Question

$0.020$g of selenium vapour at equilibrium occupying a volume of $2.463mL$ at $1$ atm and ${27}^{o}C$. The selenium is in a state of equilibrium according to reaction.
$3{Se}_2\left(g\right)\rightleftharpoons {Se}_6\left(g\right)$
What is the degree of association of selenium?
(At.wt. of $Se=$79)

• A. $0.205$
• B. $0.315$
• C. $0.14$
• D. none of these

Answer 1

Answer: (B)

$0.315$

$\begin{array}{ccc}3S{e}_2& \leftrightharpoons & S{e}_6\\ a& & \\ a(1-\alpha )& & \frac{a\alpha }{3}\end{array}$

Molar mass of the mixture ($M_0$) = $\frac{wRT}{PV}=\frac{0.02\times 300\times 0.08205}{1\times 2.463\times {10}^{-3}}=200$

Molar mass of $S{e}_2$ = $79\times 2=158$

$\frac{\text{Observed moles at equilibrium}}{\text{Initialno of moles of reactant}}=\frac{M_T}{M_0}$

$\Rightarrow \frac{a(1-\alpha )+\frac{a\alpha }{3}}{a}=\frac{158}{200}$

$\Rightarrow \alpha =0.315$