Question

What is the vapor pressure of a solution of 44.0 g of glycerol (C3H8O3) in 500.0 g of water at 25oC? The vapor pressure of water at 25oC is 23.76 torr. Glycerol can be considered a dilute, non-volatile solvent which is behaving ideally.Type the number?

Answer 1

The vapour pressure is 23.36 Torr.

Explanation:

Raoult's law states that the vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.

In symbols,

`p = chi_1p_1^@`

where

`chi_1 =`the mole fraction of the solvent
`p` and `p_1^@` are the vapour pressures of the solution and of the pure solvent

The vapour pressure lowering is given by

`Δp = p_1^@ - p_1 = p_1^@ - chi_1p_1^@ = p_1^@(1 - chi_1)`

`1 - chi_1 = chi_2`, so,

`Δp = chi_2p_1^@`

where `chi_2` is the mole fraction of the solute.

Step 1. Calculate the moles of each substance

`"Moles of glycerol" = n_2 = 44.0color(red)(cancel(color(black)("g glycerol"))) × "1 mol glycerol"/(92.09 color(red)(cancel(color(black)("g glycerol")))) = "0.4778 mol glycerol"`

`"Moles of water" = n_1 = 500.0 color(red)(cancel(color(black)("g H"_2"O"))) × ("1 mol H"_2"O")/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "27.75 mol H"_2"O"`

`"Total moles" = n_2 + n_1 = "(0.4778 + 27.75) mol" = "28.22 mol"`

Step 2. Calculate the mole fraction of glycerol

`chi_2 = n_2/(n_1 + n_2) = (0.4778 color(red)(cancel(color(black)("mol"))))/(28.22 color(red)(cancel(color(black)("mol")))) = "0.016 93"`

Step 3. Calculate the vapour pressure lowering

`Δp = chi_2p_1^@ = "0.016 93" × "23.76 Torr" = "0.4022 Torr"`

Step 4. Calculate the vapour pressure

`p_1 = p_1^@ - Δp = "(23.76 - 0.4022) Torr = 23.36 Torr"`