2.7 Kg of each of water and acetic acid are mixed, The freezing point of the solution will be - x ° C . Consider the acetic acid does not dimerise in water, nor dissociates in water x = (nearest integer) [Given : Molar mass of water =18 g mol -1 , acetic acid .=60 g mol -1 ] K f H 2 O : 1.86 K kg mol -1 K f acetic acid : 3.90 K kg mol -1 freezing point : H 2 O =273 K , acetic acid .=290 K ]

Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 4 April, Shift 2 — Question 83

2.7 Kg of each of water and acetic acid are mixed, The freezing point of the solution will be $-\mathrm{x}{ }^{\circ} \mathrm{C}$ . Consider the acetic acid

does not dimerise in water, nor dissociates in water $\mathrm{x}=$ $\qquad$ (nearest integer)

[Given : Molar mass of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$ , acetic acid $\left.=60 \mathrm{~g} \mathrm{~mol}^{-1}\right]$

${ }^{\mathrm{K}_{\mathrm{f}}} \mathrm{H}_{2} \mathrm{O}: 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

${ }^{\mathrm{K}_{\mathrm{f}}}$ acetic acid : $3.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ freezing point : $\mathrm{H}_{2} \mathrm{O}=273 \mathrm{~K}$ , acetic acid $\left.=290 \mathrm{~K}\right]$

Answer: 31

Numerical answer — enter this value.

Step-by-step solution

As moles of water $>$ moles of $\mathrm{CH}_{3} \mathrm{COOH}$

water is solvent

$0-\left(\mathrm{T}_{\mathrm{F}}\right)_{\mathrm{S}}=1.86 \times \frac{2700 / 60}{2700 / 1000}$ $\left(\mathrm{T}_{\mathrm{F}}\right)_{\mathrm{S}}=-31^{\circ} \mathrm{C}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)