A) \[\frac{1}{36}M\]
B) \[\frac{1}{16}M\]
C) \[36\text{ }M\]
D) \[6M\]
Correct Answer: D
Solution :
For \[NaCl\] \[{{K}_{sp}}={{S}^{2}}\] where, S = solubility \[\Rightarrow \] \[36={{S}^{2}}\] \[\therefore \] \[S=6\,mol\,{{L}^{-1}}\] Hence, molarity \[=6[M]\]You need to login to perform this action.
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