The composition of the equilibrium mixture (Cl `hArr 2CI`), which is attained at. `1200^(@)`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 chlorin molecules dissociated into atoms: (Relative atomic mass of Kr=84)

Options

A. 0.1374
B. 0.9325
C. 0.2573
D. 0.6256

Answer

A

Explanation

Let the initial amount of `Cl_2` be n . If `alpha` is the fraction of `Cl_2` dissociated at equilibrium , we will have
`Cl_(2) hArr 2 Cl`
`n(1 - alpha) " " n (2 alpha)`
Total amount of special at equilibrium , `n_("total") = n_(Cl_2) + n_(Cl) = n (1 - alpha) + n (2 alpha) = n (1 + alpha)`
Average molar mass of the mixture at equilibrium `(M_(av))_("mix") = (n M_(Cl_2))/(n (1 + alpha)) = (M_(Cl_2))/(1 + alpha)`
According to Graham.s law of diffusion , we get `(r_("mix"))/(r_(Kr)) = [(M_(Kr))/((M_(av))_("mix"))]^(1//2) = [(M_(Kr) (1 + alpha))/(M_(Cl_2))]^(1//2)`
or `alpha = ((M_(Cl_2))/(M_(Kr))) ((r_("mix"))/(r_(Kr)))^(2) = 1 ((71)/(84)) (1.16)^(2) - 1 = 1.1374 - 1 = 0.1374`