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?

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Answer 1 · 2016-11-17T22:03:31

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

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

$P_i = chi_iP_"total"$

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

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

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

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

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