Question

Suppose that the mixture in problem 4 is at 15 OC, where the pure vapor pressures are 12.5 mmHg for water and 32.1 mmHg for ethanol. According to Raoult’s Law, the pressure of a component in a solution is equal to its pure vapor pressure times its mole fraction, that is PA = ( ) (XA). Use Raoult’s law to determine the vapor pressure of each component in the solution. Then, add them to find the total vapor pressure. Use significant figures. Show all equations and conversion factors.

Answer 1

Answer:

Explanation:

Since we are not given the mole fraction of ethanol and water; we will solve this theoretically.

Using Raoult's Law:

$P_A = (P_o)_A*X_A$

For water:

$(P)w = P_o \times \text{mole fraction of water}$

where $P_o$ of water = 12.5 mmHg

Then, the vapor pressure of water:

$(P)w = 12.5 \ mmHg \times \text{mole fraction of water}$

For ethanol:

$P_E = P_o \times \text {mole fraction of ethanol}$

and the $P_o$ of ethanol = 32.1 mmHg

Then, the vapor pressure of ethanol:

$P_E = 32.1 \ mmHg \times \text {mole fraction of ethanol}$

The total vapor pressure $T_P = P_W + P_E$

The total vapor pressure = $(12.5 \ mmHg \times \text{mole fraction of water}) + (32.1 \ mmHg \times \text {mole fraction of ethanol})$