# Question bb2a7

Apr 20, 2015

The density of your argon sample will be 18.9 g/L.

So, you know that you have a sample of argon under certain conditions of temperature and pressure. In addition to this, you know the volume the sample occupies.

THis means that you can use the ideal gas law equation to determine how many moles of gas your sample contains.

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

${n}_{\text{Ar" = (10.0cancel("atm") * 1.00cancel("L"))/(0.082(cancel("L") * cancel("atm"))/("mol" * cancel("K")) * (273.15 - 15)cancel("K")) = "0.4724 moles Ar}}$

Now use argon's molar mass to determine the mass of your sample

0.4724cancel("mole") * "39.948 g"/(1cancel("mole")) = "18.87 g"#

Since density is defined as mass per unit volume, you've got all you need to calculate it

$\rho = \frac{m}{V}$

$\rho = \text{18.87 g"/"1.00 L" = "18.87 g/L}$

Rounded to three sig figs, the answer will be

$\rho = \textcolor{g r e e n}{\text{18.9 g/L}}$