a) Mole quantity of Na2S2O3 in 96.40 mL of 0.010 solution (n = CV):
nNa2S2O3 = 0.010 M 96.40 mL = 0.964 mmole
Note: We use a unit of millimole (mmole) which is 1/1000 of mole because we measure volumes in millilitres (mmoles are more convenient to use here).
b) According to the stoichiometry of the reaction:
2 moles of Na2S2O3 react with 1 mole of I2 , then
0.95 mmole of Na2S2O3 react with x mmole of I2
x = 0.964/2 = 0.482 mmole of I2
it means 10.00 mL of iodine solution contained 0.482 mmole of I2
c) Concentration of iodine solution (n=CV; C=n/V):