Question

NO2 dimerizes to from N2O4. A sample of the mixture has an average RMM of 60. Find the percentage of dimer in the mixture.

Answer 1

rmm of NO2 is 46 and rmm of N2O4 is 92

so 46*x+92*(1-x)=60
46x + 92
-92x =60
46x=32
x=0.696
percent of N2O4 = (1-0.696)*100%=30.4%