# 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, **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

#"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.