Question

The vapour pressure of pure water at ${26}^{\circ }C$ is 25.21 torr. The vapour pressure of a solution which contains 20.0 g glucose, $C_6{H}_{12}{O}_6$, in 70 g water is :

• A. $22.5$ torr
• B. $23.4$ torr
• C. $24.4$ torr
• D. $24.5$ torr

Answer 1

The correct option is D $24.5$ torr

Number of moles of glucose $=\frac{20.0}{180}=0.1111$

Number of moles of water $=\frac{70}{18}=3.89$

Mole fraction of glucose $=\frac{0.1111}{0.1111+3.89}=0.0278$

The relative lowering in vapour pressure is equal to the mole fraction of solute.

$\frac{P^0-P}{P^0}=X⟹\frac{25.21-P}{25.21}=0.0278$

$P=25.21-(25.21\times 0.0278)=24.5torr$

Hence, the correct option is $D$