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A metal is known to form fluoride `MF_(2)`. When `10` ampere electricity is passed through a molten salt for `330` sec, `1.95 g` metal is deposited. Find out the atomic weight of metal. What will be the quantity of charge required to deposit the same mass of `Cu` from `CuSO_(4)(aq.)`? (`At. wt.` of `Cu = 63.6`)

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Correct Answer - `A=114,Q=5926.8C`.
v.f. of metal `=2`
`w=Zit`
`1.96=(E)/(96500)` it `=(Mxxit)/(v.f.xx96500)impliesM=114gm`.
Now for `Cu` `W=(63.5xx(it))/(2xx96500)implies1.95=(63.5(it))/(2xx96500)impliesit=5926.77C`

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