# 150g of NaCl completely dissolve in 1.00 kg of water at 25.0 C. The vapor pressure of pure water at this temperature is 23.8 torr. How would you determine the vapor pressure of the solution?

Nov 20, 2015

$\text{22.8 torr}$

#### Explanation:

Since you're dealing with a non-volatile solute, the vapor pressure of the solution will depend exclusively on the mole fraction of the solvent and on the vapor pressure of the pure solvent at that temperature.

Mathematically, the relationship between the mole fraction of the solvent and the vapor pressure of the solution looks like this

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

So, your goal here is to figure out how many moles of sodium chloride and of water you have in the solution.

To do that, use the two compounds' respective molar masses. For sodium chloride you'll have

150color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "2.567 moles NaCl"

For water you'll have

1.00color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"

Now, the mole fraction of water is equal to the number of moles of water divided by the total number of moles present in the solution.

${n}_{\text{total" = n_"water" + n_"NaCl}}$

${n}_{\text{total" = 55.51 + 2.567 = "58.08 moles}}$

The mole faction of water will thus be

chi_"water" = (55.51color(red)(cancel(color(black)("moles"))))/(58.08color(red)(cancel(color(black)("moles")))) = 0.956

This means that the vapor pressure of the solution will be

P_"sol" = 0.956 * "23.8 torr" = color(green)("22.8 torr")

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the mass of sodium chloride.