Scribd
Upload
51 views2 pages

Sample Problem #2

(1) The document presents a sample problem involving a countercurrent extraction system with 30 kg of solute dissolved in 100 L of water. (2) Part (a) asks how much solute would be extracted by 35 L of ether, given the distribution coefficient is 4. The solution finds 3.522 kg would be extracted. (3) Part (b) asks how much solute would remain in the raffinate of a 3-stage countercurrent extraction. The solution finds 13.23 kg would remain.

Uploaded by

Grazel MD
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
51 views2 pages

Sample Problem #2

(1) The document presents a sample problem involving a countercurrent extraction system with 30 kg of solute dissolved in 100 L of water. (2) Part (a) asks how much solute would be extracted by 35 L of ether, given the distribution coefficient is 4. The solution finds 3.522 kg would be extracted. (3) Part (b) asks how much solute would remain in the raffinate of a 3-stage countercurrent extraction. The solution finds 13.23 kg would remain.

Uploaded by

Grazel MD
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 2

Sample Problem #2:

There are 30 kg of solute in 100 L of H 2O. (a) What is the weight of solute that will be
extracted by 35 L of ether if the distribution coefficient is 4? (b) If a 3-stage
countercurrent extraction system is used, what is the weight of solute that will remain in
the raffinate?

Given:

V1 V2

Extractor 35 L ether

L0 L1
30 kg of solute in
100 L of water

(a) What is the weight of solute that will be extracted by 35 L of ether if the
distribution coefficient is 4?

Solution: K = mass of A/ mass of B 4 = 30 kg solute / 270 kg water


mass of A/ mass of C mass of A/ 126.805 kg of ether

* 35 L ether (1 m3 / 1000 L) (3623kg / 1 m3 ) = 126.805 kg

0.111 / 4 = mass of A / 126. 805 kg ether

mass of A = 126.805 kg ether (0.111)

mass of A = 3.522 kg solute


(b) If a 3-stage countercurrent extraction system is used, what is the weight of solute
that will remain in the raffinate?

Given:

V1 Vn+1
35 L ether

Extractor Extractor Extractor

L0 L1
30 kg solute in
100 L water

Solution:
L0 + Vn+1 = M

100 kg + 126. 805 kg = 226. 805 kg

L0 xao + Vn+1 yn+1 = L1 xa1 + V1 y1

Assume x = mass fraction of solute

100 (0.30) = (100 – x) + (126.805 + x)

x = 100 (0.30) + 126. 805 (0)


100 + 126.805

x = 0.1322 (126.805)

mass of solute in the raffinate = 30 kg – 16.77 kg = 13.23 kg

You might also like