Answer to Question #233506 in Chemical Engineering for Lokika

Question

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

Answer 1

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:

$1,375 → 100\% hexane\\ 1,407 → 100\% nonane\\ 1,385 → ???\\$

molar fraction of hexane:

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

molar fraction of nonane:

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

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

Part 2

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