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प्रश्न
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.
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उत्तर
\[\ce{2KMnO4 -> K2MnO4 + MnO2 + O2}\]
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.
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When heated, potassium permanganate decomposes according to the following equation :
\[\ce{2KMnO4 -> \underset{\text{solid residue}}{K2MnO4 + MnO2} + O2}\]
(a) Some potassium permanganate was heated in the 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.
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