how many electrons would be required to deposit 6.35 g of copper at the cathode during electrolysis of an aqeous solution of copper sulphate?

To deposit 1 mole of Cu at cathode from Cu²⁺SO₄​^2- solution = 2 moles of electrons are required
That is 
To deposit 63.5 g /mole of Cu = 2 moles of electrons are required

To deposit 6.35 g = $\frac{6.35}{63.5}\times 2 =\frac{2}{10}=0.2 moles$

Thus No. of electrons required  =0.2 x 6.023 x 10²³ = 1.2046 x 10²³ electrons