KVPY2011ChemistrySolutions
At 300 ~K the vapour pressure of two pure liquids, A and B are 100 and 500 ~mm Hg , respectively. If in a mixture of A and B , the vapour pressure is 300 ~mm Hg , the mole fractions of A in the liquid and in the vapour phase, respectively, are-
Options
- A1 / 2 and 1 / 10
- B1 / 4 and 1 / 6
- C1 / 4 and 1 / 10
- D1 / 2 and 1 / 6
Correct answer
D. 1 / 2 and 1 / 6
Step-by-step solution
Y _ A = P _ A ⁰ X _ A P _ A ⁰ X _ A + P _ B ⁰ X _ B = 100 1 2 100 1 2 +500 1 2 = 50 50+250 = 50 300 =1 / 6 PS = ( p _ A ⁰- p _ B ⁰ ) X _ A + p _ B ⁰300=(-400) X _ A +500 X _ A = 1 2 X _ B = 1 2