Question

At ${80^ \circ }C$, the vapour pressure of pure liquid $A$ is $520mmHg$ and that of pure liquid $B$ is $1000mmHg$. If a mixture solution of $A$ and $B$ boils at ${80^ \circ }$ and $1$ atm pressure, the amount of $A$ (mole percent) in the mixture is:
A. $50\% $
B. $54\% $
C. $32\% $
D. $44\% $

Answer 1

Hint: The measure of a material's potential to convert into a vapour or gaseous state is known as vapour pressure. As the temperature rises, so does the vapour pressure. The boiling point of a liquid is the temperature at which the vapour pressure at the liquid's surface equals atmospheric pressure.
Formula used:
${P_T} = P_A^0{X_A} + P_B^0{X_B}$
${P_T} = $Total vapour pressure
$P_A^0 = $ vapour pressure of pure liquid $A$
$P_B^0 = $ vapour pressure of pure liquid $B$
${X_A} = $ Mole fraction of solution $A$
${X_B} = $Mole fraction of solution $B$

Complete answer:
Given-
$P_A^0 = 520mmHg$
$P_B^0 = 1000mmHg$
Let mole fraction of $A$ in solution $ = {X_A}$
Let mole fraction of $B$ in solution $ = {X_B}$
Then at $1atm$ pressure that is at $760mmHg $ is
Substituting the given values in the above formula,
${P_T} = P_A^0{X_A} + P_B^0{X_B}$
$P_A^0{X_A} + P_B^0{X_B} = 760mmHg$
$P_A^0{X_A} + P_B^0(1 - {X_A}) = 760mmHg$
$520{X_A} + 1000 - 1000{X_A} = 760mmHg$
${X_A} = \dfrac{1}{2} = 50\% $

Hence, the correct option is A. $50\% $

Additional Information:
As the number of molecules in the vapour phase increases, so does the rate of condensation. When the rate of evaporation equals the rate of condensation, the process is said to be complete. The equilibrium stage is the name given to this stage. The pressure exerted by the molecules at this point, as depicted by the manometer, is known as the liquid's vapour pressure.

Note:
At normal temperature, the material with the lowest boiling point would have the maximum vapour pressure (the easiest way to reach the gas phase). The material with the lowest vapour pressure is the one with the highest boiling point.