At 300 K, the vapour pressure of a solution containing 1 mole of -hexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mm Hg of n-heptane in its pure state ____?
Answer: 600
Solution
By Raoult's law the total vapour pressure is for hexane (A) and heptane (B). With mole hexane and moles heptane, , i.e. . After adding one more mole of heptane, giving mole hexane and moles heptane, , i.e. . Subtracting the first relation from the second gives mm Hg, the vapour pressure of pure n-heptane.
JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Integer answer