Question

$100\,g$ of liquid $A$ (molar mass $140\,gmol^{- 1}$ ) was dissolved in $1000\,g$ of liquid $B$ (molar mass $180\,gmol^{- 1}$ ). The vapour pressure of pure liquid $B$ was found to be $500\,torr$ . Calculate vapour pressure of $A$ (in torr) the solution if the total vapour pressure of the solution is $475\,torr.$

Answer 1

Step I : Calculation of vapour pressure of pure liquid $A (p°_A)$ Number of moles of liquid $A$
$\left( n _{ A }\right)=\frac{ W _{ A }}{ M _{ A }}=\frac{(100 \,g )}{\left(140 \, g \, mol ^{-1}\right)}=0.7143 \,mol$
Number of moles of liquid $B$
$\left( n _{ B }\right)=\frac{ W _{ B }}{ M _{ B }}=\frac{(1000 \, g )}{\left(180 \,g \,mol ^{-1}\right)}=5.5556 \, mol$
Mole fraction of $A$
$\left( x _{ A }\right)=\frac{ n _{ A }}{ n _{ A }+ n _{ B }}$
$=\frac{(0.7143 \, mol )}{(0.7143+5.5556) mol }=\frac{0.7143}{6.2699}=0.1139$
Mole fraction of $B \left( x _{ B }\right)=1-0.1139=0.8861$
Vapour pressure of pure liquid $B , B \left( p _{ B }^{ 0 }\right)=500$ torr
Total vapour pressure of solution $( p )=475$ torr
According to the Raoult's law
$p = p _{ A }^{ o } x _{ A }+ p _{ B }^{ o } x _{ B } $
$475 \text { torr }= p _{ A }^{ o } \times(0.1139)+500 \text { torr } \times(0.8861) $
$475 \text { torr }= p _{ A }^{ o } \times(0.1139)+443.05 \text { torr } $
$p _{ A }^{ o }=\frac{(475-443.05) \text { torr }}{(0.1139)}=\frac{31.95}{0.1139} \text { torr }=280.5 \text { torr }$
Step II : Calculation of vapour pressure of $A$ in the solution $\left( p _{ A }\right)$
According to Raoult's law,
$p _{ A }= p _{ A }^{0} x _{ A }=(280.5 \text { torr }) \times(0.1139) $
$p _{ A }=32.0 \text { torr }$