JEE Main20198 Apr 2019Morning ShiftChemistrySolutionsActual
The vapour pressures of pure liquids A and B are 400 and 600 mm Hg respectively at 298 K . On mixing the two liquids, the sum of their volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in the vapour phase, respectively are
Options
- A500 m m H g , 0.5 , 0.5
- B450 m m H g , 0.4 , 0.6
- C450 m m H g , 0.5 , 0.5
- D500 m m H g , 0.4 , 0.6
Correct answer
D. 500 m m H g , 0.4 , 0.6
Step-by-step solution
Raoult’s law for ideal solution, P T o t a l = P A +   P B P T o t a l = x A × P A ∘ +   x B × P B ∘ P T o t a l = 1 2 × 400 + 1 2 × 600 = 500   m m   o f   H g Dalton’s law, y A × P T o t a l = P A   y A = x A × P A 0 P T o t a l = 1 2 × 400 500 = 0.4 y B = 1 - 0.4 = 0.6