Calculatevolume fractionhydrogen and helium in a mixturecontaining 20 %( by weight)hydrogen .

given:
ω(Н2) = 20 %
find:
φ(Н2 and Не) -?
Solution:
М(Н2) = 2 g/mol
m(Н2) = ν(Н2) • М(Н2) =  20 mol • 2 g/mol = 40 g
М(N2) = 28 g/mol
М(NН3) = 17 g/mol
Vm= 22,4 л/mol
of ν=m/М  = V/V_m  have
V=(m •V_m)/М  ;V(N₂)=(62,2 g• 22,4 lit/mol)/(28 g/mol)  =49,76 lit;
V=(m •V_m)/М  ;V(NН₃)=(37,8 g• 22,4 g/mol)/(17 g/mol)  =50,47 lit;
V = 49,76 + 50,47 = 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.