What would be the observed vapour density of a completely dissociated sample of ammonium chloride ga — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

$NH_4Cl(g)$ dissociates as $NH_4Cl(g) \rightleftharpoons NH_3(g) + HCl(g)$. Here, 1 mole gives $n=2$ moles of products. For complete dissociation, $\alpha = 1$. Using the formula $\alpha = \frac{D - d}{(n - 1)d}$, we get $1 = \frac{D - d}{d} \Rightarrow d = D - d \Rightarrow 2d = D \Rightarrow d = D/2$. Therefore, the observed vapour density is exactly half of the theoretical vapour density.