Question

What is the vapour pressure of a solution at ${100}^{\circ }C$ containing 18 g of water and 1.71 g of sucrose (non-volatile)? $\text{Given: Molecular formula of sucrose is}{C}_{12}{H}_{22}{O}_{11}$
At ${100}^{\circ }C$, vapour pressure of pure water is $1atm$.

• A. $415.3torr$
• B. $450.5torr$
• C. $756.2torr$
• D. $788.4torr$

Answer 1

The correct option is C $756.2torr$

$\text{Molecular mass of sucrose}=(12\times 12)+(22\times 1)+(11\times 16)=342gmo{l}^{-1}$

${\chi }_{sucrose}=\frac{\frac{1.71}{342}}{\frac{1.71}{342}+\frac{18}{18}}=\frac{\frac{1}{200}}{\frac{1}{200}+1}{\chi }_{sucrose}=\frac{1}{201}=4.975\times {10}^{-3}$

$At{100}^{\circ }C,P_{solvent}^{\circ }=1atm=760torr$
When a non volatile solute is added to a volatile solvent, the vapour pressure of the solution is lowered as compared to pure solvent.
For dilute solutions, relative lowering of vapour pressure is given by :
$\frac{\Delta P}{P_{solvent}^{\circ }}={\chi }_{sucrose}$

$\therefore \frac{P_{Solvent}^{\circ }-P_{solution}}{P_{solvent}^{\circ }}=4.975\times {10}^{-3}$

$\frac{760-P_{solution}}{760}=4.975\times {10}^{-3}$

$P_{solution}=756.21torr$