Chemistry, asked by BrainlyGood, 1 year ago

Two liquids A and B form an ideal solution. The solution has a vapor pressure of 700 torr at 80 deg C. It is distilled till 2/3 rd of the total solution is collected as the condensate.
The composition of the condensate is x_A' = 0.75 and that of the residue is x_A 0.30. If the vapor pressure of the residue at 80 deg C is 600 torr, Which of the following is/ are true ?

A) The composition of the original liquid was x_A = 0.6
B) The composition of the original liquid was x_A = 0.4
C) vapor pressure of pure solution of A is 2500/3 torr
D) vapor pressure of pure solution of B is 500 torr.

Answers

Answered by kvnmurty
5

Assume  X_A = 0.6

P =  0.6 P_A + 0.4 P_B = 700 torr

P" = 0.3 P_A + 0.7 P_B = 600 torr

Solving the two equations ,  P_B = 500 torr and  P_A=2500/3 torr

Let there be N total number of moles in the total solution initially.  Let   y*N  number of moles of B be present in the condensate.  So as mole fraction of A is 0.75,  there are 3yN moles of A in condensate.

So in the residue we have (0.6 - 3y) N  moles of A and (0.4 -y)N moles of B.

mole fraction in residue of A:   (0.6-3y) / [0.4 -y +0.6-3y]  =  0.30

       solving that  y = 1/6 

So initially, there are 0.6N moles of A, 0.4N moles of B.  In the condensate there are N/2 moles of A  and  N/6 moles of B.   Then in the residue there are 0.1N moles of A and 7N/30 moles of B.

Condensate size = 2/3N moles.            residue : 1N/3 moles     The data matches with given facts.

As the numbers are consistent,  that assumption is true.

================

checking by volumes.    V = volume of solution.  VA = volume per mole of A.

V/N = 0.6 VA +  0.4 VB                                 V'/N = 0.5 VA + VB/6                      V"/N = 0.1 VA + 7 VB/30

given,  V' = 2 * V"     =>    0.5 VA + VB/6 = 0.2 VA + 7/15 VB

      =>   VA = VB          =>  V = VA * N = VB * N

   Then  V' = 2/3 V   and V"  = 1/3 V   So consistent.

 

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