Question

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

• A. $\frac{1}{36}M$
• B. $\frac{1}{16}M$
• C. $36M$
• D. $6M$

Answer 1

Answer: (D)

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