# Question #119bd

##### 1 Answer

#### Explanation:

First of all, don't get distracted by the *volume* of the two gases, this particular piece of information is **not important** when it comes to **effusion**.

You don't need to know the *exact* volume of the gases, all you need to go by is the fact that you're looking at **equal volumes**.

As you know, the **rate** at which a gas effuses is **inversely proportional** to the square root of its **molar mass** -- this is known as **Graham's Law of Effusion**.

#color(blue)(|bar(ul(color(white)(a/a)"rate of effusion" prop 1/sqrt("molar mass")color(white)(a/a)|)))#

What this basically tells you is that *heavier molecules* will effuse at a **slower rate** than *lighter molecules*.

In your case, you know that a sample of hydrogen gas, **four times as rapidly** as an unknown gas, let's say

You can thus say that

#color(green)(|bar(ul(color(white)(a/a)color(black)("rate"_(H_2) = 4 xx "rate"_"U")color(white)(a/a)|)))#

Use Graham's Law of Effusion to write this as

#"rate"_ (H_2)/"rate"_ "U" = 1/sqrt(M_ ("M H"_ 2)) * sqrt(M_("M U"))#

Hydrogen gas has a molar mass of **molar mass** of gas

#(4 * color(red)(cancel(color(black)("rate"_ "U"))))/(color(red)(cancel(color(black)("rate"_"U")))) = 1/sqrt("2.016 g mol"^(-1)) * sqrt(M_("M U")color(white)(a)"g mol"^(-1))#

#4 = sqrt((M_("M U")color(red)(cancel(color(black)("g mol"^(-1)))))/(2.016color(red)(cancel(color(black)("g mol"^(-1))))))#

which will get you

#4 = sqrt(M_("M U")/2.016)#

All you have to do now is square both sides of the equation

#4^2 = (sqrt(M_("M U")/2.016))^2#

#16 = M_("M U")/2.016 implies M_("M U") = 16 * 2.016 = 32.256#

This is where the values given to you for the volume of the two gases can come in handy. You can round this off to two **sig figs**, the number of sig figs you have for the two volumes, to get

#M_("M U") = color(green)(|bar(ul(color(white)(a/a)"32 g mol"^(-1)color(white)(a/a)|)))#