Question

At 90 degree centigrade vapour pressure of toluene is 400 tour and that of sigma xylene is 150 torr what is the composition of the liquid mixture dead boys at 90 degree centigrade when the pressure is. 50 atmosphere

Answer 1

mole fraction A = (380torr - 150torr) / (400torr - 150torr) = ... the mixture since mole fraction x v.p. pure = partial pressure.

Answer 2

Explanation:

The given data is as follows.

   Vapor pressure of mixture = 0.5 atm = $0.5 atm \times \frac{760 mm Hg}{1 atm}$

                                                = 380 mm Hg

According to Raoult's law,  $P_{solution} = P_{solvent} \times X_{solvent}$

where,      $P_{solution}$ = vapor pressure of the solution or mixture

                $P_{solvent}$ = vapor pressure of the pure solvent

                      $X_{solvent}$ = mole fraction of the solvent

Also, $P_{solution} = P_{toulene} \times X_{toulene}$ + P_{xylene} \times X_{xylene}[/tex]   ...... (1)

Also, it is known that X_{toulene} + X_{xylene} = 1

so,                       $X_{xylene} = (1 - X_{toulene})$  .....(2)

Therefore, putting the given values into equation (1) using relation in equation (2) as follows.

              $P_{solution} = P_{toulene} \times X_{toulene}$ + P_{xylene} \times X_{xylene}[/tex]

              $P_{solution} = P_{toulene} \times X_{toulene}$ + P_{xylene} \times (1 - X_{toulene})[/tex]

                     380 = $400 \times X_{toulene} + 150(1 - X_{toulene})$

                     230 = $250 \times X_{toulene}$

                     $X_{toulene}$ = 0.92

Hence,    $X_{xylene} = (1 - X_{toulene})$

                                          = (1 - 0.92)

                                          = 0.08

Thus, we can conclude that the composition of the liquid mixture is 0.92 and 0.08.