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# a mixture of NO2 and N2O4 has a vapour density of 38.3 at 300 K.What is the number of moles of NO2 in 100 gm of mixture?ans-0.437please tell how?

## 2 Answers

8 years ago

The vapor density of a gas is the density of that gas relative to that of hydrogen gas.

Therefore,

Vapor density = (density of NO2/N2O4 mixture)/ (density of H2 gas)
Vapor density = (mass of NO2/N2O4 mixture)/ (mass of H2 gas)
Vapor density = (molar mass of NO2/N2O4 mixture)/ (molar mass of H2 gas)
Vapor density = (molar mass of NO2/N2O4 mixture)/ 2
Molar mass of NO2/N2O4 mixture = Vapor density * 2
= 76.6 g mol^(-1)

Let the percentage of NO2 gas be X, the percentage of N2O4 gas be 1-X

X(molar mass of NO2 gas) + (1-X)(molar mass of N2O4) = 76.6
X(46) + (1-X)(92) = 76.6
X = 33.48%

Mass of NO2 in 100g of mixture = 33.48g
Number of moles of NO2 gas = 33.48/46 = 0.728 mol

I checked thrice but it comes 0.728 mol for NO2 not 0.437!

check yr. anr. plz!

8 years ago
The vapor density of a gas is the density of that gas relative to that of hydrogen gas. Therefore, Vapor density = (density of NO2/N2O4 mixture)/ (density of H2 gas) Vapor density = (mass of NO2/N2O4 mixture)/ (mass of H2 gas) Vapor density = (molar mass of NO2/N2O4 mixture)/ (molar mass of H2 gas) Vapor density = (molar mass of NO2/N2O4 mixture)/ 2 Molar mass of NO2/N2O4 mixture = Vapor density * 2 = 76.6 g mol^(-1) Let the percentage of NO2 gas be X, the percentage of N2O4 gas be 1-X X(molar mass of NO2 gas) + (1-X)(molar mass of N2O4) = 76.6 X(46) + (1-X)(92) = 76.6 X = 33.48% Mass of NO2 in 100g of mixture = 33.48g Number of moles of NO2 gas = 33.48/46 = 0.728 mol I checked thrice but it comes 0.728 mol for NO2 not 0.437!

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