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Introduction

At 340 K an...

Question

At $340 K$ and 1 atmospheric pressure $N_{2} O_{4}$ is 66% dissociated into $N O_{2}$ . The volume of $10 g N_{2} O_{4}$ occupied under these conditions is X $l i t r e s$ . Then X is_________.

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Solution

$N_{2} O_{4} ⇌ 2 N O_{2}$
Mole before equilibrium $10$ (Let $α$ be degree of dissociation)
Mole after equilibrium $1 - α 2 α$
$∴$ Total mole at equilibrium $= 1 + α (∵ α = 0.66)$ $= 1 + 0.66 = 1.66$
1 mole of $N_{2} O_{4}$ is taken, mole at equilibrium $= 1.66$
$10 / 92$ mole of $N_{2} O_{4}$ is taken, mole at equilibrium $= \frac{1.66 \times 10}{92} = 0.18$
$∵ P V = (w / m) \cdot R T$
$1 \times V = 0.18 \times 0.0821 \times 340$
$∴ V = 5.03$ litre
Thus, volume of $10 g N_{2} O_{4}$ under above condition is $5.03 l i t r e$ .

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