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. 0.92 M
C. 0.625 M
D. 1.25 M

Answer

B

Explanation

Moles of `Ni(NO_(3))_(2)` in 500 mL of 2M `Ni(NO_(3))_(2)` is
`=(2xx500)/(1000)=1` mol
Charge `=96.5xx18xx60=104220C`
`Ni^(2+)+2e^(-)toNi`
`2xx96500`C deposits 1 mol of `Ni(NO_(3))_(2)`
`therefore104220C` will deposit=`(104220)/(2xx96500)=0.54mol`
so, moles `Ni(NO_(3))_(2)` left
`=1.0-0.54=0.46`mol
Thus, molarity of `Ni(NO_(3))_(2)` solution
`=2xx0.46=0.92mol//l`