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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
Let the charge be `n^(+)`
`therefore M^(n+)+ "ne"^(-)rarrM`
106.4g of metal require charge `=nxx96500C`
2.977g of m etal require charge
`=((nxx96500)/(106.4)xx2.977)`
`therefore (nxx96500)/(106.4)xx2.977=10800`
`n=(10800xx106.4)/(96500xx2.977)=4`
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