Mass of Urea NH$$_2$$CONH$$_2$$ required to be dissolved in 1000 g of water in order to reduce the vapour pressure of water by 25% is ______ g. (Nearest integer)
Given: Molar mass of N, C, O and H are 14, 12, 16 and 1 g mol$$^{-1}$$ respectively.

• Reduction in vapour pressure = $$25\%$$
• Let the vapour pressure of pure water $$(P°) = 100$$
• Vapour pressure of the solution $$(P) = 100 - 25 = 75$$
• Mass of solvent (water, W) $$= 1000 g$$
• Molar mass of water (M) $$= 18 g/mol$$

Formula for Urea: $$NH_2CONH_2$$

Molar Mass of Urea ($$M_u$$) $$= (2 × N) + (4 × H) + (1 × C) + (1 × O)$$
$$M_u$$ $$= (2 × 14) + (4 × 1) + 12 + 16 = 28 + 4 + 12 + 16 = 60 g/mol$$

Where $$n_u$$ is moles of solute (Urea) and $$n$$ is moles of solvent (Water).

Moles of water (n) $$= 1000 / 18 = 55.55 mol$$

Substituting into Raoult's law equation:

$$(100 - 75) / 75 = n_{u} / 55.55$$

$$25 / 75 = n_{u} / 55.55$$

$$1 / 3 = n_{u} / 55.55$$

$$n_{u} = 55.55 / 3 = 18.518$$ mol

Mass of Urea = Moles × Molar Mass

Mass $$= 18.518 × 60 =$$ 1111.11 g