If 5.71 mL of the S2O32- solution are required to reach the endpoint of the titration, what is the molarity of S2O32- in the solution? (Use Equation 2.)

The question is incomplete as some details are missing. here is the complete question ; To standardize a thiosulfate (S2O32-) solution, 10.00 mL of a  1.25×10−2 M   IO3- solutionare treated with excess I- to generate I3- according to Equation 3 on page 162 in the lab manual. If  5.71 mL of the S2O32- solution are required to reach the endpoint of the titration, what is the molarity of thiosulfate in the solution?

Using this equation: I3(aq)+2S2O32-(aq)->3I(aq) + S4O62-(aq)

Explanation:

from the equation ; ClO(aq) + 3I(aq) + H2O(I) > I3(aq) + CI(aq) + 2OH(aq) (Equation 1)

from the equation ;  I3(aq)+2S2O32-(aq)->3I(aq) + S4O62-(aq) (equation 2)

from another equation ; IO3- (aq) + 8I-(aq) + 6H+(aq) > 3I3-(aq) + 3H2O(I) (Equation 3)

The balancing of equation (1) and equation (2) gives equation 3

Given molarity of IO3- (aq) = 1.25M x 10^-2

from the reaction concentration in equation 3; 1mole of IO3- (aq) gives 3moles of I3-(aq)

therefore moles of I3-(aq) = 3 x 1.25 x 10^-2 = 3.75 x 10^-2M

from the reaction in equation 2 ; I3(aq)+2S2O32-(aq)->3I(aq) + S4O62-(aq)

for I3-(aq), C1 = 3.75 x 10^-2M, V1 = 10.0mL, n1 = 1

for S2O32-(aq), C2 = ?, V2 = 5.71mL, n2 = 2

recalling that C1V1/N1 = C2V2/N2

C2 = C1V1 x n2/V2 x n1

Substituting; C2 = 3.75 x 10^-2M x 10.0mL x 2 / 5.71mL x 1

C2 = 0.131M = is the molarity of S2O32-(aq) in the solution