JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Topic Test - Electrochemistry

  • question_answer
    On electrolyzing the molten chloride of metal \[725\text{ }ml\] of \[C{{l}_{2}}\] liberated at \[25{}^\circ C\] and 740 mm Hg pressure at anode for every one gm of metal deposite at cathode. The formula of metal chloride is (GAM of metal \[=52.01\])                        

    A) \[MCI\]

    B)  \[MC{{I}_{2}}\]

    C) \[MC{{I}_{3}}\]

    D) \[MC{{I}_{4}}\]

    Correct Answer: C

    Solution :

    [c] equivalents of metal produced = equivalents of \[C{{l}_{2}}\] gas produced
    \[MC{{l}_{x}}\] \[{{x}_{M}}={{x}_{C{{l}_{2}}}}\]
    \[x\times {{n}_{M}}=2\times {{n}_{C{{l}_{2}}}}\]
    \[x\times \frac{1}{52.01}=2\times \frac{740}{760}\times \frac{725\times {{10}^{-3}}}{0.0821\times 298}\]
    \[\Rightarrow \]   \[x=3\]              \[MC{{l}_{3}}\]

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