Question #66b16

Assuming that you're dealing with a non-volatile solute, you can say that the vapor pressure of the solution, `P_"sol"`, will depend on the mole fraction of the solvent, `chi_"water"`, and on the vapor pressure of the pure solvent, `P_"water"^@`. as given by Raoult's Law

`color(blue)(ul(color(black)(P_"sol" = chi_"water" * P_"water"^@)))" "color(darkorange)("(*)")`

Now, the mole fraction of water is defined as the ratio between the number of moles of water and the total number of moles present in the solution.

If you take `m` to be the mass of the solute in grams, you can say that the solution will contain

`m color(red)(cancel(color(black)("g"))) * "1 mole solute"/(60color(red)(cancel(color(black)("g")))) = (m/60)` `"moles solute"`

and

`180 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "9.992 moles H"_2"O"`

The total number of moles in the solution will be equal to

`(m/60)color(white)(.)"moles" + "9.992 moles" = ((m + 599.52)/60)` `"moles"`

The mole fraction of water will thus be

`chi_"water" = (9.992 color(red)(cancel(color(black)("moles"))))/(( (m + 599.52)/60)color(red)(cancel(color(black)("moles")))) = 599.52/(m + 599.52)`

You know that the vapor pressure of the solution must be `4/5"th"` the vapor pressure of the pure solvent, so

`P_"sol" = 4/5 * P_"water"^@`

Plug this into equation `color(darkorange)("(*)")` to get

`4/5 * color(red)(cancel(color(black)(P_"water"^@))) = chi_"water" * color(red)(cancel(color(black)(P_"water"^@)))`

This is equivalent to

`599.52/(m + 599.52) = 4/5`

Solve for `m` to find

`4 * (m + 599.52) = 599.52 * 5`

`m = (599.52 * 5 - 599.52 * 4)/4`

`m = 599.52/4 = 149.88`

Therefore, you can say that the mass of the solute is equal to

`color(darkgreen)(ul(color(black)("mass solute = 150 g")))`

I'll leave the answer rounded to two sig figs.