(a) 30 g of urea (M = 60 g mol−1 ) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23·8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.

(a) It is given that vapour pressure of water, 𝑃°₁ = 23.8 mm of Hg
Weight of water taken, w₁ =846 g
Weight of urea taken, w₂ = 30 g
Molecular weight of water M₁ = 18 g mol^-1
Molecular weight of urea, M₂ = 60 g mol^-1
Now, we have to calculate vapour pressure of water in the solution we take vapour pressure as P₁.
Now from Raoult's law, we have:
$\begin{array}{l}\frac{{\mathrm{P}}_1^0-{\mathrm{P}}_0}{{\mathrm{P}}_1^0}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1-{\mathrm{n}}_2}\\ \frac{{\mathrm{P}}_1^0-{\mathrm{P}}_0}{{\mathrm{P}}_1^0}=\frac{\frac{{\mathrm{m}}_2}{{\mathrm{M}}_2}}{\frac{{\mathrm{m}}_1}{{\mathrm{M}}_1}-\frac{{\mathrm{m}}_2}{{\mathrm{M}}_2}}\\ \Rightarrow \frac{23.8-{\mathrm{P}}_1}{23.8}=0.0105\\ \Rightarrow {\mathrm{P}}_1=23.5501\mathrm{mm}\text{ of }\mathrm{Hg}\end{array}$
Hence, the vapour pressure of water in the given solution is 23.5501 mm of Hg and its relative lowering is 0.0105.

(b)