# Question 28663

Dec 3, 2016

${\text{46.5 g mol}}^{- 1}$

#### 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, ${\text{CH}}_{4}$, is $2.9$ times as fast as the rate of effusion of an unknown gas. Right from the start, this tells you that the molar mass of methane must be smaller than the molar mass of the unknown gas.

If you take ${M}_{\text{M}}$ to be the molar mass of the unknown gas, you can say that you have

"rate"_ ("CH"_ 4)/"rate"_ "unknwon" = (1/sqrt(M_ ("M CH"_ 4)))/(1/sqrt(M_"M"))

But since

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{rate"_ ("CH"_ 4) = 2.9 * "rate"_"unknown}}}} \to$ 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 ${M}_{\text{M}}$

M_"M" = 2.9^2 * M_("M CH"_4)

The molar mass of methane is equal to ${\text{16.04 g mol}}^{- 1}$, which means that the molar mass of the unknown gas will be

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.