Question

When heated, potassium permanganate decomposes according to the following equation:

\[\ce{2KMnO4 -> \underset{solid residue}{K2MnO4 + MnO2} + O2}\]

Some potassium permanganate was heated in test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.

Answer 1

\[\ce{2KMnO4 -> K2MnO4 + MnO2 + O2}\]

Loss in mass = 1.32 g = 1 lit of oxygen

Vapour density of gas = `"Wt. of certain volume of gas"/"Wt. of same volume of H"_2`

= `1.32/0.0825`

= 16 g

Molecular weight = 2 × Vapour density

= 2 ×16

= 32 g

∴ Relative molecular mass of oxygen is 32 g.