At $2 5^{\circ} C$ the vapour pressure of pure water is $23.8$ torr. Find the vapour pressure of a solution of $10 g$ urea (molar mass $60 g mo l^{- 1}$ ) in $90 g$ water. Use the ideal Raoult's law expression.

Answer: C

n_{2} = 10/60 = 0.1667

mol and

n_{1} = 90/18 = 5.00

mol, so

x_{2} = 0.1667/5.1667 = 0.0323

P = P^{0} (1 - x_{2}) = 23.8 \times 0.9677 = 23.03

torr. The value 0.77 torr is just the lowering, not the final pressure.