SolutionHard
Question
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).
Options
A.72.0 kPa
B.36.1 kPa
C.96.2 kPa
D.144.5 kPa
Solution
Mole fraction of Heptane = 
XHeptane = 0.45
∴ Mole fraction of octane = 0.55 = Xoctane
Total pressure =
= (105 × 0.45) + (45 × 0.55) kPa
= 72.0 KPa
Mole fraction of Heptane =
XHeptane = 0.45
∴ Mole fraction of octane = 0.55 = Xoctane
Total pressure =
= (105 × 0.45) + (45 × 0.55) kPa
= 72.0 KPa
XHeptane = 0.45
∴ Mole fraction of octane = 0.55 = Xoctane
Total pressure =
= (105 × 0.45) + (45 × 0.55) kPa
= 72.0 KPa
Mole fraction of Heptane =
XHeptane = 0.45
∴ Mole fraction of octane = 0.55 = Xoctane
Total pressure =
= (105 × 0.45) + (45 × 0.55) kPa
= 72.0 KPa
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