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 g mol -1 for ethylene glycol 62, K f of water =1.86 K kg mol -1 )

Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 1 February, Shift 2 — Question 79

Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at $-24^{\circ} \mathrm{C}$ is $\qquad$ kg (Molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ for ethylene glycol $62, \mathrm{~K}_{\mathrm{f}}$ of water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \times$ molality

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

$\mathrm{W}=14880 \mathrm{gm}$

$=14.880 \mathrm{~kg}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)