The number of electrons required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate will be (Given: Atomic mass of Cu= 63.5 u, NA is Avogadro constant)

Amount of Cu to be deposited at the cathode is $n=\frac{m}{M}=\frac{6.35 \mathrm{g}}{63.5 \mathrm{g} {\mathrm{mol}}^{-1}}=0.10 \mathrm{mol}$

Since 2 electrons are required to convert ${\mathrm{Cu}}^{2+}$ ion into Cu, the amount of electrons required to deposit 0.10mol of Cu will be 0.20 mol. The number of electrons required will be (0.20mol) $N_{\mathrm{A}}$ i.e. $\left(N_{\mathrm{A}}/5\right)\mathrm{mol}$.