2.8   The vapour pressure of pure liquids A and B are  450  and  700\; mm \; Hg respectively, at 350\; K . Find out the composition of the liquid mixture if total vapour pressure is 600\; mm\; Hg. Also find the composition of the vapour phase.

Answers (1)

Let the composition of liquid A (mole fraction) be xA.

So mole fraction of B will be xB = 1 - xA.

Given that,    P^{\circ}_A = 450\ mm\ of\ Hg\ ;\ P^{\circ}_B = 700\ mm\ of\ Hg

Using Raoult’s law ,

                                    p_{total} = p^{\circ}_A\ x_A\ +\ p^{\circ} _B\ (1-x_A)

Putting values of ptotal and vapour pressure of pure liquids in the above equation, we get :

                                    600 =   450.xA  +    700.(1 -  xA)

or                           600 - 700  =  450xA - 700xA

or                                  xA   =   0.4

and                               xB    =   0.6

Now pressure in vapour phase : 

                                P_A = p^{\circ}_A\ x_A

                                        =   450(0.4)  = 180 mm of Hg

                                P_B = p^{\circ}_B\ x_B

                                        =    700(0.6)   = 420 mm of Hg

           Mole\ fraction\ of\ liquid\ A = \frac{P_A}{P_A\ + P_B }                                                            

                                                                    = \frac{180}{180\ + 420 } = 0.30

And mole fraction of liquid B = 0.70

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