An adsorbent, which is a modified clay, is used to separate an organic compound A from anaqueous solution. One kilogram of the solution containing 10% of the organic was treatedwith varying amounts of the clay in a number of laboratory tests. The following data werecollected. Knowing that Y = 0. 173X1.36, where Ygram solute per gram solute-free solvent=gram solute per gram clay, and X =Gram clay used% A in the solutionin equilibrium159.1407.81606.931005.661354.762103.38325 4502.44 1.48(a) How much of the adsorbent is required to recover 90% of the solute from 1000 kg of10% solution?(b) If 678 kg of the adsorbent is used per 1000 kg of the solution, how much of theSolute is recovered if the treatment is done in two stages, using half of the clay in eachstage?

(a) To recover 90% of the solute from 1000 kg of 10% solution, we can use the following equation: Y…

(a) How much of the adsorbent is required to recover 90% of the solute from 1000 kg of 10% solution? (b) If 678 kg of the adsorbent is used per 1000 kg of the solution, how much of the Solute is recovered if the treatment is done in two stages, using half of the clay in each stage?

(a) How much of the adsorbent is required to recover 90% of the solute from 1000 kg of 10% solution? (b) If 678 kg of the adsorbent is used per 1000 kg of the solution, how much of the Solute is recovered if the treatment is done in two stages, using half of the clay in each stage?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

An adsorbent, which is a modified clay, is used to separate an organic compound A from an
aqueous solution. One kilogram of the solution containing 10% of the organic was treated
with varying amounts of the clay in a number of laboratory tests. The following data were
collected. Knowing that Y = 0. 173X1.36, where Y
gram solute per gram solute-free solvent
=
gram solute per gram clay, and X =
Gram clay used
% A in the solution
in equilibrium
15
9.1
40
7.81
60
6.93
100
5.66
135
4.76
210
3.38
325 450
2.44 1.48
(a) How much of the adsorbent is required to recover 90% of the solute from 1000 kg of
10% solution?
(b) If 678 kg of the adsorbent is used per 1000 kg of the solution, how much of the
Solute is recovered if the treatment is done in two stages, using half of the clay in each
stage?

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