Question

The vapor pressure of pure water at 25 degrees Celsius is 23.8 torr. What is the vapor pressure of a solution prepared by dissolving 18.0 g of glucose (molecular weight = 180.0g/mol) in 95.0 g of water?

Answer 1

`"23.4 torr"`

Explanation:

For solutions that contain non-volatile solutes, the vapor pressure of the solution can be determined by using the mole fraction of the solvent and the vapor pressure of the pure solvent at the same temperature.

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

• `P_"sol"` is the vapor pressure of the solution
• `chi_"solvent"` is the mole fraction of the solvent
• `P_"solvent"^@` is the vapor pressure of the pure solvent

In your case, you know that the vapor pressure of pure water at `25^@"C"` is equal to `23.8` torr. This means that all you have to do is determine the mole fraction of water in the solution.

As you know, mole fraction is defined as the number of moles of a component of a solution divided by the total number of moles present in that solution.

Use glucose and water's respective molar masses to determine how many moles of each you have

`18.0color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.0color(red)(cancel(color(black)("g")))) = "0.100 moles glucose"`

and

`95.0color(red)(cancel(color(black)("g"))) * "1 mole water"/(18.015color(red)(cancel(color(black)("g")))) = "5.273 moles water"`

The total number of moles present in the solution will be

`n_"total" = n_"glucose" + n_"water"`

`n_"total" = 0.100 + 5.273 = "5.373 moles"`

This means that the mole fraction of water will be

`chi_"water" = (5.273color(red)(cancel(color(black)("moles"))))/(5.373color(red)(cancel(color(black)("moles")))) = 0.9814`

Finally, the vapor pressure of the solution will be

`P_"sol" = 0.9814 * "23.8 torr" = color(green)("23.4 torr")`

The answer is rounded to three sig figs.