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  4. For NaCl, the Ksp=36 mol 2 L-2, the molar concentration of it will be

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Q. For $NaCl$, the $K_{sp}=36\, mol\,{}^{2}\, L^{-2}$, the molar concentration of it will be

Chhattisgarh PMTChhattisgarh PMT 2006

Solution:

For $NaCl K_{sp}=S^{2}$ where,
$S =$ solubility
$\Rightarrow 36=S^{2}$
$\therefore S=6\, mol\, L^{-1}$
Hence, molarity $=6[M]$