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AMU Medical AMU Solved Paper-1999

  • question_answer
    Solubility of lead chloride is K. Then the number of moles of \[C{{l}^{-}}\]ionsin 1000 ml of saturated lead chloride solution will be

    A)  \[{{\left( \frac{K}{4} \right)}^{1/3}}\]

    B)  \[{{(2K)}^{1/3}}\]

    C)  \[{{K}^{1/2}}\]

    D)  \[\frac{K}{2}\]

    Correct Answer: A

    Solution :

    : For lead chloride\[(PbC{{l}_{2}})\] \[{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\] \[{{K}_{sp}}=x.{{(2x)}^{2}}\] \[{{K}_{sp}}=x.4{{x}^{2}}\] \[x={{\left( \frac{K}{4} \right)}^{1/3}}\]


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