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`)

Options

A. 57.02
B. 148
C. 228
D. 114

Answer

D

Explanation

i=10 amp , t = 330 sec , w = 1.95 gm , E = ?
`W = (ixxt xx E)/(96500)`
`therefore E = (W xx 96500)/(ixxt) = (1.95 xx 96500)/(10 xx 330) = 57.02`
Molten salt is `MF_(2)`
Valency of fluorine = `(-1) xx 2 = -2`
`therefore` Valency of M= + 2
`E = ("At. mass")/("valency")`
`therefore` At. mass = `E xx` valency
`= 57.02 xx 2 = 114.04 = 114`