# Question #a9c80

Sep 4, 2017

From this data, we can say there is liquid present, approximately $5.2 g$.

#### Explanation:

This is a pretty complex gas law question.

The vapor pressure of a liquid is when liquid and gas phase transition are in equilibrium.

For the purposes of this question I've converted your pressure atm:

$P = 1.138$ atm

$1.138 a t m \cdot 1.0 L = n \cdot \frac{0.08026 L \cdot a t m}{m o l \cdot K} \cdot 300 K$

$\therefore n \approx 4.6 \cdot {10}^{-} 2 m o l$

$M M \left[C C {l}_{3} F\right] = \frac{137.5 g}{m o l}$

$4.6 \cdot {10}^{-} 2 m o l \cdot \frac{137.5 g}{m o l} \approx 6.325 g$ of trichloroflouromethane gas.

Subtract the mass of the gas present from the mass of the liquid...