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JEE Main & Advanced Chemistry Redox Reactions / रेडॉक्स अभिक्रियाएँ Question Bank Critical Thinking

  • question_answer
    The conductivity of a saturated solution of \[BaS{{O}_{4}}\] is \[3.06\times {{10}^{-6}}\,oh{{m}^{-1}}\,c{{m}^{-1}}\] and its equivalent conductance is \[1.53\,oh{{m}^{-1}}\,c{{m}^{-1}}\,equivalen{{t}^{-1}}\]. The \[{{K}_{sp}}\] of the \[BaS{{O}_{4}}\] will be [KCET 1996]

    A)                 \[4\,\times {{10}^{-12}}\]            

    B)                 \[2.5\times {{10}^{-9}}\]

    C)                 \[2.5\times {{10}^{-13}}\]            

    D)                 \[4\times {{10}^{-6}}\]

    Correct Answer: D

    Solution :

           \[\lambda m=\frac{1000\,K}{S}=\frac{1000\times 3.06\times {{10}^{-6}}}{S}=1.53\]            \[S=2\times {{10}^{-3}}\frac{mol}{litre}\]                                 \[{{K}_{sp(BaS{{O}_{4}})}}={{S}^{2}}={{(2\times {{10}^{-3}})}^{2}}=4\times {{10}^{-6}}\].


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