A current of 0.96 A is passed for 3 hours between Ni electrodes in 0.5 L of 92 M solution `Ni(NO_(3))_(2)` . The molarity of the solution after electrolysis is would be -

Options

A. `0.92` M
B. `0.625` M
C. `0.22` M
D. `1.25` M

Answer

A

Explanation

`Q = I xx t = 9.65 xx 3 xx 60 = 104220 ` C .
`Ni^(2+) + 2e^(-) to Ni`
`2 xx 96500 C ` deposit 1 mole of `Ni^(2+)` ,
`therefore 104220 C` will deposit `Ni^(2+)`
`= (1)/(2 xx 96500) xx 104220 = 0.54` mole
Originally 0.5 L of 2 M solution contain
`Ni^(2+) = 1` mole .
Now 0.5 L will contain
`Ni^(2+) = 1 - 0.54 = 0.46` mole .
Hence molarity = `(n)/(V(L)) = (0.46)/(0.5) = 0.92 mol L^(-1)`.