# Question a1be4

Sep 28, 2017

Here's what I got.

#### Explanation:

Your tool of choice here will be Graham's Law of diffusion, which states that the rate at which a gas diffuses is inversely proportional to the square root of its molar mass.

$\text{rate of diffusion} \propto \frac{1}{\sqrt{{M}_{M}}}$

Now, you know a certain quantity of oxygen gas diffuses in $\text{18 s}$ and that the same quantity of an unknown gas diffuses in $\text{45 s}$.

If you take $n$ to represent the number of moles of the two gases, you can say that you have

${\text{rate O"_2 = (n color(white)(.)"moles")/"18 s" = (n/18)color(white)(.)"mol s}}^{- 1}$

${\text{rate gas X" = (n color(white)(.)"moles")/"45 s" = (n/45)color(white)(.)"mol s}}^{- 1}$

Now, you know that you have

"rate O"_ 2 prop 1/sqrt(M_ ("M O"_ 2))

"rate gas X" prop 1/sqrt(M_ "M X")

This means that you can write

"rate O"_ 2/"rate gas X" = (1/sqrt(M_ ("M O"_ 2)))/(1/sqrt(M_ "M X"))

This is equivalent to

( (color(red)(cancel(color(black)(n)))/18)color(red)(cancel(color(black)("mol s"^(-1)))))/((color(red)(cancel(color(black)(n)))/45)color(red)(cancel(color(black)("mol s"^(-1))))) = sqrt(M_ "M X")/sqrt(M_ ("M O"_ 2))

45/18 = sqrt(M_ "M X"/M_ ("M O"_ 2))

If you take the molar mass of oxygen gas as

M_ ("M O"_ 2) = "32 g mol"^(-1)

you can say that you have

$\frac{45}{18} = \sqrt{{M}_{\text{M X"/"32 g mol}}^{- 1}}$

Square both sides of the equation

${\left(\frac{45}{18}\right)}^{2} = {\left(\sqrt{{M}_{\text{M X"/"32 g mol}}^{- 1}}\right)}^{2}$

to get

${45}^{2} / {18}^{2} = {M}_{\text{M X"/"32 g mol}}^{- 1}$

Finally, rearrange to find the molar mass of the unknown gas

M_"M X" = 45^2/18^2 * "32 g mol"^(-1) = color(darkgreen)(ul(color(black)(2.0 * 10^2 color(white)(.)"g mol"^(-1))))#

The answer is rounded to two sig figs.