200mL of pure oxygen is subjected to electric discharge,15% of oxygen is converted into ozone.the colume of ozonized oxygen is:

20 m L

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30 m L

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190 m L

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80 m L

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Solution

The correct option is A $20 m L$
15% of 200 ml is
$\frac{200 \times 15}{100} = 30 m l$
$1.5 O_{2} \to O_{3}$
30 ml of $O_{2}$ will give $\frac{30}{1.5}$ ml of $O_{3}$
$= 20 m l$ of $O_{3}$

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Similar questions

Q. A silent electric discharge is passed through pure oxygen. The volume of the gas is reduced by

\frac{100}{3} %

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Q. At T K. 100 litre of dry oxygen is present in a sealed container. It is subjected to silent electric discharge, till the volumes of oxygen and ozone become equal. What is the volume (in litre) of ozone formed at T K?

When 20 mL of a sample of ozonised oxygen is passed through silent electric discharge, 15 mL of pure ozone gas is obtained. What was the volume of ozone in the original sample?

60

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200mL of pure oxygen is subjected to electric discharge,15% of oxygen is converted into ozone.the colume of ozonized oxygen is:

The correct option is A 20mL15% of 200 ml is 200×15100=30ml1.5O2→O330 ml of O2 will give 301.5 ml of O3=20ml of O3

The correct option is A $20 m L$
15% of 200 ml is
$\frac{200 \times 15}{100} = 30 m l$
$1.5 O_{2} \to O_{3}$
30 ml of $O_{2}$ will give $\frac{30}{1.5}$ ml of $O_{3}$
$= 20 m l$ of $O_{3}$