Question

The vapour pressure of pure water is $23.5\, mm\, Hg$. Then, the vapour pressure of an aqueous solution which contains $5$ mass percent of urea is (Molar mass of urea is $60$).

• A. 23 mm Hg
• B. 18 mm Hg
• C. 31 mm Hg
• D. 35 mm Hg

Answer 1

Answer: (A)

Mass of urea $=5 g$
Mass of water $=100-5=95 g$
Number of moles of urea $=\frac{5}{60}=0.083$
Number of moles of water $=\frac{95}{18}=5.278$
$\therefore $ Total number of moles $=5.278+0.083=5.361$
Mole fraction of solvent $=\frac{5.278}{5.361}$
$\begin{aligned} P _{ s } &=\text { Mole fraction of solvent } \times P ^{\circ}=\frac{5.278}{5.361} \times 23.5 \\ &=23.14 mm Hg \approx 23 mm Hg \end{aligned}$