calculate the time for deposit 1.17g of ni at cathode when a current of 5a was passed through the solution of ni(no3)2?

Molar mass of Ni = 58.70 g/mol
Number of moles of Ni deposited = (1.17 g) / (58.70 g/mol) = 0.01993 mol

Ni²⁺(aq) + 2e⁻ → Ni(s)
To deposit 1 mole of Ni, 2 moles of e⁻ is passed.
Number of e⁻ passed = (0.01993 mol) × 2 = 0.03986 mol

1 mole of e⁻ carries 96500 C of charges.
Amount of charges passed = (96500 C/mol) × (0.03986 mol) = 3846 C

Time taken = (3846 C) / (5 A) = 769 s = 12.8 min

moles Ni:

1.17 g / 58.6934 g/mol = 0.019934 mol

The relevant chemical equation:

Ni^2+ + 2e- ---> Ni

moles electrons required:

0.019934 mol * 2 = 0.039868 mol

Coulombs of charge in the electrons:

0.039868 mol * 96485 C/mol = 3846.664 C

Time required:

5 A = 5 C/s

3846.664 C / 5 C/s = 769.3328 s

Rounding off to three sig figs gives us 769 s for the answer.