Chemistry Calculations?

Question

15.0 cm^3 of a bleach sample was diluted to 250 cm^3 with deionised water. 25.0 cm^3 portions of the solution were treated with excess acidified potassium iodide solution and then titrated with 0.1 M sodium thiosulphate solution. The average titre was found to be 25.2 cm^3. Calculate the concentration of sodium hypochlorite in the original bleach sample.

NaOCl + 2 KI + H2SO4 --> I2 + H2O + NaCl + K2SO4

2Na2S2O3 + I2 --> Na2S4O6 + 2NaI

Answer 1

1 mole NaOCl --> 1 mole I2 which reacts with 2 moles Na2S2O3

0.10 M Na2S2O3 x 25.2 mL = 2.52 millimoles Na2S2O3 (0.00252 moles Na2S2O3)

0.00252 moles Na2S2O3 is equivalent to 0.00126 moles NaOCl (0.00252 / 2)

The 25.0 mL portions of the diluted NaOCl = 1/10 of the total NaOCl (25.0/250.0 mL) which equals 1/10 of the 15.0 mL

NaOCl sample.

1.500 mL x Molarity = 1.26 millimoles

Molarity of the original NaOCl = 0.840 M

Answer 2

I am editing out my answer ...

Ronald got  the same answer ,

& got it first