Question

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

• A. $72.0\, kPa$
• B. $36.1\, kPa$
• C. $96.2\, kPa$
• D. $144.5\, kPa$

Answer 1

Answer: (A)

$72.0\, kPa$

Answer 2

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 $= \sum X_{i}P_{i}^{0}$ $= \left(105 \times 0.45\right) + \left(45 \times 0.55\right) kP_{a}$ $= 72.0 \,KPa$