CuSO4 has Cu2+ ion. So, for 1 mole of CuSO4, we will have 1 mole of Cu2+ and for that we will need 2F charge to completely reduce it to Cu. Cu2+ + 2e- ----> Cu So by given condition,2F charge reduce 1 mole of Cu2+ ion So 0.5F charge will reduce,. 1/2*0.5 = 0.25 mole of Cu2+ ionSo , no. Of ion = 0.25* 6.022* 10^23 = 1.5055* 10^23