Question #28663
1 Answer
Explanation:
The idea here is that the rate of effusion of a given gas is inversely proportional to the square root of its molar mass, as given by Graham's Law of Effusion
#color(blue)(ul(color(black)("rate" prop 1/sqrt(M_"M"))))#
This basically means that more massive molecules, i.e. heavier molecules, will have a slower rate of effusion that lighter molecules.
In other words, the bigger the molar mass of a gas, the slower its rate of effusion will be when compared to the rate of effusion of a lighter gas.
Now, you know that the rate of effusion of methane,
If you take
#"rate"_ ("CH"_ 4)/"rate"_ "unknwon" = (1/sqrt(M_ ("M CH"_ 4)))/(1/sqrt(M_"M"))#
But since
#color(blue)(ul(color(black)("rate"_ ("CH"_ 4) = 2.9 * "rate"_"unknown"))) -># methane effuses#2.9# times as fast as the unknown gas
you will have
#(2.9 * color(red)(cancel(color(black)("rate"_ "unknown"))))/color(red)(cancel(color(black)("rate"_ "unknown"))) = sqrt(M_ "M")/sqrt(M_ ("M CH"_ 4)) = sqrt(M_ "M"/M_ ("M CH"_ 4))#
Rearrange to solve for
#M_"M" = 2.9^2 * M_("M CH"_4)#
The molar mass of methane is equal to
#M_"M" = 2.9 * "16.04 g mol"^(-1) = color(darkgreen)(ul(color(black)("46.5 g mol"^(-1))))#
I'll leave the answer rounded to three sig figs.
As predicted, the molar mass of the unknown gas is bigger than the molar mass of methane, which is why it effuses at a slower rate than methane.