Question

The density of a 0.0122 M KMnO4 is 1.037 g/mL. Suppose 26.35 g of 0.0122 M KMnO4 are required to titrate 1.072 g of a household H2O2 solution. Except for answers to (a) and (e), use E notation. Do not include units, just enter numerical values.

a) calculate the ml of MnO4- added to reach the endpoint.

b) calculate the moles of MnO4- added to reach the endpoint.

c) calculate the number of moles of H2O2 in the sample.

d)Calculate the number of grams of H2O2 in the sample.

e)Calculate the %m/m H2O2 in the household H2O2 solution

Answer 1

Ans :

a)

Density = mass / volume

putting values :

1.037 g/mL = 26.35 g / V

V = 25.41 mL

so mL of MnO4- = 2.541E+1

b)

Moles of MnO4- = molarity x volume (L)

putting values :

= 0.0122 M x 0.02541 L

= 0.00031 mol

So number of moles of MnO4- = 3.1E-4

c)

2 moles MnO4- react with 5 moles H2O2

so 0.00031 mol MnO4- will react with : ( 0.00031 mol x 5 mol) / 2 mol

= 0.000775 mol H2O2

so mol H2O2 = 7.75E-4

d)

Number of grams H2O2 in sample = mol x molar mass

= 0.000775 mol x 34.0147 g/mol

= 0.0264 g

so mass of H2O2 = 2.64E-2

e)

5m/m H2O2 = ( mass H2O2 / mass sample) x 100

= (0.0264 g / 1.072 g ) x 100

= 2.46