# Question 3ccd4

Oct 18, 2017

${\text{1.6 g L}}^{- 1}$

#### Explanation:

Your tool of choice here is the ideal gas law equation, which looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

Now, the problem wants you to find the density of nitrogen dioxide in a volume of $\text{150 L}$, at a temperature of

${28}^{\circ} \text{C" = 28^@"C" + 273.15 = "301.15 K}$

and a pressure of $\text{0.85 atm}$.

As you know, the density of a substance, $\rho$, tells you the mass of exactly $1$ unit of volume of said substance. This implies that you can calculate the density of a substance by dividing the mass of a given sample, let' say $m$, and the volume it occupies, $V$.

$\textcolor{b l u e}{\rho = \frac{m}{V}}$

Now, notice that the ideal gas law equation uses the number of moles of gas, $n$. As you know, the number of moles of a substance can be expressed as a ratio between the mass of a given sample, $m$, and the molar mass of the substance, let's say ${M}_{M}$.

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

Plug this into the ideal gas law equation to get

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

All you have to do now is to rearrange this equation in order to find an expression for the density of the gas.

$P = \frac{\textcolor{b l u e}{m}}{{M}_{M} \cdot \textcolor{b l u e}{V}} \cdot R T$

$P \cdot {M}_{M} = \frac{\textcolor{b l u e}{m}}{\textcolor{b l u e}{V}} \cdot R T$

This means that you have

$P \cdot {M}_{M} = \textcolor{b l u e}{\rho} \cdot R T$

which gets you

$\textcolor{b l u e}{\rho} = \frac{P \cdot {M}_{M}}{R T}$

Finally, plug in your values to find

rho = (0.85 color(red)(cancel(color(black)("atm"))) * "46.0 g" color(red)(cancel(color(black)("mol"^(-1)))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 301.15 color(red)(cancel(color(black)("K")))) = color(darkgreen)(ul(color(black)("1.6 g L"^(-1))))#

The answer is rounded to two sig figs.

Notice that you didn't need to know the volume of the gas in order to be able to find its density.