Question

A solution containing 8.44 g of sucrose in 100 g of water has a vapour pressure 4.56 mm of Hg at 273 K. If the vapour pressure of pure water is 4.58 mm of Hg at the same temperature, calculate the molecular weight of sucrose. 

Answer 1

`P_A^circ` = 4.56 mm of Hg

P_A = 4.56 mm of Hg

`(P_A^circ - P_A)/P_A^circ = X_B = (n_B)/(n_B + n_A)`

`(4.58 - 4.56)/4.58 = (8.44/M_B)/(100/18 + 8.44/M_B)`

`0.02/4.58 = 8.44/M_B xx (18 M_B)/((100 M_B + 151.92))` 

`0.02/4.58 = (8.44 xx 18)/(100 M_B + 151.92)`

`100 M_B + 151.92 = (8.44 xx 18 xx 4.58)/0.02`

M_B = 346.38 amu

Answer 2

Given: Mass of sucrose (w₂) = 8.44 g

Mass of water (w₁) = 100 g = 0.1 kg

Vapour pressure of pure water (P⁰) = 4.58 mm Hg

Vapour pressure of solution (P) = 4.56 mm HgP

Molar mass of water = 18 g/mol

Use Raoult’s law for lowering of  vapour pressure

`(P^0 - p)/P^0 = n_"solute"/n`

`(4.58 - 4.56)/4.58 = n_"solute"/(100/18)`

= `(8.44//M)/5.56`

`0.02/4.58` = 0.00437

`n_"solvent" = 100/18`

= 5.56 mol

`8.44/M = 0.00437 xx 5.56`

`8.44/M = 0.0243`

`M = 8.44/0.0243`

= 347.33 g/mol

∴ The molecular weight of sucrose is 347.33 g/mol.