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 the answers to (a) and (e), use E notation.

(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 nuber of grams of H₂O₂  in the sample

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

Answer 1

a)

25.41ml

Explanation

Density of KMnO4 solution = 1.037g/ml

mass of KMnO4 solution = 26.35g

volume of KMnO4 solution = 26.35g/1.037g/ml = 25.41ml

b)

3.10×10^-4 mol

Explanation

moles of MnO4^- added = (0.0122mol/1000ml)×25.41ml = 3.100×10^-4mol

c)

7.750×10^-4mol

Explanation

the reaction between H2O2 and KMnO4 is

5H2O2 + 2MnO4^- + 6H⁺ -------> 5O2 + 2Mn²⁺   + 8H2O

stoichiometrically, 2moles of MnO4^- react with 5moles of H2O2

No of moles of MnO4^- consumed = 3.100 ×10^-4

No of moles of H2O2 present in the sample = (5/2)×3.100×10^-4 = 7.750×10^-4 mol

d)

2.637×10^-2g

Explanation

molar mass of H2O2 = 34.02g/mol

no of moles of H2O2 = 7.750×10^-4 mol

mass of H2O2 present in the sample = 34.02g/mol × 7.750×10^-4mol = 2.637×10^-2g

e)

2.460%

Explanation

% mm = (mass of H2O2/mass of sample)×100

= (2.637×10^-2g/1.072g)×100

= 2.460%