# A 283.3-g sample of #X_2(g)# has a volume of 30 L at 3.2 atm and 27°C. What is element #X#?

##### 1 Answer

Chlorine,

#### Explanation:

Your strategy here is to apply the **ideal gas law** equation to find the number of *moles* of gas present in your sample.

In order to be able to identify the unknown element, you need to know its **molar mass**. A substance's molar mass tells you the mass of **one mole** of that substance.

The important thing to notice here is that you're given a sample of

Since you know the mass of your sample, you can use the number of moles it contains to determine the element's molar mass.

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "# , where

*number of moles* of gas

*universal gas constant*, usually given as

**absolute temperature** of the gas

Rearrange the equation to solve for

#PV = nRT implies n = (PV)/(RT)#

Plug in your values - **do not** forget to convert the *temperature* of the gas from *degrees Celsius*, to *Kelvin*

#n = (3.2 color(red)(cancel(color(black)("atm"))) * 30color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 27)color(red)(cancel(color(black)("K")))) = "3.896 moles"#

So, if **one mole** will have a mass of

#1color(red)(cancel(color(black)("mole"))) * "283.3 g"/(3.896color(red)(cancel(color(black)("moles")))) = "72.7 g"#

Since **one mole** of

Now, a molecule of **two atoms** of *element* **half** that of the molecule

#M_M = 1/2 * "72.7 g mol"^(-1) = "36.25 g mol"^(-1)#

The closest match to this value is the molar mass of *chlorine*, *chlorine*,