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  • 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.

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.

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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

Posted by

Devendra Khairwa

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