Question

Two liquids $A$ and $B$ form ideal solutions. At $300K$, the vapour pressure of a solution containing $1$ mole of $A$ and $3$ moles of $B$ is $550mm$ $Hg$. At the same temperature, if one more mole of $B$ is added to this solution, the vapour pressure of the solution increases by $10mm$ $Hg$. The vapour pressure of $A$ and $B$ in their pure states are respectively:

• A. $p_A^o=600mm$ $Hg$ and $p_B^o=400mm$ $Hg$
• B. $p_A^o=500mm$ $Hg$ and $p_B^o=560mm$ $Hg$
• C. $p_A^o=450mm$ $Hg$ and $p_B^o=650mm$ $Hg$
• D. $p_A^o=400mm$ $Hg$ and $p_B^o=600mm$ $Hg$

Answer 1

The correct option is D $p_A^o=400mm$ $Hg$ and $p_B^o=600mm$ $Hg$

Let, $P_A^o$ and $P_B^o$ be the pure liquid pressure of A and B respectively.

Case - I:

$n_A=1mol;n_B=3mol;P_1=550$

So, $x_A=\frac{n_A}{n_A+n_B}=\frac{1}{1+3}=0.25$

$x_B=1-x_A=1-0.25=0.75$

Now,

$P_T=x_A{P}_A^o+x_B{P}_B^o$

$550=0.25P_A^o+0.75P_B^o$ ------- eqn. 1

Case - II:

$n_A=1mol;n_B=4mol;P_1=560$

So, $x_A=\frac{n_A}{n_A+n_B}=\frac{1}{1+4}=0.20$

$x_B=1-x_A=1-0.20=0.80$

Now,

$P_T=x_A{P}_A^o+x_B{P}_B^o$

$560=0.20P_A^o+0.80P_B^o$ ------- eqn. 2

Solving eqn. 1 and eqn. 2 we get

$P_A^o=400mm$ and $P_B^o=600mm$