Question

The vapour density (hydrogen=1) of a mixture consisting of NO2 and N2O4 is 38.3 at 26.7 degree Celcius. Calucute the number of moles of NO2in 100 g of the mixture. Please Explain!!!

Answer 1

density is defined as, d=m/V

100g/(38.3g/mL)=2.61096 mL

seems like very little, but since you didn't provide units, i can only speculate

PV=nRT

(1) P(0.00261L)=n(0.08206)(299.85)

ma= mass of NO2
mb= mass of N2O2

since, 100g=ma+mb -> ma=100g-mb

na= moles of NO2; nb= moles of N2O2
Ma=molar mass of NO2; Mb=molar mass of N2O2

(2) n(T)= na+nb = ma/Ma + mb/Mb = (100g-mb)/Ma + mb/Mb


P(0.00261L)=n(0.08206)(299.85)

->P/n=constant

grouping values as 1 constant:

\[\frac{ P }{\frac{ (100g-m _{N _{2} O _{2}}) }{ 46g/mol } +\frac{ m _{N _{2}O _{2}} }{ 60g/mol }}=9427.46 \frac{ atm }{ mol }\]


thats about where i get stuck, it seems like you would another value, like pressure, mole fraction or something.

Answer 2

The answer is 0.437 mole!