SolutionHard
Question
At 40oC, vapour pressure in Torr of methanol and ethanol solution is P = 119x + 135 where x is the mole fraction of methanol. Hence
Options
A.vapour pressure of pure methanol is 119 Torr
B.vapour pressure of pure ethanol is 135 Torr
C.vapour pressure of equimolar mixture of each is 127 Torr
D.mixture is completely immiscible
Solution
P = 119 x + 135
x = 1 for pure methanol.
so P0methanol = 119 + 135 = 254 Torr
But for pure ethanol x = 0
so P0ethanol = 135 Torr
x = 1 for pure methanol.
so P0methanol = 119 + 135 = 254 Torr
But for pure ethanol x = 0
so P0ethanol = 135 Torr
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