10 grams of copper sulphate solution is electrolysed using 0.1 Faraday. Weight of cupric ions left after the electrolysis is

mass of Cu present in 10 gm of CuSO₄ =$\frac{63.5\times 10}{159.5}=3.98 gm$

0.1 F deposits = 3.125 gm

Cu⁺² remained = 3.98 - 3.125 = 0.856 gm $\cong$0.8 gm