# What is the density of NO_2 in a 3.50 L tank at 780.0 torr and 37°C?

Jul 11, 2016

The density of ${\text{NO}}_{2}$ under these conditions is is 1.85 g/L.

#### Explanation:

We can use the Ideal Gas Law to solve this problem.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $\text{moles" = "mass"/"molar mass}$ or $n = \frac{m}{M}$, we can write

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

We can rearrange this to

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

But $\text{density"= "mass"/"volume}$ or color(brown)(|bar(ul(color(white)(a/a)ρ = m/Vcolor(white)(a/a)|)))" "

PM = ρRT and

color(blue)(|bar(ul(color(white)(a/a)ρ = (PM)/(RT)color(white)(a/a)|)))" "

P = 780.0 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "1.026 atm"
$M = \text{46.01 g/mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(37 + 273.15) K" = "310.15 K}$

ρ = (1.026 color(red)(cancel(color(black)("atm"))) × "46.01 g"·color(red)(cancel(color(black)("mol"^"-1"))))/("0.082 06" color(red)(cancel(color(black)("atm")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 310.15 color(red)(cancel(color(black)("K")))) = "1.85 g/L"