Question #119bd
1 Answer
Explanation:
First of all, don't get distracted by the volume of the two gases, this particular piece of information is not important when it comes to effusion.
You don't need to know the exact volume of the gases, all you need to go by is the fact that you're looking at equal volumes.
As you know, the rate at which a gas effuses is inversely proportional to the square root of its molar mass -- this is known as Graham's Law of Effusion.
#color(blue)(|bar(ul(color(white)(a/a)"rate of effusion" prop 1/sqrt("molar mass")color(white)(a/a)|)))#
What this basically tells you is that heavier molecules will effuse at a slower rate than lighter molecules.
In your case, you know that a sample of hydrogen gas,
You can thus say that
#color(green)(|bar(ul(color(white)(a/a)color(black)("rate"_(H_2) = 4 xx "rate"_"U")color(white)(a/a)|)))#
Use Graham's Law of Effusion to write this as
#"rate"_ (H_2)/"rate"_ "U" = 1/sqrt(M_ ("M H"_ 2)) * sqrt(M_("M U"))#
Hydrogen gas has a molar mass of
#(4 * color(red)(cancel(color(black)("rate"_ "U"))))/(color(red)(cancel(color(black)("rate"_"U")))) = 1/sqrt("2.016 g mol"^(-1)) * sqrt(M_("M U")color(white)(a)"g mol"^(-1))#
#4 = sqrt((M_("M U")color(red)(cancel(color(black)("g mol"^(-1)))))/(2.016color(red)(cancel(color(black)("g mol"^(-1))))))#
which will get you
#4 = sqrt(M_("M U")/2.016)#
All you have to do now is square both sides of the equation
#4^2 = (sqrt(M_("M U")/2.016))^2#
#16 = M_("M U")/2.016 implies M_("M U") = 16 * 2.016 = 32.256#
This is where the values given to you for the volume of the two gases can come in handy. You can round this off to two sig figs, the number of sig figs you have for the two volumes, to get
#M_("M U") = color(green)(|bar(ul(color(white)(a/a)"32 g mol"^(-1)color(white)(a/a)|)))#
