Question

A solution of sucrose has been prepared by dissolving 68.4 g of sucrose in one kg of water. Calculate the following:

1. The vapour pressure of the solution at 298 K.
2. Osmotic pressure of the solution at 298 K.
3. Freezing point of the solution.

(Given: Vapour pressure of water at 298 K = 0.024 atm, K_f for water = 1.86 K kg mol⁻¹, R = 0.0821 L atm mol⁻¹ K⁻¹)

Answer 1

Given: Mass of sucrose = 68.4 g

Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 g/mol

Mass of water = 1 kg

Temperature T = 298 K

i. Vapour pressure of water at 298 K (P°) = 0.024 atm

K_f = 1.86 K kg mol⁻¹

R = 0.0821 L atm mol⁻¹

`"Moles of sucrose" = "Mass of sucrose"/"Molar mass of sucrose"`

Moles of sucrose = `68.4/342`

= 0.2 mol

Moles of water = `1000/18`

= 55.56 mol

Mole fraction of water `X_"water" = 55.56/(55.56 + 0.2)`

= `55.56/55.79`

= 0.9964 

By using Raoult’s law,

`P_"solution" = P^circ * X_"solvent"`

Vapour pressure `P_"solution" = 0.024 xx 0.9964`

= 0.0239 atm

ii. Osmotic pressure (π) = CRT

= `n/V RT`

Let volume ≈ mass of water = 1 L

`pi = 0.2/1 xx 0.0821 xx 298`

= 4.90 atm

iii. Freezing point of the solution

ΔT_f ​= K_f​ ⋅ m

Molality (m) = `0.2/1`

= 0.2 mol/kg

ΔT_f ​= 1.86 × 0.2

= 0.372 K

T_f ​= 273 − 0.372

= 272.63 K