Solution (Chemistry)

Question

Vapour pressure of liquid A and B at 298K is 300mm of Hg and 450mm of Hg. Calculate the mole fraction of A in the mixture. Given total vapour pressure of the solution is 405mm Hg

Answer 1

It is given that,

\(P_A^0\) = 300 mm of Hg

\(P_B^0\) = 450 mm of Hg

P_Total = 405 mm of Hg

From Raoults law, we have

P_total = P_A + P_B

P_total = \(P_A^0 X_A + P_B^0(1 - X_A)\)

P_total = \(P_A^0 X_A + P_B^0-P_B^0X_A\)

P_total = \((P_A^0 - P_B^0)X_A + P_B^0\)

405 = (300 − 450)X_A + 450

−45 = −150X_A

X_A = 0.3

Therefore,

X_B = 1 − X_A

= 1 − 0.3

= 0.7

\(P_A= P_A^0 X_A\)

= 300 × 0.3

= 90mm of Hg

\(P_B = P_B^0 \times X_B\)

= 450 × 0.7

= 315mm of Hg

Now in the vapour phase:

Mole fraction of liquid A,

\(A = \frac{P_A}{(P_A + P_B)}\)

\(= \frac{90}{90 + 315}\)

\(= 0.22\)

Thus mole fraction of liquid A is 0.22.