# Question #5f0d5

May 19, 2017

$0.124 m o l {N}_{2}$

#### Explanation:

We'll use the ideal-gas equation for this problem.

First we must convert each unit into the appropriate unit for the equation:

$347 m L \frac{1 L}{{10}^{3} m L} = 0.347 L$

$6680 t o r r \frac{1 a t m}{760 t o r r} = 8.79 a t m$

${27}^{o} C + 273 = 300 K$

Now, we'll use the ideal-gas equation, and solve it for quantity (in $m o l$):

$n = \frac{P V}{R T}$

$n = \frac{\left(8.79 a t m\right) \left(0.347 L\right)}{\left(0.08206 \frac{L - a t m}{m o l - K}\right) \left(300 K\right)} = 0.124 m o l {N}_{2}$