Question

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

• A. $00.12mmHg$
• 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 $P^0=23.8mmHg$

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

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