Active mass of 5.6 lit of N₂ at STP is
1)22.4 M
2)0.25 M
3)M
4)4 M

No. of moles in 22.4 l of N₂ = 1 mol
No of moles in 5.6 l = $\frac{1}{22.4}\times 5.6=0.25 \mathrm{mol}$
 

Active mass = conc. in mol L^-1

​If you consider the volume of container as1 l then, 

active concentration  = 0.25 mol in 1 L or 0.25 M

Answer is (2)