Introduction:
This question is related to the concept of equilibrium, vapour density, and decomposition of N2O4 to NO2. Vapour density is an important parameter that helps in determining the molecular weight of the gas. It is defined as the ratio of the mass of a given volume of gas to the mass of an equal volume of hydrogen gas at the same temperature and pressure.

Given Information:
N2O4 decomposes to 60% into NO2 at 90C.

Solution:
To find the vapour density of the equilibrium mixture, we need to first calculate the mole fraction of N2O4 and NO2. Let x be the mole fraction of N2O4 and (1-x) be the mole fraction of NO2.

At equilibrium,
N2O4 ⇌ 2NO2

The equilibrium constant for this reaction is given by:
Kp = P(NO2)² / P(N2O4)

At equilibrium, the partial pressure of N2O4 is equal to xP and the partial pressure of NO2 is equal to (1-x)P. Substituting these values in the above equation, we get:

Kp = [(1-x)P]² / [xP]

Solving for x, we get:

x = [P² / (4Kp + P²)]

At 90C, the vapour density of N2O4 is 46 g/mol and the vapour density of NO2 is 30 g/mol. The molecular weight of the equilibrium mixture can be calculated as:

Molecular weight = x(46 g/mol) + (1-x)(30 g/mol)

Substituting the value of x, we get:

Molecular weight = [46 P² / (4Kp + P²)] + [30 (4Kp) / (4Kp + P²)]

Simplifying the above equation, we get:

Molecular weight = [(184Kp + 46P²) / (4Kp + P²)]

Therefore, the vapour density of the equilibrium mixture can be calculated as:

Vapour density = Molecular weight / 2

Conclusion:
In conclusion, the vapour density of the equilibrium mixture can be calculated using the above formula. Knowing the vapour density helps in determining the molecular weight of the gas.

Initial conc. of N2O4=1,,,
initial concept.of NO2=0
final conc.of N2O4 1-0.6=0.4
final conc.of No2=2×0.6=1.2
total conc =0.4+1.2=1.6
molar mass of N2O4=92 g/mol
final molar mass 92/1.6=230/4=57.5g/mol
so vapour density=57.5/2=28.5