Question

How do you calculate the vapor pressure at 50°C of a coolant solution that is 52.0:48.0 ethylene glycol-to-water by volume? At 50.0°C, the density of water is 0.9880 g/mL, and its vapor pressure is 92 torr.

The liquid used in automobile cooling systems is prepared by dissolving ethylene glycol (HOCH2CH2OH) in water. Ethylene glycol has a molar mass of 62.07 g/mol and a density of 1.115 g/mL at 50.0°C.

Answer 1

The vapour pressure is 61 Torr.

Explanation:

At 50 °C, the vapour pressure of ethylene glycol (EG) is negligible.

Thus, we have a solution of a nonvolatile solute (EG) in a volatile solvent (water).

If the solution is ideal, we can use Raoult's Law:

The partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

In symbols,

`color(blue)(bar(ul(|color(white)(a/a)p_i =chi_ip_i^@color(white)(a/a)|)))" "`

where

`p_i =` the partial pressure of component `i`
`chi_i =` the mole fraction of component `i`
`p_i^@ =` the vapour pressure of the pure component

Assume that we have 100 mL of the solution.

Then we have 52 mL of EG and 48 mL of water.

Step 1. Calculate the moles of water

`"Mass of H"_2"O" = 48.0 color(red)(cancel(color(black)("mL H"_2"O"))) × ("0.9880 g H"_2"O")/(1 color(red)(cancel(color(black)("mL H"_2"O")))) = "47.42 g H"_2"O"`

`"Moles of H"_2"O" = 47.42 color(red)(cancel(color(black)("g H"_2"O"))) × ("1 mol H"_2"O")/(18.02 color(red)(cancel(color(black)("H"_2"O")))) = "2.632 mol H"_2"O"`

Step 2. Calculate the moles of EG

`"Mass of EG" = 52.0 color(red)(cancel(color(black)("mL EG"))) × "1.115 g EG"/(1 color(red)(cancel(color(black)("mL EG")))) = "57.98 g EG"`

`"Moles of EG" = 57.98 color(red)(cancel(color(black)("g EG"))) × "1 mol EG"/(62.07 color(red)(cancel(color(black)("g EG")))) = "0.9341 mol EG"`

Step 3. Calculate the mole fraction of water

Let water be component 1 and EG be component 2.

`n_text(tot) = n_1+ n_2 = "(2.363 + 0.9341) mol = 3.566 mol"`

`chi_1 = n_1/n_text(tot) = (2.362 color(red)(cancel(color(black)("mol"))))/(3.566 color(red)(cancel(color(black)("mol")))) = 0.6624`

Step 4. Calculate the vapour pressure over the solution

`p_text(vap) = p_1 = chi_1p_1^@ = "(0.6624 × 92) Torr = 61 Torr"`