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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • 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)        \[36\,\,M\]        

    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\[=6\,\,[M]\]


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