A current of 0.75 A is passed through an acidic solution of ${\mathrm{CuSO}}_4$ for 10 minutes. The volume of oxygen liberated at anode (at STP) will be 

Volume equivalent of O₂ at STP = 5.6 lit

Coulombs passed = 0.75 $\times$10 $\times$60 =450 C

96500 Coulombs = 5600 ml

450 Coulombs = ?

volume of O₂ liberated at STP =$\frac{450\times 5600}{96500}=$26.1ml