JEE Main201910 Jan 2019Morning ShiftChemistrySolutionsActual
Liquids A and B form an ideal solution in the entire composition range. At 350 K , the vapour pressure of pure A and pure B are 7 × 10 3 Pa and 12 × 10 3 Pa , respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is:
Options
- Ax A = 0.4 ; x B = 0.6
- Bx A = 0.76 ; x B = 0.24
- Cx A = 0.28 ; x B = 0.72
- Dx A = 0.37 ; x B = 0.63
Correct answer
C. x A = 0.28 ; x B = 0.72
Step-by-step solution
Since, P A 0 = 7 × 10 3 P a ; P B 0 = 12 × 10 3 P a and X A = 0.4 , X B = 1 − 0.4 = 0.6 According to Raoult's Law: P T = P A 0 X A + P B 0 X B P T = 7 × 10 3 0.4 + 12 × 10 3 0.6 P T = 10 × 10 3 P a Now, P A 0 X A = P T x A x A = P A 0 X A P T = 7 × 10 3 × 0.4 10 4 ⇒ 0.28 x B = 1 - x A ⇒ 1 - 0.28 ⇒ 0.72