Wednesday
October 22, 2014

Homework Help: Chemistry

Posted by Marcus on Friday, April 26, 2013 at 8:24pm.

To illustrate the calculation, assume that 21.95 mL of a thiosulfate solution is required to reduce the iodine liberated by 20.85 mL of 0.0200M KIO3 and that 18.63 mL of the same thiosulfate solution reacts with the iodine replaced by 20.00 mL of an unknown solution of sodium hypochlorite.

(This is in the sample calculation given in my lab manual, for reference.)

And the question I'm given in my lab manuel is:

See the question in your lab manual. Suppose that you determine that the molarity of your sodium thiosulfate solution is 0.100 M. Suppose that 10.02 mL of the same thiosulfate solution reacts with the iodine replaced by 20.00 mL of an unknown solution of sodium hypochlorite. Calculate the weight, in grams, of available Cl in a liter of the unknown solution. Enter your answer as a decimal number with 3 decimal places.

This is what I did.

Since M thio = 0.100,
2 x M OCl x 0.0200L = 0.100 M x .01002L

M OCl = .02505

0.02505 mol NaOCl x 1 mole Cl x 35.45 g Cl x 1000 mg = 888.0225 mg Cl/L =888.023 ppm

But it says the answer is incorrect. So what would I be doing wrong here?

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