# The volume of 9.00 grams of carbon dioxide vapor at STP is 11.2 L. How would I figure that out?

##### 1 Answer
May 23, 2018

Just as an exercise...., and I realize this is an OLD question, first you should specify $\text{STP}$..

#### Explanation:

All sorts of standards exist across different curricula that specifies standard conditions. At the moment it really is a dog's breakfast, and the situation is much more confused than when I was a student.

Following this site, which may disagree with your curriculum, $\text{STP"=273.15*K, 1*"bar}$...$\text{1 bar} \equiv 0.987 \cdot a t m$...I myself like using atmospheres, as it is an intuitive unit of pressure, and I suspect that most chemists of my and earlier generations would agree.

And so we solve the Ideal Gas equation....$P V = n R T$

$V = \frac{n R T}{P} = \frac{\frac{9.00 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 273.15 \cdot K}{0.987 \cdot a t m}$

And I make this...$V = 4.65 \cdot L$...so I think there are unspecified data in your question....or I (or the question) might have totally ballsed it up.