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Q.

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

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a

2

b

1

c

3

d

4

answer is C.

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Detailed Solution

Since m=It M/Fve we have

ve=ItMFm=(10A)(160.83s)135gmol−196500Cmol−1(0.75g)=3

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