A mixture of argon (Ar) and nitrogen `(N_(2))` has a density of 1.40 g `L^(-1)` at STP. Mole fraction of `N_(2)` in the mixture is (Given atomic mass of Ar=40)

Answer

`0.7`

Explanation

Suppose mole fraction of `N_(2)` in the mixture = x
Then mole fraction of Ar = `(1-x)`
` :. ` Average molar mass of the mixture
` = 28 x + 40 (1-x) = 40 - 12 x `
`d=(PM)/(RT) = (("1 atm")(40-12x)"g mol"^(-1))/(0.0821 " L atm K"^(-1) "mol"^(-1)xx273 K )`
` = (40-12x)/(22.4) " g L"^(-1)`
` :. (40 - 12x)/(22.4)= 1.40 ` (Given )
or `40-12 x = 31.4 or 12 x = 8.6 or x = 0.7`