Tuesday
September 1, 2015

Homework Help: chemistry

Posted by angela on Wednesday, August 4, 2010 at 8:17pm.

A stock solution's hypochlorite activity was determined as follows: . -
2.00g of liquid sodium hypochlorite (NaOCl) was weighed into a flask and diluted with 50mL deionized water.10mL of 10% sulfuric acid and 10mL of 10% potasssium iodide were added, followed by titration with 0.1N sodium thiosulfate (Na2S2O3) until colorless.
% sodium hypochlorite
= (mL Na2S2O3)(0.1N)(0.03722)/2.00gNaOCl X 100

[The factor 0.03722 was derived by multiplying molecular weight of NaOCl (74.45g/mol) by 0.001 (mL to L conversion) and dividing by 2(mols of thiosulfate per hypochlorite).]

Please explain why we need to multiply by 0.1N? Thanks.

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