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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    If the solubility product of lead iodide \[(Pb{{l}_{2}})\] is \[3.2\times {{10}^{-8}},\] then its solubility in moles/litre will be                 [MP PMT 1990]

    A)                 \[2\times {{10}^{-3}}\]  

    B)                 \[4\times {{10}^{-4}}\]

    C)                 \[1.6\times {{10}^{-5}}\]              

    D)                 \[1.8\times {{10}^{-5}}\]

    Correct Answer: A

    Solution :

               \[{{K}_{sp}}=4{{S}^{3}}\]                                 \[4{{S}^{3}}=3.2\times {{10}^{-8}}\] ; \[S=2\times {{10}^{-3}}M\].


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