A metal is know to form fluoride `MF_(2)`. When `10A` of electricity is passed through a molten sat for `330sec, 1.95g` of metal is deposited. Find the atomic weight of M. what will be the quantity electricity required to deposit the same mass of `Cu` form `CuSO_(4)`?

Answer

`A = 114, Q = 5926, 8C`

Explanation

`M^(+2) + 2e^(-) rarr M rArr n_(e^(-)) = (2xx 1.95)/(M)`
`rArr n_(M) = (1.95)/(M)`
`=(2 xx 1.95)/(M) = (10 xx 330)/(96500)`
`M = 114`
`Cu^(+2) +2e^(-) rarr Cu`
`n_(e^(-)) = (2 xx 1.95)/(63.5) n_(Cu) = (1.95)/(63.5)`
`n_(E) = (Q)/(96500)`
`Q = 5926.8C`