Question
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be
-
350
-
300
-
360
-
700
Solution
A.
350
Let the vapour pressure of pure ethyl alcohol be P,
According to Raoult’s law 290 = 200 × 0.4 + P × 0.6
P = 290-80/0.6 = 350 mm Hg
