# A current of $9.65A$ is passed for 3h between nickel electrodes in 0.5L of a 2M solution of $Ni(NO_3)_2$.The molarity of solution after electrolysis would be

$\begin{array}{1 1}0.46M\\0.625M\\0.92M\\1.25M\end{array}$

We have $Q=It=(9.65A)(3\times 60\times 60s)=104220C$
Amount of Ni deposited =$\large\frac{104220C}{2\times 96500Cmol^{-1}}$$=0.54$ mol
Amount of $Ni^{2+}$ left behind=1 mol-0.54 mol=0.46 mol
Molarity of solution =0.46mol/0.5L=0.92$molL^{-1}$