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.