ( C_2H_6O_2 ) is mixed with 600 g of water. C…

The correct answer is Option (1) → 2.25 K

Depression in freezing point is a colligative property: for a dilute solution it is proportional to the molality of the solute particles.

    \( \displaystyle \Delta T_f = i\,K_f\,m \)

Ethylene glycol is a non-electrolyte; it does not dissociate, so it gives 1 particle per molecule and \( i = 1 \).

Molar mass of \( \ce{C2H6O2} \):

    \( \displaystyle M = 2(12) + 6(1) + 2(16) = 62\ \mathrm{g\,mol^{-1}} \)

Moles of solute and molality (600 g of water = 0.600 kg):

    \( \displaystyle n = \frac{45}{62} = 0.7258\ \mathrm{mol} \)

    \( \displaystyle m = \frac{0.7258}{0.600} = 1.2097\ \mathrm{mol\,kg^{-1}} \)

Depression in freezing point:

    \( \displaystyle \Delta T_f = 1 \times 1.86 \times 1.2097 = 2.25\ \mathrm{K} \)

So the freezing point is lowered by 2.25 K.

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Depression of Freezing Point – “45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution.”