Vapour pressure of an equimolar mixture of benzene and toluene at a given temperature was found to be 80 mm Hg . If vapour above the liquid phase is condensed in a beaker, vapour pressure of this condensate at the same temperature was found to be 100 mm Hg . If the pure state vapour pressure of benzene and toluene is respectively x and y . Then determine value of x+2 y 50 :

Step-by-step solution

aligned & X_B=0.5 X_T=0.5 & p_ total =P_B^ X _ B + P _ T ^ X _ T & 80=0.5 P _ B ^ +0.5 P _ T ^ aligned 160= P _ B ^ + P _ T ^ ...(1) aligned & y_B= P_B^ X_B p_ total = P_ B ^ 0.5 80 = P_B^ 160 & y_T= P_T^ x_T p_ total = P_T^ 0.5 80 = P_T^ 160 & p_ total ^ =100=P_B^ y_B+P_T^ y_T aligned aligned & =100=P_B^ P_B^ 160 +P_T^ P_T^ 160 & P_B^ ^2 +P_T^ ^2 =100 160 aligned From (1) Eq. aligned & (P_B^ +P_T^ )=160 & (P_B^ +P_T^ )^2=160^2=P_B^ 0^2 +P_T^ 0^2 +2 P_B^ P_T^ & =P_B^ ^2 +P_T^ ^2 +2 P_B^ P_T^ & =100 160+2 P _ B ^ P