Solutions Question 2
Question 2 - 24 January - Shift 2
The Total pressure observed by mixing two liquid $A$ and $B$ is $350 mm Hg$ when their mole fractions are 0.7 and 0.3 respectively.
The Total pressure becomes $410 mm Hg$ if the mole fractions are changed to 0.2 and 0.8 respectively for $A$ and $B$. The vapour pressure of pure $A$ is $mm Hg$. (Nearest integer)
Consider the liquids and solutions behave ideally.
Show Answer
Answer: (314)
Solution:
Let V.P. of pure $A$ be $P_A^{0}$
Let V.P of pure $B$ be $P_B^{0}$
When $X_A=0.7 \& X_B=0.3$
$P_s=350$
$\Rightarrow P_A^{0} \times 0.7+P_B^{0} \times 0.3=350\quad$…….(i)
When $X_A=0.2 \& X_B=0.8$
$P_s=410$
$\Rightarrow P_A^{0} \times 0.2+P_B^{0} \times 0.8=410\quad$……..(ii)
Solving (i) and (ii)
$P_A^{0}=314 mm Hg$
$P_B^{0}=434 mm Hg$
$=(314)$
