25cm3 of a commercial bleach NAOCl is diluted to 250 cm3 . A 25 cm3 of the diluted solution is added to an excess of potassium iodide solution and titrated against 0.200 M sodium thiosulphate solution. The volume used was 18.5 cm3. What I s the concentration of sodium chlorate in the bleach in mol dm3 and grams per 100 cm3?


OCL- + 2I + 2H+ = I2 + Cl + H2O
MY WORKING

MOlarity of sodium thiosulphate = 0.0037 moles I got this by saying in 1000 = 0.200 moles.. so in 18.5 cm3 I got out 0.0037 moles.
The molarity = 0.0037 moles/ 18.5 x 10-3 = 0.2 M

from the equation 2 moles ofd sodium thiosulphate = 0.0037 moles
1 mole of sodium chlorate = 0.0037/2 = 0.00185 moles

molarity of sodium chlorate = 0.00185moes/25 x 10 - 3 = 0.074 M

MOlar mass of NAOCl = 74.5

g dm-3 = molar mass x mol-dm3
= 74.5 x0.074
= 5.513 g dm-3

In 100 g dm-3 I just divided that value by 1000 x 100 and got 0.55 g cm-3

I haven't gone through your work, but I believe the working equation is:

NaOCl + 2KI + H2O = NaCl + I2 + 2KOH

or, ignoring spectator ions (K+ and Na+)

OCl- + 2I- = Cl- + I2 + 2OH-

Make sure you balance charge as well as atoms.