0.75 g of a metal (atomic mass= 135 u) was deposited at cathode when 10 A current is passed through metallic salt solution for 160.83 s. The charge number of metallic cation is

0.75 g of a metal (atomic mass= 135 u) was deposited at cathode when 10 A current is passed through metallic salt solution for 160.83 s. The charge number of metallic cation is

  1. A

    1

  2. B

    2

  3. C

    3

  4. D

    4

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    Solution:

    Since m=It M/Fve we have

    ve=ItMFm=(10A)(160.83s)135gmol196500Cmol1(0.75g)=3

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