Question

The vapour pressure of pure water at $26^{\circ} C$ is $25.5$ torr.. The vapour pressure of a solution which contains $20.0$ glucose, $\left( C _6 H _{12} O _6\right)$, in $100 \,g $ water (in torr) is? (report the answer by rounding off to the nearest integer)

Answer 1

Using Roult's law equation
$\frac{ P _{ A }^0- P _{ A }}{ P _{ A }^0}=\frac{ n _{ B }}{ n _{ A }+ n _{ B }} $
Or $\frac{ P _{ A }^0- P _{ A }}{ P _{ A }}=\frac{ n _{ B }}{ n _{ A }}$
$P _{ A }=$ the vapour pressure of the pure solution
$P _{ A }^0=$ the vapour pressure of the pure component
$n _{ B }$ : moles of solute
$n _{ A }$ : moles of solvent
$\frac{ P _{ A }^0- P _{ A }}{ P _{ A }}=\frac{ n _{ B }}{ n _{ A }}$
$\frac{25.5- P _{ A }}{ P _{ A }}=\frac{\frac{20}{180}}{\frac{100}{18}} $
$\frac{25.5- P _{ A }}{ P _{ A }}=\frac{20 \times 18}{100 \times 180}=0.02 $
$25.5- P _{ A }=0.02 P _{ A } $
$1.02 P _{ A }=25.5$
$P _{ A }=25.0$