Ratio of active masses of $22 g C O_{2}, 3 g H_{2}$ and $7 g N_{2}$ in a gaseous mixture:

22 : 3 : 7

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0.5 : 3 : 7

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1 : 3 : 1

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1 : 3 : 0.5

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Solution

The correct option is D 1 : 3 : 0.5
Let the volume of gaseous mixture is $V$
$C O_{2} = \frac{22}{44} = 0.5 m o l e s$

$H_{2} = \frac{3}{2} = 1.5 m o l e s$

$N_{2} = \frac{7}{28} = 0.25 m o l e s$

Total moles = $0.5 + 1.5 + 0.25 = 2.25$

Ratio of active masses $= 0.5 / 2.25 : 1.5 / 2.25 : 0.25 / 2.25 = 1 : 3 : 0.5$

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