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Standard XII

Physics

Applying NLM along 3 Axis

At 200oC a ...

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 of the vapour at $20^{o} C$ 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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Solution

The correct option is D $33$ %
$A (g) \to 2 B (g) + C (g)$ $V . D = 30$
$t = 0$ $a$ $0$ $0$ $↪$ molecular weight $= 30 \times 2 = 60$
$t = t_{e q}$ $a - x$ $2 x$ $x$
${M . W}_{m i x} = \frac{t o t a l w e i g h t}{n o . o f m o l e s} = \frac{I n i t i a l w e i g h t}{n o . o f m o l e s}$ ( $∵$ law of concentration of mass)
$= \frac{a \times 100}{a + 2 x} = 60$
$\frac{a}{a + 2 x} = 3 / 5 \Rightarrow 5 a = 3 a + 6 x$
$2 a = 6 x$
$x = a / 3$
% dissociation $x / a = 1 / 3 \Rightarrow 33$ %

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At 200oC a compound A (molar mass =100g mol−1) in the vapour phase dissociates asA(g)⟶2B(g)+C(g). The vapour density of the vapour at 20oC is found to be 30. The percent dissociation of the compound A at 200oC will be:

The correct option is D 33% A(g)→2B(g)+C(g) V.D=30 t=0 a 0 0 ↪ molecular weight =30×2=60t=teq a−x 2x xM.Wmix=total weightno. of moles=Initial weightno. of moles (∵ law of concentration of mass)=a×100a+2x=60aa+2x=3/5⇒5a=3a+6x2a=6xx=a/3% dissociation x/a=1/3⇒33%

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 of the vapour at $20^{o} C$ is found to be $30$ . The percent dissociation of the compound $A$ at $200^{o} C$ will be: