Question

If the vapour pressure of pure water at $25^\circ C$ is $23.8 \, mmHg,$ then calculate the vapour pressure lowering caused by the addition of $100 \, g$ of sucrose (molecular mass $= 342 \, \text{g/mol}$ ) to $1000 \, g$ of water

• A. $00.12 \, mmHg$
• B. $0.125$ $mmHg$
• C. $1.15$ $mmHg$
• D. $1.25$ $mmHg$

Answer 1

Answer: (B)

Given molecular mass of sucrose $=342$

Moles of sucrose $=\frac{100}{342}=0.292$ mole

Moles of water $N=\frac{1000}{18}=55.5$ moles and

Vapour pressure of pure water $P^{0}=23.8 \, mmHg$

According to Raoult’s law $\frac{\Delta P}{P^{0}}=\frac{n}{n + N}\Rightarrow \frac{\Delta P}{23.8}=\frac{0.292}{0.292 + 55.5}$

$\Delta P=\frac{23.8 \times 0.292}{55.792}=0.125 \, mmHg$ .