If 96500 coulomb electricity is passed through CuSO4 solution, it will liberate

At the cathode

Cu²⁺(aq)+2e⁻→ Cu(s)

At the anode 4OH⁻(aq) →2H₂​O(l)+O₂​(g)+4e−

Faraday’s constant = 96500C/mol

To deposit 1g of Cu we need 2 X 96500 C 
So, 96500 C will deposit 0.5 mole = 0.5 x 63.5= 31.76g.