AP EAMCET201923 Apr 2019Morning ShiftChemistrySolutionsActual
At (T( ~K ) ), the vapour pressures of pure liquids (A ) and (B ) are (100 ~mm ) and (160 ~mm ) respectively. An ideal solution is formed by mixing 2 moles of (A ) and 3 moles of (B ) at the same temperature. The mole fraction of (A ) and (B ) in the vapour state respectively are
Options
- A(0.706,0.294 )
- B(0.294,0.706 )
- C(0.40,0.60 )
- D(0.60,0.40 )
Correct answer
B. (0.294,0.706 )
Step-by-step solution
Key Idea Vapour pressure of solution, (p_ total =p_A+p_ B^ = _A p_A^ + _B p_B^ [ p_A= _A p_A^ ] ) Also vapour pressure of component (1, p₁=y₁ p_ total ) where (y ), is the mole fraction of component 1 in vapour phase. Given, Vapour pressure of pure liquid (A, p_A^ =100 ~mm ) Vapour pressure of pure liquid (B, p_B^ =160 ~mm ) ( ) Total vapour pressure of solution (=p_A+p_B ) ( aligned p_ total & = _A p_A^ + _B p_B^ = 2 5 100+ 3 5 160 & =40+96=136 ~mm aligned ) Also ( p_A=y_A p_ total ) where, (y_A ) is the mole frac