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10800 C of electricity passed through th...

`10800 C` of electricity passed through the electrolyte deposited `2.977g` of metal with atomic mass `106.4 g mol^(-1)`. The charge on the metal cation is

A

`+4`

B

`+3`

C

`+2`

D

`+1`

Text Solution

Verified by Experts

The correct Answer is:
a
`M^(n+)+n e^(-) rarr M`
Mole of metal deposited `=(2.977)/(106.4)=0.27mol`
`1 mol `of `M` deposited = `n `Faraday
`=nxx96500C.`
`0.027 mol =nxx965--xx0.027`
`:. Nxx96500xx0.027=10800C`
`:. n=4`
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