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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 4

  • question_answer
    If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is \[-0.0558{}^\circ C\], formula of complex is [\[{{K}_{f}}\]of water = 1.86 K kg \[\text{mo}{{\text{l}}^{-1}}\]]

    A) \[[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}\]

    B) \[[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl\]

    C) \[[Co{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]\]

    D) \[[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}\]

    Correct Answer: A

    Solution :

    [a] :\[\Delta {{T}_{f}}={{K}_{f}}\times i\times m\] \[0.0558=1.86\times i\times 0.01\Rightarrow i=3\] Given complex is a strong electrolyte. For strong electrolyte, i = n n = No. of ions = 3 ions Therefore, formula of the complex is \[[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}\].


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