At `27^(@)C`, hydrogen is leaked through a tiny hole into a vessel for 20 min. Another unknown gas at the same temperature and pressure as that of hydrogen is leaked through same hole for 20 min. After the effusion of the gases the mixture exerts a pressure of 6 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3L. What is the molecular weight of the unknown gas?

Options

A. 1088
B. 10.88
C. 108.8
D. None of these

Answer

A

Explanation

`n_("Total") = (PV)/(RT) = (6 xx 3)/(0.082 xx 300) = 0.73, n_(X) = n_("Total") -0.7 = 0.03`
Using Graham’s law of Diffusion `(n_(H_(2))//t)/(n_(X)//t) = (0.7)/(n_(X)) = sqrt((M_(X))/(M_(H_(2)))) = sqrt((M_(X))/(2))`
`M_(X) = 2 ((0.7)/(n_(X)))^(2) = 1088`