The vapour density of PCl_5 is 104.16,but whe | vedclass

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(B) The formula for the degree of dissociation $(\alpha)$ in terms of theoretical vapour density $(D_t)$ and observed vapour density $(D_o)$ is given

Vedclass · Accepted Answer

$68$

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(B) The formula for the degree of dissociation $(\alpha)$ in terms of theoretical vapour density $(D_t)$ and observed vapour density $(D_o)$ is given by: $\alpha = \frac{D_t - D_o}{D_o 	imes (n - 1)}$ For the dissociation reaction: $PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g)$,the number of moles of products $(n)$ is $2$. Substituting the values: $\alpha = \frac{104.16 - 62}{62 	imes (2 - 1)}$ $\alpha = \frac{42.16}{62} \approx 0.68$ Therefore,the degree of dissociation is $0.68 	imes 100 = 68\%$.

The vapour density of $PCl_5$ is $104.16$,but when heated to $230 \, ^\circ C$,its vapour density decreases to $62$. What is the degree of dissociation of $PCl_5$ at this temperature (in $\%$) ?

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