The vapour pressure of pure ethanol at 20 deg C is 44 mm HG . Alcohol contains 45 % ethanol and 55% water by weight. What would be partial vapour pressure of ethanol above a shot glass containing 200 ml of alcohol in closed room at 20 deg C at 765 mm HG? If this shot glass with alcohol is not covered and let it to evaporate , what would be concentration of ethanol in PPM? MW ethanol = 46.07 amu and MW water = 18.02 amu?

Answer 1

First, we need to determine the amount of ethanol and water in the shot glass containing 200 ml of alcohol.

Amount of ethanol = 200 ml x 0.45 = 90 ml
Amount of water = 200 ml x 0.55 = 110 ml

Convert the amounts to grams:
Mass of ethanol = 90 ml x 0.789 g/ml = 70.41 g
Mass of water = 110 ml x 1.0 g/ml = 110 g

Now, we can calculate the partial vapour pressure of ethanol above the shot glass:

Partial pressure of ethanol = (Mass of ethanol / Total mass) x Vapour pressure of pure ethanol

Partial pressure of ethanol = (70.41 g / (70.41 g + 110 g)) x 44 mmHg
Partial pressure of ethanol = (70.41 g / 180.41 g) x 44 mmHg
Partial pressure of ethanol ≈ 17.03 mmHg

Next, if the shot glass is uncovered and allowed to evaporate, the concentration of ethanol in parts per million (PPM) can be calculated using the following formula:

Concentration of ethanol in PPM = (Mass of ethanol / Total mass) x (1000000)

Concentration of ethanol in PPM = (70.41 g / 180.41 g) x 1000000
Concentration of ethanol in PPM ≈ 390133.20 PPM

Therefore, the partial vapour pressure of ethanol above the shot glass at 20 deg C in a closed room at 765 mmHg would be approximately 17.03 mmHg and the concentration of ethanol in parts per million (PPM) if the shot glass is uncovered and left to evaporate would be approximately 390133.20 PPM.