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Vapour density definition-Vapour density calculation, Vapour density of the mixture of gases: Definition, derivation, examples, practice problems & FAQs

Vapour density definition-Vapour density calculation, Vapour density of the mixture of gases: Definition, derivation, examples, practice problems & FAQs

Hom many of you have thrown smoke bombs while playing virtual games like Call of Duty or Counter-Strike? Why did you use it? It gave you temporary relief from the opposition player's vision who feels choked and suffocated.

Have you wondered what could have occurred in order to feel this sensation?

Or what could be the constituents of such a dangerous mixture? 

One of the common ingredients used in smoke bombs is bleaching powder. Bleaching powder is used as a smoke bomb because it releases chlorine gas. The vapour density of chlorine gas is approx 2.44 times higher than normal breathable atmospheric gas in other words, we can say chlorine gas is approx 2.44 times heavier than normal breathable atmospheric gas. Chlorine gas settles down due to its higher weight than air, a person can feel suffocation.


Table of contents

  • Vapour density
  • Degree of dissociation
  • Vapour density of mixture
  • Practice problems
  • Frequently asked questions-FAQs

Vapour density:


We know, according to Avogadro’s hypothesis, All gases having equal volume have the same number of gaseous molecules (not atoms) at same temperature and pressure.

[at the same Temperature and Pressure, the molar volume of gas = molar volume of H2 gas ]


Degree of dissociation (a)


Vapour density of mixture

Relation between Vapour density and degree of dissociation (),

A(g) ⇋ nB(g)

Initial moles C 0

Moles at equilibrium C - C𝛼 nC𝛼

Where 𝛼 is the degree of dissociation

Total moles at equilibrium = C - C𝛼 + nC𝛼 =C[1+(n-1)𝛼]

Assume, all calculation for an ideal gas, we can use, PV = nRT

Also, PM = dRT

Where, P = pressure

V = volume

N = number of moles

R = universal gas constant

T = temperature

M = Molar mass

d = density



Practice problems:

Q 1. Calculate vapour density of nitrogen gas?

a. 14
b. 18
c. 28
d. 7

Answer: (A)


Q 2. Find the ratio between the vapour density of methane gas and oxygen gas?


Answer: (A)
Methane gas chemical formula = CH4


Q 3. In the dissociation of N2O4 into NO2, (1 + 𝝰) values with the vapour densities ratio (Dd) is as given by: [𝝰 - degree of dissociation, D - vapour density before dissociation, d - vapour density after dissociation]






Answer: (A)


Represents a straight line equation y = mx + c, where m = slope and c = intercept

Q 4. The degree of dissociation is 0.5 at 800 K and 2 atm for the gaseous reaction 

PCl5 (g) ⇌ PCl3 (g) + Cl2 (g). Assuming ideal behaviour of all the gases. Calculate the density of equilibrium mixture at 800 K and 2 atm.

a. 208.5
b. 104.25
c. 69.5
d. 52.125

Answer: (C)
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)

Molar mass of PCl5 = 208.5 g mol-1

Vapour density PCl5(D) = 104.25

Let vapor density of mixture at equilibrium = d


d = 69.5

Frequently asked questions-FAQs

Q 1. What is the unit of vapour density?
Vapour density is a unitless quantity because the density of any gas with respect to hydrogen gas is termed vapour density.


You should not confuse with formula Molar mass divided by 2.

Q 2. Can we calculate vapour density with respect to gas other than hydrogen?
according to standard vapour density is calculated with respect to the density of hydrogen gas. If we calculate with respect to another gas then it is termed relative density. 

Q 3. What is the relation between % dissociation and degree of dissociation?
degree of dissociation = 1

If, % dissociation is 20 %, means its degree of dissociation = 1

Q 4. Are all compounds have varying degrees of dissociation like 5 % to 80 % according to condition?
No, for the same compound degree of dissociation vary only for weak electrolyte but strong electrolytes have always a degree of dissociation = 1. 

Related Topics



Avagadro’s Hypothesis

Parts per million

Atomic Mass

Definition of Equivalent weight


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