[lenin]

In order to standardise a solution of sodium thiosulphate, a solution potassium iodate was made by dissolving 0.980g in water to give exactly 250cm^3 of solution. when 20.00cm^3 of this solution was added to an access of acidified potassium iodide solution, 21.35cm^3 of the sodium thiosulphate solution was requires to react with the liberated iodine. what was the concentration of the original sodium thiosulphate solution?

edit: ok equation is 2S2O3 +I2 -->2I- +S406 + I2

moles of iodate is ...don't know

dont know, confused

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

Start with reaction equations.

[Arkcon]

edit: ok equation is 2S2O3 +I2 -->2I- +S406 + I2

moles of iodate is ...don't know

dont know, confused

OK, a complete balanced equation will help you more.  You've been given grams of sodium iodate, and an amount of water its dissolved in, and amount of that solution used.  You can figure out moles of sodium iodate, and the balanced equation will relate that to moles of the other reactants.