Question

For $NaCl$ , the $K_{sp}=36mo{l}^2{L}^{-2}$ , 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 $=6[M]$