Question:
Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer:
Vapour pressure of heptane (p10) = 105.2 Bar
Vapour pressure of octane (p20) = 46.8 kPa
Mass of heptane = 26g
Molar mass of heptane (C7H16)= 7x 12 +16x1 =100 g/mol
Number of moles of heptanes = 26/100 = 0.26 mol
Mass of octane =35g
Molar mass of octane (C8H18) = 8×12 + 18×1 = 114 gram / mol
Number of moles of octane = 35/114 =0.31 mol
Mole fraction of heptanes = (0.26/(0.26+0.31)) =0.456
Mole fraction of octane = 0.31/(0.26+0.31) =0.544
Partial pressure of heptane = 105.2×0.456=47.97 kPa
Partial pressure of octane =46.8×0.544 = 25.46 kPa
Total pressure = 47.97+25.46 =73.43 kPa
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