Question

A solution was prepared by dissolving 27.0 g sucrose in 600. g of water. The solution was found to have a density of 1.02 g/mL at25.0oC. Using this and any other information you might need, predict the boiling point andthe freezing point of this solution. Additionally, findthe vapour pressure and the osmotic pressure of this solution at 25.0oC.

Answer 1

- Boiling point of solution: 100.06°C

- Freezing point of solution: - 0.24°C

- Vapour pressure of solution: 23.74 mmHg

- Osmotic pressure: 3.13atm

Step-by-step explanation:

In this excersise, we have to apply all the colligative properties (vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure)

Generally, they work with molality (mole of solute in 1kg of solvent)

Sucrose is C₁₂H₂₂O₁₁ (molar mass: 342.3 g/m)

Mass / Molar mass = Mole

27 g / 342.3 g/m = 0.0788 m

In 600 g ___ 0.0788 mole

In 1000g ___ (1000 . 0.0788)/600 = 0.131 m

Total moles = Mol from sucrose + Mol from water

600 g / 18 g/m = 33.3 mole

0.0788 + 33.3 = 33.412 total moles in solution

With this information we can know, mole fraction from solute.

0.0788 / 33.412 = 2.36x10⁻³

Total mass = 27 g + 600 g = 627 g

Using density, we can know the volume

1.02 g/ml = solution mass / solution volume

1.02 g/ml = 627 g / solution volume

627 g / 1.02g/ml = solution volume → 614.7 mL

Let's calculate molarity: (mol/L)

0.6147 L ___ 0.0788 mole of solute

1L _______ (0.0788 / 0.6147) = 0.128 M

T° of solution: 25°C

Absolute T° = T°C + 273 → 298K

Sucrose is non electrolytic so Van't Hoff factor is 1, we won't consider it. Now we can apply all the colligative properties.

- OSMOTIC PRESSURE

π = M . R . T

π = 0.128 mol/L . 0.082 L.atm/ mol.K . 298K

π = 3.13 atm

- BOILING POINT ELEVATION

ΔT = Kb . m

ΔT = T° boling solution - T° boling pure solvent

Kb = Ebulloscopic constant (For water is 0.512 °C/m)

m = molality

T° boiling solution - 100°C = 0.512°C/m . 0.131 m

T° boiling solution = ( 0.512°C/m . 0.131 m) + 100°C → 100.06°C

- FREEZING POINT DEPRESSION

ΔT = Kf . m

ΔT = T° freezing pure solvent - T° freezing solution

Kf = Cryscopic constant (For water is 1.86 °C/m)

m is molality

0°C - T°freezing solution = 1.86°C/m . 0.131m

- T°freezing solution = (1.86°C/m . 0.131m) + 0°C → - 0.24°C

- VAPOUR PRESSURE LOWERING

ΔP = P° - Xₙ

ΔP = Vapour pressure of pure solvent - Vapour pressure of solution

P° = Vapour pressure of pure solvent (23.8 mmHg)

Xₙ = mole fraction for solvent

23.8 mmHg - Vapor pressure of solution = 23.8 mmHg . 2.36x10⁻³

- Vapor pressure of solution = (23.8 mmHg . 2.36x10⁻³) - 23.8mmHg

Vapor pressure of solution = 23.74 mmHg