# To allocate each other by partial distillation, a mixture of A and B has gone through distillation and the distillate has been cold down and distillated again.Then again the second distillate also...

To allocate each other by partial distillation, a mixture of A and B has gone through distillation and the distillate has been cold down and distillated again.Then again the second distillate also has been distillated.Like this, they have got the 4th distillate.

The mole fractions ratio of the 4th distillate is *A : B = 8 : 1*.The pure vapour pressure of A is *900 torr* and pure vapour pressure of B is *600 torr*.

**Find the mole fractions ratio of the initial mixture.**

Please help me on this.

### 1 Answer | Add Yours

Let the mole fraction of A in the initial mixture was x, then mole fraction of B was (1-x)

`P^o_A` =900 torr. and `P^o_B` =600 torr.

According to Raoult's law, for the first distillate, we can write

`P_T_1 = x*P^o_A +(1- x)*P^o_B `

`=x*900 +(1- x)*600=300(3x+2-2x)= 300(x+2)`

Mole-fraction of A in the first distillate `= (x*900)/(300(x+2))`

`=(3x)/(x+2)`

and, mole-fraction of B in the first distillate

`=((1-x)*600)/(300(x+2))= (2(1-x))/(x+2)`

For the second distillate,

`P_T_2 = (3x)/(x+2)*900 +(2(1-x))/(x+2)*600`

`=300(9x+4-4x)/(x+2)= 300(5x+4)/(x+2)`

Mole-fraction of A in the second distillate

`=((3x)/(x+2)*900)/(300(5x+4)/(x+2))=(9x)/(5x+4)`

And, mole-fraction of B in the second distillate

`=(2(1-x)/(x+2)*600)/(300(5x+4)/(x+2))= (4(1-x))/(5x+4)`

For the third distillate,

`P_T_3 = (9x)/(5x+4)*900 +(4(1-x))/(5x+4)*600 `

`=300(27x+8-8x)/(5x+4)= (300(19x+8))/(5x+4)`

Mole-fraction of A in the third distillate

`=((9x)/(5x+4)*900)/(300(19x+8)/(5x+4))=(27x)/(19x+8)`

And, mole-fraction of B in the third distillate

`=((4(1-x)*600)/(5x+4)) /(300(19x+8)/(5x+4))=8(1-x)/(19x+8)`

For the fourth distillate,

`P_T_4 = (27x)/(19x+8)*900 +(8(1-x))/(19x+8)*600 `

`=300(81x+16-16x)/(19x+8)= (300(65x+16))/(19x+8)`

Mole-fraction of A in the fourth distillate

`=((27x)/(19x+8)*900)/((300(65x+16))/(19x+8))=(81x)/(65x+16)`

and,

mole-fraction of B in the fourth distillate

`=((8(1-x)*600)/(19x+8))/((300(65x+16))/(19x+8))= (16(1-x))/(65x+16)`

Ratio of mole-fraction of A to mole fraction of B`=((81x)/(65x+16))/((16(1-x))/(65x+16))=(81x)/(16(1-x))`

By condition, `(81x)/(16(1-x))=8/1`

`rArr x=0.612`

and, `(1-x)=0.388`

**Therefore, ratio of mole-fraction of A to mole fraction of B in the initial mixture was 0.612/0.388, or 1.58.**

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Thank a lot!