How would you calculate the vapor pressure of a solution made by dissolving 88.2 g of urea (molar mass = 60.06 g/mol) in 303 mL of water at 35°C? Vapor pressure of water at 35 degrees is 42.18 mm Hg.

Since urea is a non-volatile substance, the vapor pressure above the solution will only contain water vapor.

This means that you can use the mole fraction of water and the vapor pressure of pure water at `35^@"C"` to determine the vapor pressure of the solution.

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

`chi_"water"` - the vapor pressure of water
`P_"water"^@` - the vapor pressure of pure water

Now, in order to determine the mole fraction of water, you will need to know the number of moles of water you have in that sample. To do that, use water's density at `35^@"C"`, which is listed as being equal to `"0.994 g/mL"`

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

`303color(red)(cancel(color(black)("mL"))) * "0.994 g"/(1color(red)(cancel(color(black)("mL")))) = "301.2 g"`

Use water's molar mass to find the number of moles you have in this many grams of water

`301.2color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "16.72 moles water"`

Now use urea's molar mass to determine how many moles you have in that sample of urea - you will need this number for the mole fraction of water!

`88.2color(red)(cancel(color(black)("g"))) * "1 mole urea"/(60.06color(red)(cancel(color(black)("g")))) = "1.469 moles urea"`

The mole fraction of water is defined as the number of moles of water divided by the total number of moles present in solution

`n_"total" = n_"water" + n_"urea"`

`n_"total" = 16.72 + 1.469 = "18.19 moles"`

This means that you have

`chi_"water" = (16.72color(red)(cancel(color(black)("moles"))))/(18.19color(red)(cancel(color(black)("moles")))) = 0.9192`

Therefore, the vapor pressure of the solution will be

`P_"sol" = 0.9192 * "42.18 mmHg" = color(green)("38.8 mmHg")`

The answer is rounded to three sig figs.