Question

A solution has a 1:4 mole ratio of Pentane to Hexane. the vapour pressure of the pure hydrocarbons at 20°C is 440 mm of Hg for Pentane and 120 mm of Hg for Hexane. The mole fraction of Pentane in the vapour phase would be?

Answer 1

Step 1: Given data

Ratio of Pentane to Hexane= 1:4

Partial pressure of Pentane= 440 mm Hg

Partial pressure of Hexane= 120 mm Hg

Step 2: Calculating total pressure

Mole fraction of Pentane is:

$\frac{1}{1+4}=\frac{1}{5}$

Mole fraction of Hexane is:

$\frac{4}{1+4}=\frac{4}{5}$

$\therefore$The total pressure of the solution is:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{s}}={\mathrm{X}}_{\mathrm{p}}{\mathrm{P}}_{\mathrm{p}}^0+{\mathrm{X}}_{\mathrm{h}}{\mathrm{P}}_{\mathrm{h}}^0 \\ {\mathrm{P}}_{\mathrm{s}}=\frac{1}{5}\times 440+\frac{4}{5}\times 120 \\ {\mathrm{P}}_{\mathrm{s}}=184\mathrm{nm} \end{gathered}$$

Step 3: Calculating mole fraction

The mole fraction of Pentane in the vapour phase is:

$$\begin{gathered} \frac{{\mathrm{X}}_{\mathrm{p}}{\mathrm{P}}_{\mathrm{p}}}{{\mathrm{P}}_{\mathrm{s}}}=\frac{{\displaystyle \frac{440}{5}}}{184}=\frac{{\displaystyle 88}}{184} \\ =0.478 \end{gathered}$$

Therefore, The mole fraction of Pentane in the vapour phase would be 0.478.