Question

Calculate the vapour pressure lowering

caused by the addition of 68.4 g of

sucrose(molecular weight=342g) to 500 g

of water if the vapour pressure of pure

water at 25⁰c is 20.0 mm of hg​

Answer 1

We have to find the vapor pressure lowering caused by the addition of 68.4 g of sucrose (Mol wt 342g) to 500g of water if the vapor pressure of pure water at 25°C is 20 mm Hg.

solution : initial vapor pressure ( vapor pressure of pure water) , P₀ = 20 mm

mass of sucrose = 68.4 g

molar mass of sucrose = 342 g/mol

so no of moles of sucrose , n₁ = 68.4/342 = 0.2 mol

mass of water = 500g

molar mass of water = 18g/mol

so no of moles of water , n₂ = 500/18 = 27.78 mol

now using concept,

lowering of vapor pressure = ∆P/P₀ = mole fraction of solute

⇒∆P/P₀ = n₁/(n₁ + n₂)

⇒∆P/20 = 0.2/(27.78 + 0.2) = 0.2/27.98

⇒∆P = 20 × 0.2/27.98 = 0.143 mm Hg

Therefore the vapor pressure lowering caused by addition of sucrose is 0.143 mm Hg of solution.