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Standard XII

Chemistry

Effective Equilibrium Constant

At 200 o C ...

Question

At $200^{o} C$ a compound ' $A$ ' (molar mass $= 100 g {m o l}^{- 1}$ ) in the vapour phase dissociates as $A (g) ⟶ 2 B (g) + C (g)$ .
The vapour density at $200^{o} C$ is found to be $30$ . The percent dissociation of the compound $A$ at $200^{o} C$ will be?

33

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66

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87

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100

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Solution

The correct option is A $33$ %
The given vapour density(d) is the observed one.
$A (g) \to 2 B (g) + C (g)$
$% α = \frac{D - d}{d (y - 1)} \times 100$
where,
D is calculated vapour phase i.e.
$100 X 2 = 200.$
y is number of moles of product per mole reactant.
$% α = \frac{100 - 30}{30 \times 2}$
$% α = 33 %$
Option A is correct answer

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At 200oC a compound 'A' (molar mass =100gmol−1) in the vapour phase dissociates as A(g)⟶2B(g)+C(g).The vapour density at 200oC is found to be 30. The percent dissociation of the compound A at 200oC will be?

The correct option is A 33%The given vapour density(d) is the observed one.A(g)→2B(g)+C(g)%α=D−dd(y−1)×100where,D is calculated vapour phase i.e.100X2=200.y is number of moles of product per mole reactant.%α=100−3030×2%α=33%Option A is correct answer

At $200^{o} C$ a compound ' $A$ ' (molar mass $= 100 g {m o l}^{- 1}$ ) in the vapour phase dissociates as $A (g) ⟶ 2 B (g) + C (g)$ .
The vapour density at $200^{o} C$ is found to be $30$ . The percent dissociation of the compound $A$ at $200^{o} C$ will be?