Question

Calculate the vapor pressure of water above a solution prepared by dissolving 28.5 g of glycerin 1C3H8O32 in 125 g of water at 343 K

Answer 1

Vapour pressure of the solution is 223.8mmHg

Step-by-step explanation:

Based on Raoult's law, vapour pressure of a solution decreases related to the vapour pressure of the pure solvent following the equation:

`P_(solution) = X_(solvent)P_(solvent)^0`

Where P is vapour pressure and X mole fraction of each related substance.

Vapour pressure of pure water is 233.8mmHg

To find Mole fraction of the solution you need to find moles of glycerin and moles of water, mole fraction of water will be:

`(n_(water))/(n_(water)+n_(glycerin))`

Moles of water (Molar mass: 18.02g/mol):

125g × (1mol / 18.02g) = 6.937 moles water

Moles of glycerin (Molar mass: 92.09g/mol):

28.5g × (1mol / 92.09g/mol) = 0.309 moles glycerin

Thus, mole fraction of water, X, is:

`(6.937mol)/(6.937mol+0.309mol)`

= 0.957

Replacing in Raoult's law:

`P_(solution) = 0.957*233.8mmHg`

P = 223.8mm Hg

Vapour pressure of the solution is 223.8mmHg