Q17 / 90 · JEE Main 2024 (01 Feb Shift 2) · Chemistry

Mass of ethylene glycol (antifreeze) to be added to $18.6 kg$ of water to protect the freezing point at $- 24 ° C$ is $kg$ (Molar mass in ${gmol}^{- 1}$ for ethylene glycol $62, K_{f}$ of water $= 1.86 K kg {mol}^{- 1}$ )

Correct answer: 15

Solution

Given, ${ΔT}_{f} = - 24^{0} C$

Molar mass of Ethylene glycol= $62$

${ΔT}_{f} = i K_{f} \times$ molality

$24 = (1) \times 1.86 \times \frac{W}{62 \times 18.6}$

$W = 14880 gm$

$= 14.880 kg$

Hence, the correct answer is 15.