An unknown compound A dissociates at $500^{\circ} C$ to give products as follows:
$A (g) ⇌ B (g) + C (g) + D (g)$
Vapour density of the equilibrium mixture is 60 when it dissociates to the extent to $20^{%}$ . What will be the molecular weight (in grams) of Compound A

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Solution

$A (g) ⇌ B (g) + C (g) + D (g)$
$α = 0.2$ Vapour density =60
$\Rightarrow M_{o b s} = 120 Δ n_{g} = 3 - 1 = 2$
$α = \frac{M_{T h} - M_{o b s}}{M_{o b s} (n - 1)}$
$0.2 = \frac{M_{T h} - 120}{120 (3 - 1)}$
$M_{T h} = (0.2 \times 240) + 120$
$= 48 + 120$
= 168

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