JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(\mathrm{K}_{\text {sp }}\) for \(\mathrm{Cr}(\mathrm{OH})_3\) is \(1.6 \times 10^{-30}\). What is the molar solubility of this salt in water?
- A \(\frac{1.8 \times 10^{-30}}{27}\)
- B \(\sqrt[5]{1.8 \times 10^{-30}}\)
- C \(\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}\)
- D \(\sqrt[2]{1.6 \times 10^{-30}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{Cr}(\mathrm{OH})_3 \rightleftharpoons \underset{ \mathrm{s}}{\mathrm{Cr}^{3+}}(\mathrm{aq})+\underset{3 \mathrm{~s}}{3 \mathrm{OH}^{-}(\mathrm{aq})} \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{s}(3 \mathrm{~s})^3 \\ & 1.6 \times 10^{-30}=27 \mathrm{~s}^4 \\ &…
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