Question

The vapour pressure of two pure liquids A and B are 50 and 40 torr respectively. If 8 moles of A is mixed with x moles of B, then vapour pressure of solution obtained is 48 torr. What is the value of x?

Answer 1

We have

The vapour pressure of two pure liquids $A$ and $B$ are $50$ and $40$ torr respectively.

Now, according to question

We have to $8mol$ of $A$ is mixed with $x$ mole of $B$

Now, from the Raults law

$P_T=p_A^{\circ }+P_B^{\circ }$

$=P_A\times x_A+P_B\cdot \left(1-x_A\right)$

$=P_A\times x_A+P_B-P_B\times x_A$

$=\left(P_A-P_B\right)x_A+P_B$

Now,

$\Rightarrow 48=\left(50-40\right)x_A+40$ [Given form the the question the vapour pressure of soution ]

Hence.

$x_A=0.8$

and,

$x_B=0.2$.