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In the reaction, ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g),$ 50 grams of ${CaCO}_{3}$ is allowed to dissociate in 22.4 lit vessel at $819^{0} c$ . If 50% of ${CaCO}_{3}$ is left at equilibrium, active masses of ${CaCO}_{3}, CaO and {CO}_{2}$ at equilibrium respectively are

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25g ; 14 g ; 1/22.4 mol/lit

1 ; 1 ; 1/89.6 mol/lit

25 ; 14 ; 1/89.6 mol/lit

1 ; 1 ; 1

answer is B.

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Detailed Solution

Given

{n_{CaCO_{3}}}\; initially\;taken\;=\;\frac{50}{100}=0.5

T = (819 + 273) K

50% of calcium carbonate dissociated at equilibrium

\large \mathop {CaC{O_3}}\limits^{1mole} \left( s \right)

At equilibrium, $[solid] = [{CaCO}_{3}] = [CaO] = 1;$

At equilibrium, $n_{{CO}_{2}} = 0.25, V_{vessel} = 22.4 lit;$

\therefore Active\;mass\;=\frac{0.25}{22.4}=\frac{1}{89.6}M

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