Physical Chemistry Problem Help?

Molar mass of N₂O₄ = (14.0×2 + 16.0×4) g/mol = 92.0 g/mol
Molar mass of NO₂ = (14.0 + 16.0×2) g/mol = 46.0 g/mol

Consider the final reaction mixture at equilibrium:
Pressure, P = 0.597 bar
Density, m/V = 1.477 g dm⁻³
Gas constant, R = 0.08314 bar L mol⁻¹ K⁻¹
Temperature, T = (273 + 25) K = 298 K
molar mass, M = ? g/mol

PV = nRT and n = m/M
Then, PV = (m/M)RT
M = (m/V)RT/P

Average molar mass = 1.477 × 0.08314 × 298 / 0.597 g/mol = 61.3 g/mol

Let α be the degree of dissociation (0 ≤ α ≤ 1).
                     N₂O₄(g) ⇌ 2NO₂(g)
Initial (mol):       1              0
Change (mol):  -α           +2α
Eqm (mol)       1 - α          2α

Average molar mass:
[92.0(1 - α) + 46.0(2α)] / [(1 - α) + 2α] = 61.3
(92.0 - 92.0α + 92.0α) / (1 + α) = 61.3
92.0 / (1 + α) = 91.3
92.0 = 61.3 + 61.3α
61.3α = 30.7
The degree of dissociation, α = 0.5

At equilibrium:
𝑃N₂O₄ = 0.597 × (1 - α)/(1 + α) bar = 0.597 × (1 - 0.5)/(1 + 0.5) bar = 0.199 bar
𝑃NO₂ = 0.597 × 2α/(1 + α) bar = 0.597 × 2(0.5)/(1 + 0.5) bar = 0.398 bar

Kp = 𝑃NO₂²/𝑃N₂O₄ = (0.398² bar)/(0.199 bar) = 0.796 bar