Answer:
0.2188 bar.
Explanation:
We can use the Ideal Gas Law to solve this problem.
pV = nRT
However, [tex]n = \frac{ m}{ M}[/tex], so
[tex]pV = \frac{ m}{ M}RT[/tex]
[tex]p = \frac{mRT }{ MV}[/tex]
In this problem,
T = (100.0 + 273.15) K = 373.15 K
[tex]p = \frac{\text{5.000 g}\times \text{0.083 14 bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{373.15 K}}{ 70.91 \text{g}\cdot\text{mol}^{-1}\times \text{10.000 L}} = \text{0.2188 bar}[/tex]
