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.
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