JEE Main 2021 · ChemistryNumerical answerNumerical valueHardCalculation

JEE Main 24 February 2021, Shift 1, Chemistry Q55

Question 55 of 90 in this shift, Chemistry question 25 of 30, Section B.

When 9.45 g of

\mathrm{ClCH_2COOH}

is added to 500 mL of water, its freezing point drops by

0.5^\circ\mathrm{C}

. The dissociation constant of

\mathrm{ClCH_2COOH}

x \times 10^{-3}

. The value of

x

is __________. (Rounded off to the nearest integer) [

K_{f(\mathrm{H_2O})}

=1.86

\mathrm{K\ kg\ mol^{-1}}

]

Official answer

NTA final key.

Chapter: Solutions
Topic: Colligative Properties
Idea tested: Calculating Ka from Freezing Point Depression

Same idea in other shifts

Asked 4× in all

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29 Jun 2022, Shift 1 · Q831.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid…MediumNumerical value

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 108 of the 174 JEE Main shifts held from 2017 to 2026.