(molecular mass = M) dissociates: (g) ⇌ (g) + (g). If total pressure is P and density is d at temperature T K, the degree of dissociation of is:

Question

$PCl_5$ (molecular mass = M) dissociates: $PCl_5$(g) ⇌ $PCl_3$(g) + $Cl_2$(g). If total pressure is P and density is d at temperature T K, the degree of dissociation of $PCl_5$ is:

Prashant Kotian · Accepted Answer

Correct answer: B. Step 1 - Set up the ICE table For $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$, starting with 1 mol, let $\alpha$ = degree of dissociation: | | PCl5 | PCl3 | Cl2 | |--|--|--|--| | Initial | 1 | 0 | 0 | | Change | -α | +α | +α | | Equilibrium | 1-α | α | α | Total moles at equilibrium: $n_{\text{total}} = 1 + \alpha$ Step 2 - Relate observed molar mass to degree of dissociation Total mass is conserved = M (initial molar mass of PCl5). So: $$M_{\text{obs}} = \frac{M}{1 + \alpha} \implies \alpha = \frac{M}{M_{\text{obs}}} - 1$$ Step 3 - Express observed molar mass in terms of density From the ideal gas equation in density form ($P = \frac{dRT}{M_{\text{obs}}}$): $$M_{\text{obs}} = \frac{dRT}{P}$$ Step 4 - Solve for α $$\alpha = \frac{M}{M_{\text{obs}}} - 1 = \frac{M}{\frac{dRT}{P}} - 1 = \boxed{\frac{PM}{dRT} - 1}$$ Step 5 - Explain each option * **Option (A) PM/dRT**: Incorrect. This equals $M/M_{\text{obs}} = 1+\alpha$, not $\alpha$ itself. * **Option (B) PM/dRT - 1**: Correct. Equals $M/M_{\text{obs}} - 1 = \alpha$. * **Option (C) dRT/PM - 1**: Incorrect. This equals $M_{\text{obs}}/M - 1 = 1/(1+\alpha) - 1$, which is negative. * **Option (D) dRT/PM**: Incorrect. This equals $M_{\text{obs}}/M = 1/(1+\alpha)$.

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