JEE Main 2020 · ChemistryMultiple choiceSingle correctMediumCalculation

JEE Main 9 January 2020, Shift 2, Chemistry Q27

Question 27 of 75 in this shift, Chemistry question 2 of 25, Section A.

The solubility product of

\mathrm{Cr(OH)_3}

at 298 K is

6.0\times10^{-31}

. The concentration of hydroxide ions in a saturated solution of

\mathrm{Cr(OH)_3}

will be :

(1) $(18\times10^{-31})^{1/4}$ Official answer
(2) $(4.86\times10^{-29})^{1/4}$
(3) $(2.22\times10^{-31})^{1/4}$
(4) $(18\times10^{-31})^{1/2}$

Official answer

Option 1

NTA final key (Jan 2020).

Chapter: Equilibrium
Topic: Solubility Product
Idea tested: Solubility-Ksp Relationship

Same idea in other shifts

Asked 8× 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 108 of the 174 JEE Main shifts held from 2017 to 2026.