Question #12834

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

Expert's answer

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%

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%

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