# Problem regarding vapour density of a mixture

• tbn032
The mole fraction of ##N_2O_4## in the gas mixture would now be ##(1+\frac{x}{100})^3=1+\frac{1}{100}+\frac{2}{100}=53.33%f

#### tbn032

The vapour density of N204 at certain temperature is 30. Calculate the percentage of dissociation of N204 at this temperature. N2O4(g)⇌2NO2(g)?
I am unable to understand the concept behind vapour density of the mixture.
Currently, I understand that
2 x vapour density=molar mass.
Vapour density =
mass of n molecules of gas ÷ mass of n molecules of hydrogen.
Vapour density
= molar mass of gas ÷ molar mass of H₂.
I am unable to apply the above formula because a mixture does not have molar mass.
And I am also not able to understand that if
2 x vapour density=molar mass
then in the question, molar mass of N204 at the certain temperature given would be 60 instead of 92.

mixture does not have molar mass.

True, but it has an apparent molar mass, one that directly relates to the gas density. "Molar mass of the gas that would have given density".

2 x vapour density=molar mass

No idea where you got it from, but it is wrong. Check the units.

Imagine 1 mole of a gas of molar mass mM at STP. What is its mass? What is its volume? What is its density?

Can you reverse the process? Calculate the apparent molar mass from the density?

Note:

The vapour density of N204 at certain temperature is 30

doesn't make much sense. 30 of what? Miles per hour? Pounds per gallon? Ounces per pint? Again: units are crucial part of the information.

doesn't make much sense. 30 of what? Miles per hour? Pounds per gallon? Ounces per pint? Again: units are crucial part of the information.
"Vapour density is the density of a vapour in relation to that of hydrogen. It may be defined as mass of a certain volume of a substance divided by mass of same volume of hydrogen.

Vapour density = mass of n molecules of gas / mass of n molecules of hydrogen.
Vapour density = molar mass of gas / molar mass of H2
vapour density = molar mass of gas / 2.016
vapour density = 1⁄2 × molar mass
(and thus: molar mass = ~2 × vapour density)
For example, vapour density of mixture of NO2 and N2O4 is 38.3.
Vapour density is a dimensionless quantity."
https://en.wikipedia.org/w/index.php?title=Vapour_density&oldid=1082641758

OK, I would call that a relative vapour density, but English is my second language, so there can be some differences in the nomenclature. Ideas behind are identical though.

In general when dealing with density I feel like it is much safer to use absolute values, measured in mass/volume units. While they may require an additional step during some types of calculations they are unambiguous.

Have you moved ahead with the question?

My reactions are the same as Borek. Not sure this is not an unnecessarily pedantic concept and definition that one could well get through chemistry without.
To treat it how it deserves what is this vapour density for 100% dissociated N2O4, I.e. 100% NO2?

For where does the 2 come from, look up the definition in Wikipedia. You will note that it is actually 2.016 - all to do with the exact atomic mass given to the hydrogen atom which is not what it once was, not exactly 1

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Suppose you started out with one mole of ##N_2O_4## and x% dissociated to ##NO_2##. Then you would have ##1-\frac{x}{100}## moles of ##N_2O_4## left, and ##\frac{2x}{100}## moles of ##NO_2## would have formed. So, the total number of moles present would now be ##1+\frac{x}{100}##.

Based on this, what would now be the mole fraction of ##N_2O_4## in the gas mixture? Of ##NO_2##? In terms of x, what would be the average molecular weight of the gas mixture?