Sucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure at `25^@"C"` of the `"1.5-m"` sucrose solution. Assume that the solution behaves ideally. The vapor pressure of pure water at `25^"C"` is `"23.8 torr"` ?

The idea here is that when you're dealing with a nonvolatile solute, you can calculate the vapor pressure of the solution by using the mole fraction of the solute and the vapor pressure of the pure solvent at the given temperature--think Raoult's Law here.

`color(blue)(ul(color(black)(P_"sol" = chi_"sucrose" * P_ ("H"_ 2"O")^@)))`

Here

• `P_"sol"` is the vapor pressure of the solution
• `chi_"sucrose"` is the mole fraction of sucrose
• `P_ ("H"_ 2"O")^@` is the vapor pressure of water at `25^@"C"`

Now, you know that your solution has a molality of `"1.5 m"`, which means that you get `1.5` moles of sucrose, the solute, for every `"1 kg"` of water, the solvent.

To make the calculations easier, let's pick a sample of this solution that contains exactly `"1 kg"` of water. Consequently, this sample will also contain `1.5` moles of sucrose.

Use the molar mass of water to convert the mass of water, which is equal to `10^3` `"g"`, to moles

`10^3 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "55.51 moles H"_2"O"`

The mole fraction of sucrose is calculated by dividing the number of moles of sucrose by the total number of moles present in the solution.

`chi_"sucrose" = (1.5 color(red)(cancel(color(black)("moles"))))/((1.5 + 55.51)color(red)(cancel(color(black)("moles")))) = 0.0263`

You can thus say that the vapor pressure of the solution will be equal to

`P_"sol" = 0.0263 * "23.8 torr" = color(darkgreen)(ul(color(black)("0.63 torr")))`

The answer is rounded to two sig figs, the number of sig figs you have for the molality of the solution.