2KMnO₄ ⟶ K₂MnO₄ + MnO₂ + O₂

K₂MnO₄ + MnO₂ is the solid residue.

Potassium permanganate was heated in a 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 H₂ under the same conditions of temperature and pressure has a mass of 0.0825 g. Calculate the relative molecular mass of oxygen.

2KMnO₄ ⟶ K₂MnO₄ + MnO₂ + O₂

Loss in mass = 1.32 g = 1 lit of oxygen

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{1.32}{0.0825} \\[0.5em] = 16 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 16 = 32 g

Hence, relative molecular mass of oxygen is 32 g