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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    For \[NaCl,\]the \[{{K}_{sp}}=36\,\,mo{{l}^{2}}{{L}^{-2}},\] the molar concentration of it will be

    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 \[=6[M]\]


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