$K M n O_{4}$ decomposes on heating according to the equation-

$2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$

On heating $K M n O_{4}$ , 1 litre of oxygen was collected at room temperature and 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.

16 g

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8 g

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32 g

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24 g

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Solution

The correct option is C
32 g

Vapour density of oxygen = $\frac{W e i g h t o f 1 l i t r e o f g a s O_{2}}{W e i g h t o f 1 l i t r e o f H_{2}}$

Weight of 1 litre of oxygen = 1.32 g [since on collecting 1 litre of $O_{2}$ the loss in mass is 1.32 g]

Vapour density of oxygen = $\frac{1.32}{0.0825}$ = 16

Relative molecular mass of oxygen = 2 $\times$ Vapour density = 32 g.

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Similar questions

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

Some 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 hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.

Q. Some 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 hydrogen under the same conditions of temperature and pressure has a mass of

0.0825 g

.
Calculate the relative molecular mass of oxygen.

When heated, potassium permanganate decomposes according to the following equation:
2KMnO₄ $\overset{}{\to}$ K₂MnO₄ + MnO₂ + O₂
(a) Some 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 hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
(b) Given that the molecular mass of potassium permanganate is 158 g, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres). (K = 39, Mn = 55, O = 16)

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

$2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$

Given that the molecular mass of potassium permanganate is $158$ , what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of $15.8$ g of potassium permanganate? (Molar volume at room temperature is $24$ liters.)

Q. The molecular mass of