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JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    When\[0.1\,\,mol\,\,CoC{{l}_{3}}\], \[{{(N{{H}_{3}})}_{5}}\] is treated with excess of \[AgN{{O}_{3}},\,\,0.2\,\,mole\] of \[AgCl\] are obtained. The conductivity of solution will correspond to

    A) \[1:3\]electrolyte            

    B) \[1:2\]electrolyte

    C) \[1:1\]electrolyte            

    D)  \[3:1\]electrolyte

    Correct Answer: B

    Solution :

    Formation of \[0.2\] mole of \[AgCl\] from \[0.1\] mole of complex means that there are two ionisable\[Cl\]. Hence, formula is \[[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}]\,\,i.e.,\,\,1:2\] type electrolyte.


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