# How many molecules of #CO_2# gas at 713 mmHg and 56.7°C are present in 10.5 L?

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to

*use the***ideal gas law**equation to determine how many**moles**of gas you have under those conditions for pressure and temperature*use***Avogadro's number**to determine how many molecules of carbon dioxide would be found in that many moles

So, the ideal gas law equation looks like this

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

*universal gas constant*, usually given as

**always** expressed in *Kelvin*

The important thing to be aware of here is that the units you have for pressure, temperature, and volume **must match** those used in the expression of the universal gas constant.

You will thus need to convert the pressure of the gas from *mmHg* to *atm* and its temperature from *degrees Celsius* to *Kelvin*.

So, rearrange the ideal gas law equation to solve for

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

#n = (713/760 color(red)(cancel(color(black)("atm"))) * 10.5color(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 + 56.7)color(red)(cancel(color(black)("K"))))#

#n = "0.36375 moles CO"_2#

Now, **one mole** of any substance contains exactly **Avogadro's number**.

This means that your sample of carbon dioxide will contain

#0.36375 color(red)(cancel(color(black)("moles CO"_2))) * (6.022 * 10^(23)"molec. CO"_2)/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)(2.19 * 10^(23)"molec. CO"_2#

The answer is rounded to three sig figs.