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Degree of Dissociation

0.020 g of se...

Question

$0.020 g$ of selenium vapours at equilibrium occupying a volume of $2.463 mL$ at $1 atm$ and $27^{\circ}$ C.
The selenium is in a state of equilibrium according to the reaction
$3 S e_{2} (g) ⇌ S e_{6} (g)$
What is the degree of association of selenium?
(Atomic weight of Se = 79 amu)

0.205

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0.315

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0.14

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0.51

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Solution

The correct option is B 0.315
$3 S e_{2} (g) ⇌ S e_{6} (g)$
a - a $α$ $\frac{a α}{3}$

$\frac{o b s e r v e d m o l e s a t e q u i l i b r i u m}{i n i t i a l m o l e s o f r e a c t a n t} = \frac{M_{i n i t}}{M_{e q}}$

$\frac{a (1 - α) + \frac{a α}{3}}{a}$ = $\frac{M_{i n i t}}{M_{e q}}$

$(1 - α) + \frac{α}{3}$ = $\frac{M_{i n i t}}{M_{e q}}$

$M_{i n i t} = M_{S e_{2}} = 2 \times 79 = 158$

Molar mass of mixture $M_{e q} = \frac{W \times R \times T}{P \times V}$
$M_{e q} = \frac{0.02 \times 0.0821 \times 300}{1 \times 2.463 \times 10^{- 3}}$
$M_{e q} = 200$
$∴ α = 0.315$

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