JCECE Medical JCECE Medical Solved Paper-2013

  • question_answer
    When \[\text{0}\text{.1}\,\text{mol}\,\text{CoC}{{\text{l}}_{\text{3}}}{{\text{(N}{{\text{H}}_{\text{3}}}\text{)}}_{\text{5}}}\] is treated with excess of \[\text{AgN}{{\text{O}}_{\text{3}}}\text{,0}\text{.2}\,\text{mole}\]of\[\text{AgCl}\] 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 AgCI from 0.1 mole of complex means that there are two ionizable Cl. Hence, formula is\[[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}]\]i.e., 1:2 type electrolyte. 


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