Question

SCOREBOARD OVERVIEW PERFORMANCE LEADERBOARD 7 72 Chemistry English The vapour pressure of a solution of liquid A and liquid B is 600 torr. The mole fraction of component A in liquid phase is 0.7. The vapour pressure of pure A and B are respectively Question Type: Single Correct Type 1 300 torr, 130 torr 2 1300 torr, 130 torr 3 300 torr, 1300 torr​

Answer 1

Given:

Total vapour pressure of solution liquid A and liquid B = 600 torr

Mole fraction of component A = 0.70

To find:

Vapour pressure of pure liquid A and pure liquid B.

Solution:

It is given, total vapour pressure of solution of liquid A and liquid B,                        

                                 $P_T$ = 600 torr

                  Mole fraction of component A, $x_A$ = 0.7

We know that,  Sum of mole fraction of component A and mole fraction of

                         component B is equal to 1.

i.e.         $x_A + x_B = 1$

             0.7  +  $x_B$  =  1

                    $x_B = 1 - 0.7$

                    $x_B = 0.3$

so, the mole fraction of component B is 0.3.

Now, the formula used to calculate the vapour pressure of pure liquid is as follows,

                     $p = xP_T$

here, p = vapour pressure of pure liquid

         x = mole fraction

        $P_T$ = total vapour pressure of solution of liquid

now, vapour pressure of pure liquid A is calculated as,

            $p_A = x_AP_T$

by putting all the values, we get

           $p_A = 0.7 \times 600$

           $p_A$ = 420 torr

vapour pressure of pure liquid B is calculated as,

          $p_B = x_BP_T$

by putting all the values, we get

          $p_B = 0.3 \times 600$

           $p_B$ = 180 torr

Therefore, Vapour pressure of pure A is 420 torr.

                  Vapour pressure of pure B is 180 torr.