# Question #133a5

##### 1 Answer

#### Explanation:

The thing to remember about a gas' **rate of effusion** is that it's **inversely proportional** to the square root of its molar mass - this is known as *Graham's Law*

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

*Effusion* is simply the term used to describe how gas molecules escape through a small hole. This means that the *rate of effusion* will depend on how *"massive"* the gas molecules are.

More specifically, the fact that the rate of effusion is *inversely proportional* to the square root of the mass of its particles tells you that the **heavier** a gas particle is, the **slower** it will effuse, i.e. escape through a small hole in its container.

Similarly, **lighter** gas particles will effuse at a **faster** rate when compared with heavier gas particles.

So, look up the molar masses of neon,

#"For Ne: " "rate"_(Ne) prop 1/sqrt("20.18 g mol"^(-1))#

#"For C"_4"H"_10: " rate"_text(butane) prop 1/sqrt("58.12 g mol"^(-1))#

This means that you can compare the rates of effusion for the two gases by dividing these two expressions

#"rate"_(Ne)/"rate"_text(butane) = 1/sqrt("20.18 g mol"^(-1)) * sqrt("58.12 g mol"^(-1))/1#

#"rate"_(Ne)/"rate"_text(butane) = sqrt( (58.12 color(red)(cancel(color(black)("g mol"^(-1)))))/(20.18color(red)(cancel(color(black)("g mol"^(-1)))))) = sqrt(58.12/20.18)#

#"rate"_(Ne)/"rate"_text(butane) = 1.6971#

So, this tells you that neon will effuse at a rate that is **faster** than the rate of effusion of butane.

Since you know that the rate of effusion of neon is equal to

#"rate"_text(butane) = "rate"_text(Ne)/1.6971#

#"rate"_text(butane) = "800. m s"^(-1)/1.6971 = "471.39 m s"^(-1)#

Rounded to three **sig figs**, the number of sig figs you have for the rate of effusion of neon, the answer will be

#"rate"_"butane" = color(green)("471 m s"^(-1))#

So, does this result make sense?

Since butane has heavier molecules, it makes sense that it will effuse at a *slower* rate compared with neon.