# Question #a1be4

##### 1 Answer

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 **the same quantity** of an unknown gas diffuses in

If you take

#"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**.