Answer to Question #233506 in Chemical Engineering for Lokika

Question #233506

A 150 moles of liquid mixture containing 60 mol% n-hexane (more volatile) and n-heptane is to be batch distilled at latm total pressure. The relative volatility of n-hexane in the mixture is 2.36.


i)If 50 moles is distilled, what is the average composition of the distillate and the composition of the liquid left in the still?


ii)If the accumulated vapor is 90% n-hexane, calculate the amount of distillate


1
Expert's answer
2021-09-07T23:57:54-0400

Part 1

The Refractive index is a unidimensional measurement that describes how fast light travels through the material. It is used to see the liquid purity of compounds.

In this problem you have in the beginning a 50:50 solution oh hexane: nonane and realize distillation.

In distillation, the collect is enriched with the most volatile substance (hexane). So you will have in the final sample more hexane than nonane.

Assuming the refractive index varies linearly with mole fraction, you will have:

"1,375 \u2192 100\\% hexane\\\\\n\n1,407 \u2192 100\\% nonane\\\\\n\n1,385 \u2192 ???\\\\"

molar fraction of hexane:

"100 - \\frac{(1,385 - 1,375) }{ (1,407 - 1,375)} \u00d7 100 = 69 \\%"

molar fraction of nonane:

"100 - \\frac{(1,407 - 1,385) }{ (1,407 - 1,375)} \u00d7 100 = 31 \\%"

The relation of the distillating fraction is (69:31) of (hexane:nonane)


Part 2

"\\frac{31}{100}* \\frac{10}{100}* 100=3.1 \\%"


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