A current of `96.5` A is passed for 18 min between nickel electrodes in 500 mL solution of `2M Ni(NO_(3) ) _(2)` . The molarity of solution after electrolysis would be:

Options

A. `0.46` M
B. `0.92M`
C. `0.625M`
D. `1.25M`

Answer

B

Explanation

`Ni(NO_(3))_(2)` in 500 mL of `2M Ni(NO)_(2)`
`(2 xx 500)/(1000)= 1mol`
`96.5 xx 18 xx 60= 104220C`
`Ni^(2+) + 2e^(-) rarr Ni^(2) xx 96500`
`Ni(NO_(3))_(2)` 104220C will deposit `(104220)/(2 xx 96500)= 0.54` mL
`1-0.54 = 0.46`
Molarity is `2 xx 0.46 = 0.92` mol/L