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

Dec 27, 2015

$2.19 \cdot {10}^{23}$

#### 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

$\textcolor{b l u e}{P V = n R T} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, 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 $n$, the number of moles of carbon dioxide present in this sample

$P V = n R T \implies n = \frac{P V}{R T}$

$n = \left(\frac{713}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 10.5color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 56.7)color(red)(cancel(color(black)("K}}}}\right)$

$n = {\text{0.36375 moles CO}}_{2}$

Now, one mole of any substance contains exactly $6.022 \cdot {10}^{23}$ molecules of that substance - this is known as 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.