Question

At $80{}^{o}C$, the vapour pressure of pure liquid ${}^{\prime }{A}^{\prime }$ is $520\text{mm Hg}$ and that of pure liquid ${}^{\prime }{B}^{\prime }$ is $1000\text{mm Hg}$. If a mixture of ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ boils at $80{}^{o}C$ and $1\text{atm}$ pressure, the amount of ${}^{\prime }{A}^{\prime }$ (in mole percent) in the mixture is:
($1\text{atm}=760\text{mm Hg}$)

Answer 1

At $1$ atmospheric preesure the boiling point of mixture is $80{}^{o}C$.
At boiling point the vapour pressure of the mixture, $P_T=1\text{atm}=760\text{mm Hg}$
Given,
$\text{Vapour pressure of pure liquid}{}^{\prime }{A}^{\prime },P_A^o=520\text{mm Hg}$
$\text{Vapour pressure of pure liquid}{}^{\prime }{B}^{\prime },P_B^o=1000\text{mm Hg}$
Again, $P_T=P_A^0{\chi }_A+P_B^o{\chi }_B$, we get
Where, ${\chi }_A,\text{and}{\chi }_B$ are mole fractions of $A$ and $B$.
So, $P_T=520{\chi }_A+1000(1-{\chi }_A)$
$\Rightarrow 760=520{\chi }_A+1000-1000{\chi }_A$
$\Rightarrow 480{\chi }_A=240$
$\Rightarrow {\chi }_A=\frac{240}{480}=0.5=50\text{mole percent}$