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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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answer is C.

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Detailed Solution

Since $m = I t M / F v_{e}$ we have

$v_{e} = \frac{I t M}{F m} = \frac{(10 A) (160.83 s) (135 g {mol}^{- 1})}{(96500 C {mol}^{- 1}) (0.75 g)} = 3$

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