1.

A gaseous mixture of O_(2) and X containing 20% (mole %) of X, diffused through a small hole in 234 seconds while pure O_(2) takes 224 seconds to diffuse through the same hole. Find the molecular weight of X.

Answer»

SOLUTION :We have ,
`(t_("mix"))/(t_(O_(2))) = sqrt((M_("mix"))/(M_(O_(2))))`
`(234)/(224) = sqrt((M_("mix"))/(32))`,
`therefore M_("mix") = 34.921`
As the mixture contains 20% (mole %) of X, the molar ration of `O_(2)` and X may be represented as 0.8 N : 0.2n, n being the total no. of moles.
`therefore M_("mix") = (32 XX 0.8 n + M_(x) xx 0.2 n)/(n) = 34.921`
`therefore M_(x)` (mol. wt. of X) = 46.6.


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