Question

5 moles of sucrose (molecular mass = 342 g mol⁻¹) dissolved in 1000 g of water (molecular mass = 18 g mol⁻¹). Vapour pressure of pure water at 298 K = 4.57 mm Hg.

What will be the vapour pressure of sucrose solution at the same temperature?

Options

• 0.419 mm Hg

• 6.570 mm Hg

• 4.190 mm Hg

• 0.657 mm Hg

Answer 1

4.190 mm Hg

Explanation:

To calculate the vapour pressure of the sucrose solution, we will use Raoult’s Law:

`P_"solution" = P_"water"^circ xx chi_"solvent"`

⁣`chi_"solvent"` = Mole fraction of water

Moles of sucrose (solute) = 5 mol

Mass of water = 1000 g

Molar mass of water = 18 g/mol

Moles of water = `1000/18` = 55.56 mol

`chi_"water" = 55.56/(55.56 + 5)`

= `55.56/60.56`

= 0.9174

`P_"solution" = 4.57 xx 0.9174`

= 4.19 mm Hg