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?

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Solution

## Step 1: Given dataRatio of Pentane to Hexane= 1:4Partial pressure of Pentane= 440 mm HgPartial pressure of Hexane= 120 mm HgStep 2: Calculating total pressureMole fraction of Pentane is:$\frac{1}{1+4}=\frac{1}{5}$Mole fraction of Hexane is:$\frac{4}{1+4}=\frac{4}{5}$$\therefore$The total pressure of the solution is:${\mathrm{P}}_{\mathrm{s}}={\mathrm{X}}_{\mathrm{p}}{\mathrm{P}}_{\mathrm{p}}^{0}+{\mathrm{X}}_{\mathrm{h}}{\mathrm{P}}_{\mathrm{h}}^{0}\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{\mathrm{s}}=\frac{1}{5}×440+\frac{4}{5}×120\phantom{\rule{0ex}{0ex}}{\mathrm{P}}_{\mathrm{s}}=184\mathrm{nm}$Step 3: Calculating mole fractionThe mole fraction of Pentane in the vapour phase is:$\frac{{\mathrm{X}}_{\mathrm{p}}{\mathrm{P}}_{\mathrm{p}}}{{\mathrm{P}}_{\mathrm{s}}}=\frac{\frac{440}{5}}{184}=\frac{88}{184}\phantom{\rule{0ex}{0ex}}=0.478$Therefore, The mole fraction of Pentane in the vapour phase would be 0.478.

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