Question

Vapour pressure of pure water at 25 degree Celsius is 23. 62 mm what will be the vapour pressure of a solution of 1.5 gram of Urea in 50 gram water

Answer 1

vapor pressure of pure water , P = 23.62mm

Let us assume that vapor pressure of solution is p'

given,

weight of Urea = 1.5g

molecular weight of urea = 60 g/mol

so, mole of Urea = 1.5/60 = 1/40 = 0.025

and weight of water = 50gm

molecular weight of water = 18g/mol

so, mole of water = 50/18 = 2.77

using lowering of vapor pressure formula,

i.e., lowering of vapor pressure = mole fraction of solute.

(23.62 - p')/23.62 = 0.025/(2.77 + 0.025)

or, 0.025 × 23.62 = (2.795) × (23.62 -p')

or, 0.59 = 66 - 2.795p'

or, p' = 23.40 mm

hence , required answer is 23.40mm