1. Tardigrade
  2. Question
  3. Chemistry
  4. 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).

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Q. 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 $= 114\, g \,mol^{-1}$).

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Solution:

Mole fraction of Heptane $= \frac{25 /100}{\frac{25}{100}+\frac{35}{114}} = \frac{0.25}{0.557} = 0.45$
$X_{Hep \,tane} = 0.45 .$
$\therefore $ Mole fraction of octane $= 0.55 = X_{octane}$
Total pressure $= \sum X_{i}P_{i}^{0}$
$= \left(105 \times 0.45\right) + \left(45 \times 0.55\right) kP_{a}$
$= 72.0 \,KPa$