A mixture of an o2 and n2 o4 has a vapour density 38.3 at 300 kelvin
Answers
Answer:
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
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Explanation:
Answer:
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
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Explanation: