Â
2Cu2+ + 4I- --> 2Cul + I2
2Na2S2O3 + I2 ----------> 2NaI + Na2S4O6
no of moles of Na2S2O3 = molarity * volume in L
                                       Â
= 0.1005*0.025Â Â = 0.0025125moles
2moles of Na2S2O3 react with 1 moles of I2
0.0025125 moles of Na2S2O3 react with =
1*0.0025125/2Â Â = 0.00125625 moles of I2
2Cu2+ + 4I- --> 2Cul + I2
1 moles of I2 obtained from 2 moles of Cu+2
0.00125625 moles of I2 obtained from = 2*0.00125625 = 0.0025125
moles of Cu+2
mass of Cu+2Â Â = no of moles * atomic mass
of Cu
                      Â
= 0.0025125*63.5Â Â = 0.15954g of Cu
percentage of Cu  = 0.15954*100/0.1965  =
81.19%
