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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    The solubility of\[PB{{I}_{2}}\]at 25°C is \[0.7g\,{{L}^{-1}}.\] The solubility product of\[Pb{{I}_{2}}\] at this temperature is(molar mass of\[Pb{{I}_{2}}=461.2\,g\,\text{mo}{{\text{l}}^{-1}}\]).   JEE Main Online Paper (Held On 26-May-2012)  

    A) \[1.40\times {{10}^{-9}}\]                            

    B)                        \[0.14\times {{10}^{-9}}\]

    C)                        \[140\times {{10}^{-9}}\]                             

    D)                        \[14.0\times {{10}^{-9}}\]

    Correct Answer: D

    Solution :

                     \[\underset{s}{\mathop{Pb}}\,\underset{2s}{\mathop{{{I}_{2}}}}\,P{{b}^{++}}+2{{I}^{-}}\] \[{{K}_{sp}}=s\times {{(2s)}^{2}}=4{{s}^{3}}\] \[=4\times \left( \frac{0.7}{461.2} \right)=14.0\times {{10}^{-9}}\]


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