Answer to Question #299371 in General Chemistry for Irish

Question

Calculate the new boiling point of a solution containing 5.27 g of sucrose (C12H22O11) and 45 g 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 behaviour. It is:

Δt = i × K_b × m

where

Δt = the change in boiling point (the boiling point of pure water is 100°C)

i = Vant Hoff's coefficient (for nonelectrolytes, like C₁₂H₂₂O₁₁, i = 1)

K_b = the molal boiling-point elevation constant (K_b for water is 0.512°C kg mol¯¹)

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.

Therefore,

Moles of C₁₂H₂₂O₁₁ = (5.27 g) / (342.3 g mol¯¹) = 0.0154 mol

Molality of C₁₂H₂₂O₁₁ = (0.0154 mol) / (0.045 kg) = 0.342 mol kg¯¹

Thus:

Δt = i × K_b × m = (1) × (0.512°C kg mol¯¹) × (0.342 mol kg¯¹) = 0.175°C = 0.18°C

 

The boiling point of pure water is 100°C.

Therefore,

t = 100°C + 0.18°C = 100.18°C

t = 100.18°C

Answer: The new boiling point of a solution is 100.18°C