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`)

Explanation

Eq. of metal `=(i.t.)/(96500)` (i=10 ampere, t=330sec)
`(1.95)/(E)=(10 xx 330)/(96500)`
`E_("metal")=57.0`
At. Wt. of metal `=57 xx 2 ("metal is bivalent as salt is "MF_(2))`
=114.0
Also, if `W_(cu)=1.95`g, then
Eq. of Cu `=(i.t.)/(96500)`
or `(1.95)/(63.6//2)=(i.t.)/(96500)`
i.e. =Q=5917.45 coulomb