# What is the molar volume of 200.0 grams of carbon monoxide??

Dec 28, 2014

Molar volume problems must contain some information about pressure and temperature, that is the only way a value for a substance's molar volume can be determined. Here's how that works:

Starting from the ideal gas law, $P V = n R T$, let's try and find an expression for molar volume

$p V = n R T \to \frac{P V}{n} = R T \to \frac{V}{n} = \frac{R T}{P} = {V}_{m o l a r}$

So, a gas' molar volume depends on temperature and pressure. Let's assume that we are at 273.15 K and 1.00 atm, and we want to determine what volume 1 mole of a gas occupies:

$\frac{V}{1 m o l e} = \frac{0.082 \frac{L \cdot a t m}{m o l \cdot K} \cdot 273.15 K}{1.00 a t m} = 22.4 \frac{L}{m o l} = {V}_{m o l a r}$

This represents the volume 1 mole of any ideal gas occupies at STP - Standard Temperature and Pressure (273.15 K, 1.00 atm).

Let's try and determine the molar volume of $200.0 g$ of $C O$ at STP:

$\frac{V}{n} = 22.4 \frac{L}{m o l} \to V = n \cdot \frac{22.4 L}{m o l}$ - this means that the more moles of o substance you have, the bigger its volume will be at STP. Since we have

$200.0 g \cdot \frac{1 m o l e C O}{28.0 g} = 7.14$ moles, the volume occupied will be $V = 7.14 m o l e s \cdot 22.4 \frac{L}{m o l} = 160.0 L$