A) \[\frac{1}{36}M\]
B) \[\frac{1}{16}M\]
C) \[36\,\,M\]
D) \[6\,\,M\]
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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