Question

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A solution containing 40% propanol in water is to be extracted with a solvent consisting of 5% propanol in octanol. What would be the propanol concentration in the raffinate stream following a single extraction using equal masses of feed and solvent?

Answer 1

here, propanol is solute, water is carrier liquid and octanol is solvent used for liquid extraction.

x_f= propanol concentration in the feed=0.4

y_s= propanol concentration in the solvent=0.05

feed amount= F kg , solvent amount = S kg

acc. to question F=S

amount of mixture to be settled= M=F+S=2*F

x_m= propanol concentration in the mixture after mixing

note that liquid extraction consists of two parts- mising and settling.

E= amount of extract

R= amount of raffinate

y= concentration of solute in extract phase

x= concentration of solute in raffinate phase

now applying mass balance for mixing

M*x_m=F*x_f+S*ys

it implies, x_m=0.225

now applying mass balance for settling,

M*x_m= E*y+R*x

for calculation of x and y we require liquid liquid equilibrium data.

first draw the ternary diagram for it, then plot point F representing feed lying on the side of triangle,S representing the solvent lying on the otherside of triangle and then M represnting mixture. point M will lie on line FS . find the tie line which passes through the M. This line will give the concentration of propanol in extract and raffinate phase.

if we know the value of F then E and R can also be calculated.