A solution of `M(NO_(3))_(2)`was electrolysed by passing a current of 2.5 A and 3.06 g of the metal was deposited in 35 minutes. Determine the molar mass of the metal.

Explanation

Quantity of charge passed (Q) `=Ixxt=(2.5 A)xx(35xx60 s)=5250 As =5250 C`
No.of faradays of charge passed `=((5250C))/((96500C))=0.0544.`
0.0544 F of charge deposited metal (M)=3.06 g
1 F of charge deposited metal (M) `=((3.06g))/(0.0544 F)xx(1F)=56.25 g`
Equivalent mass of metal =56.25 g=56.25 amu.
Molar mass of the metal `="Equivalent mass" xx"valency"`
`=(56.25 amu)xx(2)=112.5" amu"`