SolutionMedium
Question
At 80oC, the vapour pressure of pure liquid ′A′ is 520 mm Hg and that of pure liquid ′B′ is 1000 mm Hg. If a mixture solution of ′A′ and ′B′ boils at 80oC and 1 atm pressure, the amount of ′A′ in the mixture is (1 atm = 760 mm Hg)
Options
A.52 mol percent
B.34 mol percent
C.48 mol percent
D.50 mol percent
Solution
PT = PAoXA + PBoXB
760 = 520XA + PBo (1 - XA)
⇒ XA = 0.5
Thus, mole% of A = 50%
760 = 520XA + PBo (1 - XA)
⇒ XA = 0.5
Thus, mole% of A = 50%
Create a free account to view solution
View Solution FreeMore Solution Questions
An aqueous solution (0.85%) of NaNO3 is apparently 90% dissociated at 27°C. The osmotic pressure of solution is...The boiling points of C6H6, CH3OH, C6H5NH2 and C6H5NO2 are 80°C, 65°C, 184°C and 212°C, respectively. Which of the follo...The boiling point of an azeotropic mixture of water and ethyl alcohol is less than that of theoretical value of water an...For each of the following dilute solutions, van’t Hoff factor is equal of 3, except...Water should boil at 90°C at pressure...