Question

A gaseous mixture contains 440.0 Torr of H2(g), 365.7 Torr of N2(g), and 88.7 Torr of Ar(g). Calculate the mole fraction, X, of each of these gases?

Answer 1

`chi_"H₂" = 0.4919; chi_"N₂" = 0.4089; chi_"Ar" = 0.0992`

Explanation:

We can express the mole fraction `chi_i` of an individual gas `i` in a mixture in terms of the component's partial pressure `P_i`:

`chi_i = P_i/P_"total"`

Then

`P_i = chi_iP_"total"`

`P_"H₂" = "440.0 Torr"`
`P_"N₂" = "365.7 Torr"`
`P_"Ar" = "88.7 Torr"`

`P_"total" = "(440.0 + 365.7 + 88.7) Torr" = "894.4 Torr"`

`chi_"H₂" = (440.0 color(red)(cancel(color(black)("Torr"))))/(894.4 color(red)(cancel(color(black)("Torr")))) = 0.4919`

`chi_"N₂" = (365.7 color(red)(cancel(color(black)("Torr"))))/(894.4 color(red)(cancel(color(black)("Torr")))) = 0.4089`

`chi_"Ar" = (88.7 color(red)(cancel(color(black)("Torr"))))/(894.4 color(red)(cancel(color(black)("Torr")))) = 0.0992`