$4 g$ $H_{2}$ , $32 g$ $O_{2}$ , $14 g$ $N_{2}$ and $11 g$ ${C O}_{2}$ are taken in a bulb of $500 m l$ . Which one of these has maximum active mass?

H_{2}

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O_{2}

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N_{2}

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{C O}_{2}

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Solution

The correct option is A $H_{2}$
Answer A.
$4 g$ of $H_{2} = \frac{4}{2} = 2 m o l e s$
$32 g$ of $O_{2} = \frac{32}{32} = 1 m o l e s$
$14 g$ of $N_{2} = \frac{14}{28} = 0.5 m o l e s$
$11 g$ of $C O_{2} = \frac{11}{44} = 0.25 m o l e s$
Active mass of,
$H_{2} = \frac{2 m o l e s}{0.5 L} = 4 m o l e s / L ⟶ M a x i m u m - a c t i v e - m a s s .$
$O_{2} = \frac{1}{0.5} = 2 m o l e s / L$
$N_{2} = \frac{0.5}{0.5} = 1 m o l e s / L$
$C O_{2} = \frac{0.25}{0.5} = 0.5 m o l e s / L$

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