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A synthetic mixture of nitrogen and Argo...

A synthetic mixture of nitrogen and Argon has a density of `1.4 g L^(-1)` at `0^@C`. Calculate the average molecular weight. Find out the volume percentage of nitrogen in the mixture.

Text Solution

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Molecular weight (M) can be obtained from density(d) as, `M = (dRT)/(P)`
Average molecular weight of the mixture ` = (1.4 xx 0.0821 xx 273)/(1) = 31.4`
IF the % volume fo `N_(2)` is .x.
`31.4 = (x xx 28 + (100-x)40)/(100)`
`12 x = 840 or x = 70`
THe volume percentage of `N_(2)` in the mixture = 70
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