Heptane and octane form ideal solution. At 373 K, the vapour pressures of the pure liquids are 106 kPa and 46 kPa, respectively. What will be the vapour pressure, in bar, of a mixture of 30.0 g of hep

Question

Heptane and octane form ideal solution. At 373 K, the vapour pressures of the pure liquids are 106 kPa and 46 kPa, respectively. What will be the vapour pressure, in bar, of a mixture of 30.0 g of heptane and 34.2 g of octane?

Prashant Kotian · Accepted Answer

Correct answer: D. Molar mass of heptane (C7H16) = 100 g/mol, octane (C8H18) = 114 g/mol. Moles of heptane n_hep = 30.0 / 100 = 0.30 mol. Moles of octane n_oct = 34.2 / 114 = 0.30 mol. Mole fraction of heptane X_hep = 0.30 / (0.30 + 0.30) = 0.50, and mole fraction of octane X_oct = 0.50. Using Raoult's law: P_total = P_hep^o * X_hep + P_oct^o * X_oct = 106 * 0.50 + 46 * 0.50 = 53 + 23 = 76 kPa. Since 100 kPa = 1 bar: 76 kPa = 0.76 bar.

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