Question

The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at is 23.8 mm Hg.

• A. 1.25 mm Hg
• B. 0.125 mm Hg
• C. 1.15 mm Hg
• D. 00.12 mm Hg

Answer 1

Answer: (B)

0.125 mm Hg

Answer 2

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 $P ^ { 0 } = 23.8$ mm Hg

According to Raoult’s law

$\frac { \Delta P } { P ^ { 0 } } = \frac { n } { n + N } \Rightarrow \frac { \Delta P } { 23.8 } = \frac { 0.292 } { 0.292 + 55.5 }$

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