A current of 9.65 A is passed for 3 hours between nickel electrodes in 0.5 L of a 2 M solution of Ni(NO₃)₂. The molarity of the solution after electrolysis would be (atomic mass of Ni = 58)

Number of coulombs of current passed = 9.65 $\times$3600 $\times$3 = 104220

Number of moles of electrolyte reduced = $\frac{104220\times 0.5}{96500}$=0.54

Moles of Ni(NO₃)₂ remaining after electrolysis = 1-0.54 =0.46

Concentration = 0.46 $\times$2 =0.92