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Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquidcomponents are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure of amixture of 26.0 g of heptane and 35 g of octane ? |
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Answer» SOLUTION :Molar MASS of heptane `(C_7H_16)= 100 g "MOL"^(-1)` Molar mass of octane `(C_8H_18) = 114 g "mol"^(-1)` Number of moles in 26.0 g heptane = `(26.0)/(100 g "mol"^(-1)) = 0.26 `mol Number of moles in 35.0 g octane = `(35.0)/(114 g "mol"^(-1) ) = 0.31 `mol Mole fraction of heptane ` = (0.26)/(0.26 + 0.31) = 0.456` Mole fraction of octane = 1 - 0.456 = 0.544 Partial pressure of heptane = 0.456 x 105.2 kPa = 47.97 kPa Partial pressure of octane = 0.544 x 46.8 kPa = 25.46 kPa Total pressure = 47.97 + 25.46 = 73.43 kPa. |
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