# Question bf746

Apr 12, 2017

${\text{0.90 g L}}^{- 1}$

#### Explanation:

The idea here is that you can use the molar volume of a gas at STP to calculate the density of your sample of neon under these conditions.

More often than not, you'll find STP conditions defined as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions, $1$ mole of any ideal gas occupies $\text{22.4 L}$ $\to$ this is known as the molar volume of a gas at STP.

Now, in order to find the density of the gas, you must determine two things

• the mass of the sample
• the volume it occupies

Since you already know the volume, all you have to do here is to use the molar mass of neon to convert the number of moles in your sample to grams.

You will have

1.0 color(red)(cancel(color(black)("moles Ne"))) * "20.18 g"/(1color(red)(cancel(color(black)("mole Ne")))) = "20.18 g"

The density of the gas is defined as the mass of exactly $1$ unit of volume. In your case, this means the mass of $\text{1 L}$ of neon under STP conditions.

Since you know that $\text{20.18 g}$ occupy $\text{22.4 L}$, you can say that $\text{1 L}$ will occupy

1 color(red)(cancel(color(black)("L"))) * "20.18 g"/(22.4color(red)(cancel(color(black)("L")))) = "0.90 g"#

Therefore, the density of $1$ mole of neon under STP conditions is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{density = 0.90 g L}}^{- 1}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the sample of neon.