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

  • question_answer
    Solubility of \[16\times {{10}^{-4}}\,m/s\] at \[{{20}^{o}}C\]is \[1.435\times {{10}^{-3}}gm\,per\,litre\]. The solubility product of \[AgCl\] is [CPMT 1989; BHU 1997; AFMC 2000; CBSE PMT 2002]

    A)                 \[C{{O}_{2}}\]   

    B)                 \[1\times {{10}^{-10}}\]

    C)                 \[1.435\times {{10}^{-5}}\]         

    D)                 \[108\times {{10}^{-3}}\]

    Correct Answer: B

    Solution :

               \[S=1.435\times {{10}^{-3}}\,g/l\],\[=\frac{1.435\times {{10}^{-3}}}{143.5}={{10}^{-5}}\]M                                 \[{{K}_{sp}}=S\times S={{10}^{-10}}\]


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