# Question 5f98e

Dec 2, 2016

${\text{20.2 g mol}}^{- 1}$

#### Explanation:

The thing to remember about the rate of effusion of a given gas is that it's inversely proportional to the square root of its molar mass, as given by Graham's Law of Effusion

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{rate} \propto \frac{1}{\sqrt{{M}_{M}}}}}}$

This means that heavier molecules will effuse at a slower rate than lighter molecules, i.e. gases that have bigger molar masses will effuse slower when compared with gases that have smaller molar masses, under the same conditions for pressure and temperature.

Right from the start, you can say that molar mass of the unknown gas will be smaller than ${\text{83.8 g mol}}^{- 1}$, since it effuses faster than krypton.

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

"rate Kr"/"rate unknown gas" = (1/sqrt(M_"M Kr"))/(1/sqrt(M_"M")) = sqrt(M_"M")/sqrt(M_"M Kr") = sqrt(M_"M"/M_"M Kr")

Now, the trick here is to realize that the rate of effusion is not the time given to you. More specifically, the times given by the problem tell you the time needed for equal masses of krypton and of unknown gas, let's say $m$, to effuse through the hole.

In this regard, you will have

${\text{rate Kr" = m/"87.3 s" = (m * 1/87.3) " s}}^{- 1}$

${\text{rate unknown gas" = m/"42.9 s" = (m * 1/42.9)" s}}^{- 1}$

This means that you have

"rate Kr"/"rate unknown gas" = ((color(red)(cancel(color(black)(m))) * 1/87.3) color(red)(cancel(color(black)("s"^(-1)))))/((color(red)(cancel(color(black)(m))) * 1/42.9) color(red)(cancel(color(black)("s"^(-1))))) = 42.9/87.3

Therefore, you will have

$\frac{42.9}{87.3} = \sqrt{{M}_{\text{M"/M_"M Kr}}}$

Plug in your values to find

42.9/87.3 = sqrt(M_"M"/("83.8 g mol"^(-1)))#

Rearrange to solve for ${M}_{\text{M}}$

${M}_{\text{M"/"83.8 g mol}}^{- 1} = {\left(\frac{42.9}{87.3}\right)}^{2}$

${M}_{\text{M" = (42.9/87.3)^2 * "83.8 g mol}}^{- 1}$

${M}_{\text{M" = color(darkgreen)(ul(color(black)("20.2 g mol}}^{- 1}$

The answer is rounded to three sig figs.

As predicted, the molar mass of the unknown gas is smaller than the molar mass of krypton.