[meenu]

The BP requires Hydrogen Peroxide Mouthwash to contain 5.0 to 7.0% m/v H2O2. 10.0 mL of a sample was diluted to 100.0 mL with water. To 10.0 mL of this diluted solution, 20 mL of 1 M sulphuric acid was added and then titrated with 0.0203 M potassium permanganate solution. 37.63 mL was required to reach the end-point. Given that each mL of 0.02 M KMnO4 is equivalent to 1.701 mg H2O2, calculate the % m/v H2O2 content of the mouthwash sample. Does the sample comply with the BP requirement?

few initial steps im  confused 

1- What happens on dilution ? 
( i think concentration changes if we dilute the solution from 10 to 100 mL.
 But only 10 mL of diluted  solution is taken.then what hapens to ammount of H2O2)
2 - what happens when  20 mL of 1M sulphuric acid was added 

 for titration i know steps 
volume of 0.02 M potassium permagnate solution
 0.02 M x V mL =  0.0203 M x 37.63 mL
V mL =  38.19 mL

now 1 ml of 0.02 M  KMnO4 solution = 1.701 mg H2O2
so 38.19 ml of solution = 38.19 x 1.701 mg H2O2
                              = 64.9 mg H2O2
3- How to convert this to the ammount initially present in 10 mL of sample ? (before dilution )

[AWK]

1- What happens on dilution ? 
( i think concentration changes if we dilute the solution from 10 to 100 mL.
 But only 10 mL of diluted  solution is taken.then what hapens to ammount of H2O2)
2 - what happens when  20 mL of 1M sulphuric acid was added

Just dilution. You should remember you titrate 0.1 of  H2O2 content. Titration needs an acidic environment (you use an excess of H2SO4).
Start from a balanced reaction
H2O2 + KMnO4 + H2SO4 =