The vapour density of is . When heated to , it dissociates as and the observed vapour density reduce — Chemical Equilibrium Chemistry Question
Question
The vapour density of $PCl_5$ is $104.16$. When heated to $230^\circ\text{C}$, it dissociates as $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ and the observed vapour density reduces to $62$. What is the degree of dissociation of $PCl_5$ at this temperature?
Answer: B
💡 Solution & Explanation
For $PCl_5 \rightleftharpoons PCl_3 + Cl_2$, the number of product moles $n = 2$. Using the formula $\alpha = \frac{D - d}{(n - 1)d}$, where theoretical $D = 104.16$ and observed $d = 62$. $\alpha = \frac{104.16 - 62}{(2 - 1) \times 62} = \frac{42.16}{62} \approx 0.68$. Therefore, the percentage dissociation is $68\%$.
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