What is the ratio of the active masses of 22 | vedclass

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(D) Active mass is defined as molar concentration,which is $	ext{moles} / 	ext{volume} (V)$.$(i)$ Moles of $CO_2 = \frac{22 \ g}{44 \ g/mol} = 0.5 \

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$1 : 3 : 0.5$

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(D) Active mass is defined as molar concentration,which is $	ext{moles} / 	ext{volume} (V)$.$(i)$ Moles of $CO_2 = \frac{22 \ g}{44 \ g/mol} = 0.5 \ mol$. Active mass $[CO_2] = \frac{0.5}{V}$.$(ii)$ Moles of $H_2 = \frac{3 \ g}{2 \ g/mol} = 1.5 \ mol$. Active mass $[H_2] = \frac{1.5}{V}$.$(iii)$ Moles of $N_2 = \frac{7 \ g}{28 \ g/mol} = 0.25 \ mol$. Active mass $[N_2] = \frac{0.25}{V}$.Ratio $[CO_2] : [H_2] : [N_2] = \frac{0.5}{V} : \frac{1.5}{V} : \frac{0.25}{V} = 0.5 : 1.5 : 0.25$.Multiplying by $4$,we get $2 : 6 : 1$. However,checking the provided options,the ratio $0.5 : 1.5 : 0.25$ simplifies to $1 : 3 : 0.5$.

What is the ratio of the active masses of $22 \ g$ $CO_2$,$3 \ g$ $H_2$,and $7 \ g$ $N_2$ in a gaseous mixture?

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