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The composition of the equilibrium mixture $(Cl_2\iff2Cl)$ which is attained at $1200^{\large\circ}C$ is determined by measuring the rate of effusion through a pinhole . It is observed that at 1.80 mm Hg pressure , the mixture effuses 1.16 times as fast as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated anto atoms.(Atomic wt of Kr = 84)


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$\large\frac{V_{mix}}{V_{Kr}} = \sqrt{\large\frac{M_{Kr}}{M_{mix}}}$
Or 1.16 = $\sqrt{(\large\frac{84}{M})}$
M = 62.425
For $Cl_2 \iff 2Cl$
$\large\frac{\text{Normal mol.wt}}{\text{Exp mol. wt.}} = 1 + \alpha$
$\large\frac{71}{62.425} = 1 + \alpha$
$\alpha = 0.137\; (or) \;13.7\%$
Hence answer is (A)
answered Apr 5, 2014 by sharmaaparna1

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