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JEE Main 28 July 2022, Shift 1, Chemistry Q84

Question 84 of 90 in this shift, Chemistry question 24 of 30, Section B.

150 g of acetic acid was contaminated with 10.2 g ascorbic acid (

\mathrm{C_6H_8O_6}

) to lower down its freezing point by

(x\times10^{-1})^\circ

C. The value of

x

is __________. (Nearest integer) [Given

K_f=3.9\ \mathrm{K\,kg\,mol^{-1}}

; molar mass of ascorbic acid

=176\ \mathrm{g\,mol^{-1}}

]

Official answer

NTA final key (2022 Session 2).

Chapter: Solutions
Topic: Colligative Properties
Idea tested: Freezing Point Depression

Same idea in other shifts

Asked 5× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.