What mass of urea (molar mass = 60 g mol ⁻¹ ) must be dissolved in 90 g of water to reduce the vapour pressure of water by 25 % ?

Step-by-step solution

Let the mass of urea be w g . Moles of urea, n_ solute = w 60 Moles of water, n_ solvent = 90 18 = 5 mol Relative lowering of vapour pressure is equal to the mole fraction of the solute: P P^0 = X_ solute Given that the vapour pressure is reduced by 25 % , the relative lowering is 0.25 . 0.25 = n_ solute n_ solute + n_ solvent 1 4 = n_ solute n_ solute + 5 n_ solute + 5 = 4 n_ solute 3 n_ solute = 5 n_ solute = 5 3 mol Mass of urea, w = 5 3 60 = 100 g Answer: 100 g