What is the vapor pressure of a #"1.00-molal"# sugar solution at #25^@"C"#? Sugar is nonvolatile, nonelectrolyte solute. The vapor pressure of water at #25^@"C"# is #"23.8 torr"#. The molar mass of water is #"18 g/mol"#

1 Answer
Jan 14, 2018

Answer:

#"23.4 torr"#

Explanation:

The idea here is that the vapor pressure of the solution will actually be lower than the vapor pressure of the pure solvent because of the presence of the nonvolatile solute #-># think Raoult's Law here.

The vapor pressure of the solution can be expressed as

#color(blue)(ul(color(black)(P_"sol" = chi_"solvent" * P_"solvent"^@)))#

Here

  • #P_"sol"# is the vapor pressure of the solution at the given temperature
  • #chi_"solvent"# is the mole fraction of the solvent in the solution
  • #P_"solvent"^@# is the vapor pressure of the pure solvent at the given temperature

Now, you know that your solution has a molality equal to #"1.00 mol kg"^(-1)#, which means that you get #1.00# mole of sugar for every #"1 kg"# of solvent.

To make the calculations easier, let's pick a sample of this solution that contains exactly #"1 kg"# of water. use the molar mass of water to convert the number of grams to moles

#1 color(red)(cancel(color(black)("kg"))) * (10^3 quad color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("1 mole H"_2"O")/(18 color(red)(cancel(color(black)("g")))) = "55.56 moles H"_2"O"#

The mole fraction of water is defined as the number of moles of water divided by the total number of moles present in your solution. Since the sample contains exactly #"1 kg"# of water, you know that also contains #1.00# mole of sugar, so the mole fraction of water will be

#chi_"solvent" = (55.56 color(red)(cancel(color(black)("moles"))))/((55.56 + 1.00)color(red)(cancel(color(black)("moles")))) = 0.98232#

You can thus say that the vapor pressure of the solution at #25^@"C"# will be equal to

#P_"sol" = 0.98232 * "23.8 torr"#

#color(darkgreen)(ul(color(black)(P_"sol" = "23.4 torr")))#

The answer is rounded to three sig figs.