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Answer 1 · 2017-09-04T17:56:27

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

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:

$P = 1.138$ $P = 1.138$ atm

$1.138 a t m \cdot 1.0 L = n \cdot \frac{0.08026 L \cdot a t m}{m o l \cdot K} \cdot 300 K$ $1.138atm*1.0L = n*(0.08026L*atm)/(mol*K)*300K$

$therefore n approx 4.6*10^-2mol$

$MM[C Cl_3F] = (137.5g)/(mol)$

$4.6*10^-2mol* (137.5g)/(mol) approx 6.325g$ of trichloroflouromethane gas.

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