Question

Ethanol is commonly referred to simply as alcohol or spirits, It is the principal type of alcohol found in alcoholic beverages. It is a psychoactive drug and one of the oldest recreational drugs used by humans. It can cause alcohol intoxication when consumed in sufficient quantity. It is a volatile, flammable, colorless liquid with the structural formula CH3CH2OH. The density of an aqueous solution containing 45 weight percent ethanol is 0.92472 g/mL.

a) A person weighing 75 kg takes a swallow of an alcoholic beverage corresponding to 25 gram of pure ethanol. It is assumed that the alcohol mixes evenly in person´s water phase, constituting 62 % of the person´s weight. Calculate the person´s ethanol mass per mille (alkoholpromille in Danish).

b) The vapor pressures of ethanol (C2H5OH) and 1-propanol (C3H7OH) at 35 oC are 100.0 mmHg and 37.6 mmHg, respectively. Assume ideal behaviour and calculate the partial pressures of ethanol and 1-propanol at 35 oC over a solution of ethanol in 1- propanol, in which the mole fraction of ethanol is 0.200.

Answer 1

b) According to Raoult's law,

Partial pressure of A = Mole fraction of A x Vapor pressure of A in its pure form

Partial pressure of ethanol

Given that

vapor pressure of ethanol = 100 mmHg

mole fraction of ethanol = 0.200

Hence,

        partial pressure of ethanol = Mole fraction x vapor pressure in pure form

                                              = 0.200 x 100 mmHg

                                              = 20 mmHg

Therefore,

partial pressure of ethanol = 20 mmHg

Partial pressure of 1-propanol:

Given that

vapor pressure of 1- propanol = 37.6 mmHg

mole fraction of 1- propanol = 1 - mole fraction of ethanol = 1 - 0.2 = 0.8

Hence,

partial pressure of 1-propanol = Mole fraction x vapor pressure in pure form

                                          = 0.800 x 37.6 mmHg

                                          = 30.08 mmHg

Therefore,

partial pressure of 1-propanol = 30.08 mmHg