Calculate vapor pressure and mole fraction?

Question

A solution of sodium chloride in water has a vapor pressure of 20.9 torr at 25 degrees C. What is the mole fraction of NaCl solute particles in this solution?
What would be the vapor pressure of this solution at 45 degrees C? The vapor pressure of pure water is 23.8 torr at 25 degrees C and 71.9 torr at 45 degrees C and assume sodium chloride exists as Na+ and Cl- ions in solution.

Answer 1

If pure water has a vapor pressure of 23.8 torr and the vapor pressure with salt added is 20.9 torr, then we know the salt has reduced the vapor pressure of the water. The formula to calculate the change in vapor pressure of a solvent, known as Raoult's Law, states:

Change in Vapor Pressure= Mole Fraction of Solute*Vapor Pressure of Pure Solvent

(23.8-20.9)=X*20.9
Solving for X:
.139 mole fraction

So far so good, but notice that NaCl yields two ions:

NaCl-->Na+ Cl-
So for each 1 mole NaCl, you have a total of two ions. Therefore, the actual mole fraction of solid NaCl is 1/2 of .139, or about .07.

At 45 Celsius:

(71.9-p)=.139*71.9

We set up the above equation with "p" representing the new vapor pressure. We now solve for it:

p=71.9-(.139*71.9)=61.923

So at 45 Celsius, with a mole fraction of .139 Na+Cl-, the vapor pressure of a salt-water solution is lowered to 61.92.