NEETSolutionHard
Question
If vapour pressures of pure liquids ′A′ & ′B′ are 300 and 800 torr respectively at 25oC. When these two liquids are mixed at this temperature to form a solution in which mole percentage of ′B′ is 92, then the total vapour pressure is observed to be 0.95 atm. Which of the following is true for this solution.
Options
A.ᐃVmix > 0
B.ᐃHmix < 0
C.ᐃVmix = 0
D.ᐃHmix = 0
Solution
According to Raoult’s law
PT = (0.08 × 300 + 0.92 × 800) torr = (24 + 736) torr = 760 torr = 1 atm
Pexp. = 0.95 atm < 1 atm
Hence solution shows -ve deviation
so ᐃHmix < 0, and ᐃVmix < 0.
PT = (0.08 × 300 + 0.92 × 800) torr = (24 + 736) torr = 760 torr = 1 atm
Pexp. = 0.95 atm < 1 atm
Hence solution shows -ve deviation
so ᐃHmix < 0, and ᐃVmix < 0.
Create a free account to view solution
View Solution FreeMore Solution Questions
The vapour pressure of the solution of two liquids A(po = 80 mm) and B(po = 120 mm) is found to be 100 mm when xA = 0.4....Heptane and octane form ideal solution. At 373 K, the vapour pressures of the pure liquids are 106 kPa and 46 kPa, respe...The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol, separately. The ...The elevation in boiling point, when 13.44 g of freshly prepared CuCl2 is added to one kg of water, is (Some useful data...A solution of ‘x’ mole of sucrose in 100 g of water freezes at − 0.2°C. As ice separates out, the freezing point goes do...