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The composition of the equilibrium mixture (CI2⇌ 2CI), which is attained at 1200oC, is determined by measuring the rate of effusion through a pin-hole. It is observed that at 1.80 mmHg pressure, the mixture effuses 1.16 times as fast as krypton effuses under the same conditions. Calculate the fraction of the chlorine molecules dissociated into atoms. (relative atomic mass of Kr = 84.)

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra
Mixture Krypton
rmix = 1.16 rKr = 1
Mmix = ? MKr = 84
We know that
rmix/rKr = √MKr/Mmin or 1.16/1 = √84/Mmix
or (1.16)2 84/Mmix =>Mmix = 84/(1.16)2 = 62.426
Determination of the composition of the equilibrium mixture/Let the fraction of CI2 molecules dissociated at equilibrium = x
CI2 ⇌ 2CI Total
Initially 1 0 1
At equilibrium 1 – x 2x 1 – x + 2x = 1 + x
∴ Total moles at equilibrium = 1 – x + 2x = 1 + 1
∵ Normal molecular mass/Experimental molecular mass = 1+x
∴ 71/64.426 = 1+ ∝
∴ ∝ = 0.137 = 13.7%
Last Activity: 11 Years ago
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