# A sample of carbon monoxide gas is collected in a 100 mL container at a pressure of 688 mmHg and a temperature of 565°C. How many grams of this gas is present this given sample?

Aug 7, 2016

The mass of $\text{CO}$ is 0.0369 g.

#### 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 $n = \text{mass"/"molar mass} = \frac{m}{M}$, we can write the Ideal Gas Law as

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

We can rearrange this to get

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

$M = \text{28.01 g·mol"^"-1}$
P = 688 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.9053 atm"
$V = \text{0.100 L}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(565 + 273.15) K" = "838.15 K}$

m = (28.01 "g"·color(red)(cancel(color(black)("mol"^"-1"))) × 0.9053 color(red)(cancel(color(black)("atm"))) × 0.100 color(red)(cancel(color(black)("L"))))/("0.0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1""mol"^"-1"))) × 838.15 color(red)(cancel(color(black)("K")))) = "0.0369 g"

∴ The mass is 0.0369 g.