Determine the volume fraction of nitrogen in the mixture thereof with ammonia , if the mass fraction of nitrogen in the mixture is 62,2 %.

given:
ω(N2) = 62,2 %
find:
φ(N2) -?
Solution:
mmixture = 100 g
m(N2) = 62,2 g, а m(NН3) = 100 g – 62,2 g = 37,8 g
М(N2) = 28 g/mol
М(NН3) = 17 g/mol
Vm= 22,4 l/mol
of  ν=m/М  = V/V_m   have
V=(m •V_m)/М  ;V(N₂)=(62,2 g• 22,4 l/mol)/(28 g/mol)  =49,76 lit;
V=(m •V_m)/М  ;V(NН₃)=(37,8 g• 22,4 l/mol)/(17g/mol)  =50,47 lit;
Vmixture = 49,76 lit + 50,47 lit  = 100,23 lit
φ(N₂)=V(N₂)/V_mixture =(49,76 lit)/(100,23 lit)  =0,50.
φ(NН₂)=V(NН₃)/V_mixture =(50,47 lit)/(100,23 lit)  =0,50.
Answer:φ(N2)  = 0,50.