Question

These are what i used.
0.0375M sodium thiosulphate , 2.5% KI .47.2 ml copper sulphate solution.
the average volume of Na2S2O3 is 19.25 cm, i have already calculated the number of moles of thiosulphate which is 7.218 * 10-4 moles of Na2S2O3 and also the number of moles of I2 that reacted with S2O32 which was calculated to be 7.218*10^-4/2 = 3.609 * 10^-4

I have to do the calculations for :

The number of Cu2+ that reacted with I- to produce the number of moles of I2 above
The number of moles of Cu2+ in the volumetric flask
The concentration of Cu2+ in the stock solution

Any kind of help would be really appreciated!

Answer 1

You started in the middle of the question. I'm just guessing that you have performed an experiment in which you determined %Cu in a sample. The equations are Cu^2+ + KI(xs) ==> CuI(s) + I2 and I suppose you titrated the liberated I2 with thiosulfate like this
2S2O3^2- + I2 ==> 2I^- + S4O6^2-
mols S2O3^2-, what you titrated with is M x L = ?. Then convert this to moles Cu.
? mols S2O3^2- x (1 mol I2/2 mol S2O3^2-) x (2 mol Cu/1 mol I2) =? mol Cu.