Question

Question #a1be4

Answer 1

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.

`"rate of diffusion" prop 1/sqrt(M_M)`

Now, you know a certain quantity of oxygen gas diffuses in `"18 s"` and that the same quantity of an unknown gas diffuses in `"45 s"`.

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

`"rate O"_2 = (n color(white)(.)"moles")/"18 s" = (n/18)color(white)(.)"mol s"^(-1)`

`"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

`45/18 = sqrt(M_"M X"/"32 g mol"^(-1))`

Square both sides of the equation

`(45/18)^2 = (sqrt(M_"M X"/"32 g mol"^(-1)))^2`

to get

`45^2/18^2 = M_ "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.