JEE Main & Advanced Sample Paper JEE Main Sample Paper-12

  • question_answer
    The conductivity of a saturated solution of \[A{{g}_{3}}P{{O}_{4}}\] is \[9\times {{10}^{-6}}S{{m}^{-1}}\]and its equivalent conductivity is \[1.50\times {{10}^{-4}}\,S{{m}^{2}}\] equivalent\[^{-1}\]. The \[{{K}_{sp}}\] of\[A{{g}_{3}}P{{O}_{4}}\] is

    A)  \[4.32\times {{10}^{-18}}\]                        

    B)  \[1.8\times {{10}^{-9}}\]

    C)  \[8.64\times {{10}^{-13}}\]                        

    D)  None of these

    Correct Answer: A

    Solution :

     \[\lambda =K\times \frac{1000}{N}\] \[1.50\times {{10}^{-4}}\times {{10}^{4}}=9\times {{10}^{-6}}\times {{10}^{-2}}\times \frac{1000}{N}\] \[N=6\times {{10}^{-5}}\] \[S=M=\frac{N}{{{n}_{f}}}=\frac{6\times {{10}^{-5}}}{3}=2\times {{10}^{-5}}\text{mol/L}\] \[A{{g}_{3}}P{{O}_{4}}=3A{{g}^{+}}+P{{O}_{4}}\] \[{{K}_{sp}}={{(3S)}^{3}}.S=27{{S}^{4}}=27\times {{(2\times {{10}^{-5}})}^{4}}=4.32\times {{10}^{-18}}\]


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