VMMC VMMC Medical Solved Paper-2002

  • question_answer
    if solubility product of \[\text{AgCl}\] is \[\text{1}{{\text{0}}^{-10}}\] (mol \[\omega t=143.35\]),   then   its   solubility  in gram/litre will be:

    A)  \[143.5\times {{10}^{-5}}\]        

    B)         \[143.5\times {{10}^{-2}}\]        

    C)  \[14.35\times {{10}^{-1}}\]        

    D)         \[1.435\times {{10}^{-6}}\]

    Correct Answer: A

    Solution :

    \[{{K}_{sp}}={{10}^{-10}}\] \[s=\sqrt{{{K}_{sp}}}=\sqrt{{{10}^{-10}}}=1\times {{10}^{-5}}\,\text{mole/litre}\] \[=1\times {{10}^{-5}}\times 143.5\,\text{g/litre}\] \[=143.5\times {{10}^{-5}}\,\text{g/litre}\]


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