# Question 9008a

Nov 4, 2015

$\text{0.00198 moles}$
$\text{0.0789 g}$

#### Explanation:

This is a pretty straightforward example of how the ideal gas law can be used to determine the number of moles of a gas when given the volume, temperature ,and pressure of the sample.

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
$V$ - its volume
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, expressed in Kelvin

Now, an important thing to notice here is that the units given to you for the temperature, pressure, and volume of the gas do not match what is used for the universal gas constant, $R$.

More specifically, you have

• pressure in torr $\to$ you need it in atm
• volume in mililiters $\to$ you need it in liters
• temperature in degrees Celsius $\to$ you need it in Kelvin

Keep this in mind when plugging in your values in the ideal gas law equation. Rearrange the equation to solve for $n$

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

$n = \left(\frac{820.0}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 50.0 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 60.0)color(red)(cancel(color(black)("K}}}}\right)$

$n = \text{0.001975 moles Ar}$

To get the mass of argon, use its molar mass, which tells you what the mass of one mole of argon is

0.001975color(red)(cancel(color(black)("moles Ar"))) * "39.948 g Ar"/(1color(red)(cancel(color(black)("mole Ar")))) = "0.078897 g Ar"

Now, you need to round both values to three sig figs, the number of sig figs you have for the temperature and volume of the gas

n = color(green)("0.00198 moles")" " and " "m = color(green)("0.0789 g")

Alternatively, you can use the ideal gas law equation to solve directly for the mass of the gas.

You know that you can write the number of moles of a gas as the ratio between mass and molar mass

$n = \frac{m}{M} _ \text{M}$

This means that you can write the ideal gas law equation as

$P V = {\overbrace{\frac{m}{M} _ \text{M}}}^{\textcolor{b l u e}{= n}} R T$

Rearranging to solve for m will get you

$m = \frac{P V}{R T} \cdot {M}_{\text{M}}$