On mixing, heptane and octane form an ideal solution. At 373 k, the vapour pressures of the two liquid components (heptane and octane) are 105 kpa and 45 kpa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol-1 and of octane =114 g mol-1) [aieee-2010] (1) 144.5 kpa (2) 72.0 kpa (3) 36.1 kpa (4) 96.2 kpa
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Mole fraction of Heptane =
100
25
+
114
35
25/100
=
0.557
0.25
=0.45
X
Heptane
=0.45
∴ Mole fraction of octane =1−0.45=0.55=X
octane
Total pressure =∑X
i
P
i
0
P
t
=(105×0.45)+(45×0.55) kPa
P
t =72.0 kPa
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