Question

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

• A. 52 mole percent
• B. 34 mole percent
• C. 48 mole percent
• D. 50 mole percent

Answer 1

Answer: (D)

$P _{ T =} P _{ A }^{\circ} X _{ A }+ P _{ B }^{\circ} X _{ B }$
Mixture solution boil at $1 \,atm =760\, mm\, Hg =$ total pressure of solution. $760=(520 X )_{ A }+1000\left(1-( X )_{ A }\right)$
$\Rightarrow 760=520 X _{ A }+1000-1000 X _{ A } \Rightarrow 480 X _{ A }=240$
$ X _{ A }=0.5$,
mole percentage of $A =50 \%$