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Properties of Water

0.020 g of se...

Question

$0.020$ g of selenium vapour at equilibrium occupying a volume of $2.463$ mL at $1$ atm and $27^{o}$ C. The selenium is in a state of equilibrium according to reaction.
$3 S e_{2} (g) ⇌ S e_{6} (g)$
What is the degree of association of selenium?
(At. wt. of $S e = 79$ )

0.205

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0.315

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0.14

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None of these

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Solution

The correct option is A $0.315$
$3 {S e}_{2} ⇌ {S e}_{6} a a (1 - α) \frac{a α}{3}$
Hence t he molar mass of mixture $(M_{o}) = \frac{w R T}{P V} = \frac{0.02 \times 300 \times 0.082}{1 \times 2.463 \times 10^{- 3}} = 200$
Therefore,
molar mass of $S e_{2} = 79 \times 2 = 158$
$\frac{o b s e r v e d m o l e s a t e q u i l i b r i u m}{i n i t i a l n u m b e r o f m o l e s} = \frac{M_{T}}{M_{o}}$
$\frac{a (1 - α) + \frac{a α}{3}}{a} = \frac{158}{200}$
$α = 0.315$

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