Verified Solution • 2025–26 Session moderate

QJEEM25Ch9e20: JEE Main 2025 Chemistry - Class 11 JEEM25Ch9e20 Solution: for is . What is the molar solubility of this salt in water?

Ksp\mathrm{K}_{\mathrm{sp}} for Cr(OH)3\mathrm{Cr}(\mathrm{OH})_3 is 1.6×10−301.6 \times 10^{-30} . What is the molar solubility of this salt in water?

Multiple Choice Options

  • A 1.8×10−305\sqrt[5]{1.8 \times 10^{-30}}
  • B 1.8×10−3027\frac{1.8 \times 10^{-30}}{27}
  • C 1.6×10−30274\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}
  • D 1.6×10−302\sqrt[2]{1.6 \times 10^{-30}}

Step-by-Step Verified Solution

Cr(OH)3( s)⇌Cr(aq)+3s+3OH(aq)−3s\mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathop {\mathrm{Cr_{(aq)}^{ + 3}}}\limits_s + \mathop {\mathrm{3OH_{(aq)}^ - }}\limits_{3s}

At eq :

amp;Ksp=(s)⋅(3 s)3=27 s4amp;27 s4=1.6×10−30amp; s=(1.627×10−30)1/4\begin{aligned}& \mathrm{K}_{\mathrm{sp}}=(\mathrm{s}) \cdot(3 \mathrm{~s})^3=27 \mathrm{~s}^4 \\& 27 \mathrm{~s}^4=1.6 \times 10^{-30} \\& \mathrm{~s}=\left(\frac{1.6}{27} \times 10^{-30}\right)^{1 / 4}\end{aligned}

Option (1)

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring