Hi guys, I need help with a sample exam question. For the first one I got .101 m

Question

Hi guys, I need help with a sample exam question. For the first one I got .101 mol but I don't know how to approach the second part.

You dissolve 12.5 g of sodium thiosulfate (Na2S2O3 . 5H2O) in enough water to make 500.0 mL of solution. You use this solution to titrate .5001 gram KIO3 solution. Remember that 1 moles KIO3 react with 6 moles Na2S2O3. 25 mL of Na2S2O3 . 5H2O reacts with the KIO3.

a) What is the concentration in mole/L of the Na2S2O3 solution?

b) A 3.00 ml bleach sample (NaOCl) is reacted with 31.52 mL of this Na2S2O3 solution. Remember that 1 ml of NaOCl reacts with 2 moles Na2S2O3 What is the concentration, in moles/liter of the NaOCl solution?

Explanation / Answer

(a) Molarmass of Na2S2O3 . 5H2O is

= (2xAt.mass of Na ) + ( 2xAt.mass of S) + ( 3xAt.mass of O) + 5[(2xAt.mass of H)+At.mass of O]

= (2x23)+(2x32)+(3x16) +5[(2x1)+16]

= 248 g/mol

Given mass of Na2S2O3 . 5H2O is 12.5 g

So number of moles , n = mass/molar mass

                                  = 12.5 / 248

                                  = 0.05 moles

So Molarity of Na2S2O3 . 5H2O is , M = number of moles / volume in L

                                                       = 0.05 mol / 0.5L

                                                       = 0.100 mol/L

(b) Given that 1 moles NaOCl react with 2 moles of Na2S2O3

                M moles NaOCl react with 0.05 moles of Na2S2O3

               M = ( 0.05x1) /2

                   = 0.025 moles of NaOCl

So Molarity of NaOCl , M = number of moles of NaOCl / volume in L

                                     = 0.025 mol / (3.00/1000)L

                                     = 8.333 mol/L

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