Glycerol (c3h8o3, 92.1 g/mol) is a nonvolatile nonelectrolyte substance. consider that you have an aqueous solution that contains 34.4 % glycerol by mass. if the vapor pressure of pure water is 23.8 torr at 25oc, what is the vapor pressure of the solution at 25oc?

The vapor pressure of the solution at 25°C is 26.01 Torr

Explanation:

This is a usual excersise of colligative properties. In this case we apply the vapor pressure lowering formula:

ΔP = Xst . P°

Where ΔP is the diferrence between  Pressure of solution - Pressure of pure solvent.

And Xst  the molar fraction.

P° is Pressure of pure solvent.

So the formula will be:

Pressure Solution - P° = Xst . P°

Pressure Solution - 23.8 Tor = Xst . 23.8 Torr

Xst : Mole fraction ( Moles of solute or solvent / Total moles)

34.4 % m/m means that in 100 g of solution I have 34.4 g of solute

If I have 34.4 g of solute and the mass of 100 g in solution, I can know the mass of solvent, and finally the moles.

100 g solution - 34.4 g solute = 65.6 g (mass of solvent)

Molar mass of water : 18 g/m

Moles of water: Mass of water / Molar mass

65.6 g / 18g/m = 3.64 moles

Moles of glycerol : Mass glycerol / Molar mass glycerol

34.4 g / 92.1 g/m = 0.373 moles

Total moles: moles of glycerol + moles of water

0.373 m + 3.64 m = 4.01 m

So Xst = 0.373 m / 4.01 m → 0.093

Pressure Solution - 23.8 Tor = 0.093 . 23.8 Torr

Xst HAVE NO UNITS

Pressure Solution = (0.093 . 23.8 Torr ) + 23.8 Tor

Pressure Solution = 26.01 Torr