Question

5 mole of liquid A and 5 mole of liquid B form an ideal solution at 25oC given  = 300 mm Hg and  = 500 mm Find composition of first bubble of vapour​

Answer 1

Answer:

complete answer is given in the attached image

Answer 2

Let us recall the concept of the vapour pressure of colligative properties from the solution chapter.

given values

vapour pressure of A= 300mmHg

vapour pressure of B= 500mmHg

number of moles of A=5

number of moles of B=5

Mole fraction of A($x_{A}$)= $\frac{ no of moles of A}{ total no of moles}$

                                   =$\frac{5}{5+5}$

                                   =0.5

Mole fraction of B($x_{B}$)=  1-$x_{A}$

                                   = 1-0.5

                                   = 0.5

By applying raoult's law

$p_{A} = p_{A}$°$x_{A}$

   = 0.5×300mmHg

    = 150mmHg

$p_{B} =p_{B}$°$x_{B}$

    = 0.5×500mmHg

    =250mmHg

Mole fraction of A at vapour phase = $\frac{p_{A} }{p_{B }+p_{A} }$

                                                          =$\frac{150}{250+150}$

                                                         = 0.375

final answer

the composition of the first bubble at vapour phase is 0.375.