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Q.
$4 \,g \,H_2, 32\, g \,O_2, 14\, g\, N_2$ and $11\,g \,CO_2$ are taken in a bulb of $500 \,ml$. Which one of these has maximum active mass?
Equilibrium
Solution:
Active mass of $H _{2}=\frac{\frac{4}{2}}{\frac{500}{1000}}=4$
Active mass of $O _{2}=\frac{\frac{32}{32}}{\frac{500}{1000}}=2$
Active mass of $N_{2}=\frac{\frac{14}{28}}{\frac{500}{1000}}=1$
Active mass of $CO _{2}=\frac{\frac{11}{44}}{\frac{500}{1000}}=0.5$