1. Tardigrade
  2. Question
  3. Chemistry
  4. The vapour density of PCl5 is 104.25 but when heated to 230°C, its vapour density is reduced to 62. The degree of dissociation of PCl5 at this temperature will be

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Q. The vapour density of $PCl_{5}$ is $104.25$ but when heated to $230°C$, its vapour density is reduced to $62.$ The degree of dissociation of $PCl_{5}$ at this temperature will be

Solutions

Solution:

$PCl_{5}(g) \rightleftharpoons PCl_{3} +Cl_{2} (g) $
$1+\alpha =\frac{104.25}{62} 1+\alpha=1.68$
$\therefore \alpha=0.68 \,or \,68\%$