Question

At 20°C the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20°C above an equimolar mixture of benzene and methyl benzene is ...... x 10^–2 (Nearest integer)

Answer 1

P°_B = 40 P°_T = 20 K_B = 0.5 = K_r

Now

Answer 2

Given,

P°_{benzene = 70 torr}  

_{& P°methyl benzene= 20 torr} 

Both are equimolar = n

For Mole fraction of benzene in vapour phase we use Dalton's law 

      P_A = X_A • P_total  ---(1)   , where 

        P_A= Partial pressure of Benzene 

 P_B= Partial pressure of methyl  benzene

       P_total = Total pressure of solution

Now, A/q  Raoult's law,

P_total = P_A + P_B   

           = x_A•P°_A + x_B•P°_B

          = n/n+n ×70 + n/n+n × 20 

        = 0.5 × 70 + 0.5 × 20

  P_total = 45 torr 

And 

       P_A = x_A•P°_A 

             = n/n+n × 70

           = 0.5 × 70

       P_A = 35 torr 

Now from equation no. (1)

     P_A = X_A • P_total 

      X_{A =} P_A/P_total 

          = 35/45

      X_A = 0.778 

         

Comments

The final answer is 78. As 0.778 = 77.8 x 10^-2 = 78 x 10^-2