Question

At 80 degree centigrade, the vapour pressure of pure benzene is 753 mm Hg and pure toluene is 290 mm Hg. Calculate the composition of liquid in mole percent which at 80 degree C is in equilibrium with the vapour containing 30 mole percent of benzene.

Answer 1

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it is 90 percent

Answer 2

Given:   pressure of  pure toluene (p01) = 290 mm Hg
              pressure of  pure benzene  (po2)= 753 mm Hg
              mole fraction of benzene in vapour state (y​2) = 30100 = 0.3
As we know, mole fraction in vapour state is given by
y2=p2ptotalwhere p2 is the pressure of benzene in the mixtureAlso p2 =p02 x2x2 is the mole fraction of benzene in liquid state.As we know, total pressure is given by ptotal = p01 + ( p02 − p01 ) x2substituting the values of  p02 and p0  1in the equations of p2 and ptotal p2 = 753 x2and ptotal =290 + (753 − 290 ) x2               = 290+463 x2substituting p2 and ptotal in the equation of y2y2=p2ptotal0.3=753 x2290+463 x2solving the equation for x20.3×290 + 0.3×463 x2= 753 x287 +138.9 x2 = 753 x2753 x2 − 138.9 x2 = 87614.1 x2= 87x2 = 87614.1     =0.14and x1 = 1−0.14           = 0.86

Therefore, the composition of benzene and toluene in liquid mixture are 0.14 and 0.86 respectively.