Answer to Question #165801 in Chemistry for Phyroe

Question

What is the boiling point of a solution of cane sugar, C12H22O11, containing 100 g of the solute per kilogram of water?

Answer 1

Solution:

A solution will boil at a higher temperature than the pure solvent. This is the colligative property called boiling point elevation.

An equation has been developed for this behavior. It is:

Δt = i × K_b × m

K_b for water is 0.52°C kg mol¯¹

i = 1 (for nonelectrolyte - C₁₂H₂₂O₁₁)

m = the molality of the solute

Molality of C₁₂H₂₂O₁₁ = Moles of C₁₂H₂₂O₁₁ / kilograms solvent

Moles of C₁₂H₂₂O₁₁ = Mass of C₁₂H₂₂O₁₁ / Molar mass of C₁₂H₂₂O₁₁

The molar mass of C₁₂H₂₂O₁₁ is 342.3 g mol^-1.

Hence,

Moles of C₁₂H₂₂O₁₁ = (100 g) / (342.3 g mol^-1) = 0.292 mol

Molality of C₁₂H₂₂O₁₁ = (0.292 mol) / (1 kg) = 0.292 mol/kg

Thus:

Δt = i × K_b × m = (1) × (0.52°C kg mol¯¹) × (0.292 mol/kg) = 0.15184°C = 0.15°C

The boiling point of pure water is 100°C.

Hence,

t = 100°C + 0.15°C = 100.15°C

t = 100.15°C

Answer: The boiling point of a solution of cane sugar (C₁₂H₂₂O₁₁) is 100.15°C.