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 contain8ng 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, let's calculate the amount of ethanol and water in the shot glass:

Total weight of 200 ml of alcohol = 200g (since density of alcohol is approximately 1 g/ml)

Weight of ethanol = 0.45 * 200g = 90g
Weight of water = 0.55 * 200g = 110g

Now let's calculate the partial vapour pressure of ethanol using Raoult's law:

Partial vapour pressure of ethanol = Xethanol * Pethanol
where Xethanol is the mole fraction of ethanol in the solution and Pethanol is the vapour pressure of pure ethanol.

Moles of ethanol = 90g / 46.07 g/mol = 1.954 mol
Moles of water = 110g / 18.02 g/mol = 6.105 mol

Total moles = 1.954 mol + 6.105 mol = 8.059 mol

Mole fraction of ethanol = 1.954 mol / 8.059 mol = 0.2426

Partial vapour pressure of ethanol = 0.2426 * 44 mmHg = 10.666 mmHg

Now, if the shot glass is uncovered and allowed to evaporate, the concentration of ethanol in parts per million (PPM) can be calculated as follows:

Concentration of ethanol in PPM = (moles of ethanol / total moles) * 1,000,000

Concentration of ethanol in PPM = (1.954 mol / 8.059 mol) * 1,000,000 = 242,600 PPM

Therefore, the partial vapour pressure of ethanol above the shot glass containing 200 ml of alcohol at 20 deg C in a closed room at 765 mmHg would be 10.666 mmHg, and the concentration of ethanol in PPM when allowed to evaporate would be 242,600 PPM.