At 200-C a compound -A- molar mass - 100 g mol-1 In the vapour phase dissociates as Ag - 2Bg - Cg- The vapour density of the vapour at 200-C is found to be 30- The percent dissociation of the compound A at 200-C will be-

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Standard XII

Chemistry

Degree of Dissociation

At 200∘C a ...

Question

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

33

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Solution

The correct option is A $33$ %
$α = \frac{D - d}{(h - 1) d}$
$A_{(g)} ⟶ 2 B_{(g)} + C_{(g)}$
$α =$ Degree of Dissociation
$D =$ Vapour density (initial)
$n =$ equilibrium Vapour Density $= 30$
$n =$ number of moles of gaseous product $= 3$
Molar Mass $= 2 D$
$D = \frac{100}{2}$
$D = 50$
$α = \frac{50 - 30}{(3 - 1) 30}$
$α = \frac{20}{2 (30)}$
$α = \frac{1}{3}$
Percentage $= 33.33 %$

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At 200∘C a compound ′A′ (molar mass =100 g mol−1) In the vapour phase dissociates as A(g)→2B(g)+C(g). The vapour density of the vapour at 200∘C is found to be 30. The percent dissociation of the compound A at 200∘C will be:

The correct option is A 33%α=D−d(h−1)dA(g)⟶2B(g)+C(g)α= Degree of Dissociation D= Vapour density (initial)n= equilibrium Vapour Density=30n= number of moles of gaseous product =3Molar Mass=2DD=1002D=50α=50−30(3−1)30α=202(30)α=13Percentage=33.33%

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