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Question:

On mixing, heptane and octane form an ideal solution. At 373K, 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 gm of heptane and 35 gm of octane will be: (molar mass of heptane = 100 g/mol and of octane = 114 g/mol) 72.0 kPa, 36.1 kPa, 96.2 kPa, 144.5 kPa

72.0 kPa

96.2 kPa

144.5 kPa

36.1 kPa

Solution:

Mole fraction of Heptane = 25/100 / (25/100 + 35/114) = 0.25 / 0.557 = 0.45
XHeptane = 0.45
Therefore, Mole fraction of octane = 1 - 0.45 = 0.55 = Xoctane
Total pressure = ΣXiP0i
Pt = (105 × 0.45) + (45 × 0.55) kPa
Pt = 72.0 kPa