Solutions Question 330
Question: The vapour pressure of two pure liquids A and B that form an ideal solution, are 400 and 800 mm of Hg respectively at a temperature $ t{}^\circ C. $ The mole fraction of A in a solution of A and B whose boiling point is $ t{}^\circ C. $ will be
Options:
A) 0.4
B) 0.8
C) 0.1
D) 0.2
Show Answer
Answer:
Correct Answer: C
Solution:
V.P. of solution at $ t{}^\circ C=760mm $ [at b.p., V.P. of solution =atompheric pressure] Thus $ =P_A^{{}^\circ }.x_{A}+P_B^{{}^\circ }.x_{B} $ or $ P=P_A^{{}^\circ }.X_{A}+P_B^{{}^\circ }.(1-x_{A}),[\therefore x_{A}+x_{B}=1] $ or $ 760=400X_{A}+800(1-X_{A}) $
$ [ \therefore P=760mm,of,Hg ] $ or $ -800+760=-400x_{A} $ or $ -40=-400x_{A} $ or $ x_{A}=\frac{40}{400}=0.1 $ Thus mole fraction of in solution is 0.1
