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If $\mathrm{K}_{\mathrm{sp}}$ is solubility product of $\mathrm{Al}(\mathrm{OH})_3$, its solubility is expressed by formula,
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The correct answer is:
$\sqrt[4]{\frac{K_{s p}}{27}}$
$\begin{array}{ll} & \mathrm{Al}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Al}_{(\mathrm{mq})}^{3+}+3 \mathrm{OH}_{(\mathrm{sq})}^{-} \\ & \text {Here, } x=1, \mathrm{y}=3 \\ \therefore \quad & \mathrm{K}_{s p}=x^x \mathrm{y}^{\mathrm{y}} \mathrm{S}^{x+y}=(1)^1(3)^3 \mathrm{~S}^{1+3}=27 \mathrm{~S}^4 \\ \therefore \quad & \mathrm{S}=\sqrt[4]{\frac{\mathrm{K}_{\mathrm{sp}}}{27}}\end{array}$
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