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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?

2 Answers

5 votes

Final answer:

The vapor pressure of the solution can be calculated using Raoult's law. First, we need to calculate the mole fraction of glycerol and water in the solution. Then, we can use the mole fraction to calculate the vapor pressure of the solution.

Step-by-step explanation:

In order to find the vapor pressure of the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

First, we need to calculate the mole fraction of glycerol in the solution. To do this, we need to convert the mass percent of glycerol to moles. Since the molar mass of glycerol is 92.1 g/mol and the solution contains 34.4% glycerol by mass, we can calculate the moles of glycerol:

Moles of glycerol = (34.4 g / 92.1 g/mol)

Next, we need to calculate the mole fraction of water in the solution. Since the solution is 34.4% glycerol, the mass percent of water is 100% - 34.4% = 65.6%. Using the molar mass of water (18.0 g/mol), we can calculate the moles of water:

Moles of water = (65.6 g / 18.0 g/mol)

Now we can calculate the mole fraction of glycerol and water:

Mole fraction of glycerol = (moles of glycerol) / ((moles of glycerol) + (moles of water))

Mole fraction of water = (moles of water) / ((moles of glycerol) + (moles of water))

Finally, we can calculate the vapor pressure of the solution using Raoult's law:

Vapor pressure of solution = (mole fraction of water) * (vapor pressure of water)

User Tal Yarkoni
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2 votes

Answer:

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

Step-by-step 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

User Jamie Rumbelow
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