# A solution of sodium chloride in water has a vapor pressure of 19.6 torr at 25 degrees Celsius. What is the mole fraction of NaCl solute particles in this solution?

Nov 21, 2015

$0.173$

#### Explanation:

The vapor pressure for a solution that contains a non-volatile solute will depend on the vapor pressure of the pure solvent at that temperature and on the mole faction of the solvent.

color(blue)(P_"sol" = chi_"solvent" * P_"solvent"^@)" ", where

${P}_{\text{sol}}$ - the vapor pressure of the solution
${\chi}_{\text{solvent}}$ - the mole fraction of the solvent
${P}_{\text{solvent}}^{\circ}$ - the vapor pressure of the pure solvent

This means that in order to be able to calculate the mole fraction of sodium chloride, you need to know what the vapor pressure of pure water is at ${25}^{\circ} \text{C}$.

You can use an online calculator to find that the vapor pressure of pure water at ${25}^{\circ} \text{C}$ is equal to about $\text{23.7 torr}$.

http://www.endmemo.com/chem/vaporpressurewater.php

So, plug in your values into the above equation and solve for ${\chi}_{\text{water}}$

${\chi}_{\text{water" = P_"sol"/P_"water}}^{\circ}$

chi_"water" = (19.6color(red)(cancel(color(black)("torr"))))/(23.7color(red)(cancel(color(black)("torr")))) = 0.827

Since the solution only contains water and sodium chloride, you can say that

${\chi}_{\text{water" + chi_"NaCl}} = 1$

This means that the mole fraction of sodium chloride is

${\chi}_{\text{NaCl" = 1 - chi_"water}}$

${\chi}_{\text{NaCl}} = 1 - 0.827 = \textcolor{g r e e n}{0.173}$