Question

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?

Answer 1

Answer:

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