Question

The vapor pressure of water is 23.76 mm Hg at 25°C.

How many grams of urea, CH₄N₂O, a nonvolatile, nonelectrolyte (MW = 60.10 g/mol), must be added to 247.9 grams of water to reduce the vapor pressure to 23.28 mm Hg ?

water = H₂O = 18.02 g/mol.

answer=_____ g urea

Answer 1

By Raoult's Law,
vapor pressure = (mole fraction of solvent)(vapor pressure of pure solvent)
23.28 mm Hg = (X)(23.76 mm Hg)
X = 0.9797

(247.9 g H2O)(mol / 18.02 g) = 13.757 mol H2O

(13.757 mol H2O)(mol solution / 0.9797 mol H2O) = 14.042 mol solution

(14.042 - 13.757) mol = 0.285 mol solute (urea)

(0.285 mol)(60.10 g / mol) = 17.1285 g