# Mass Transfer Questions and Answers – Design of Extractor for Single Stage Operations

«
»

This set of Advanced Mass Transfer Questions & Answers focuses on “Design of Extractor for Single Stage Operations”.

1. Stages in the extractor are _______________
a) Equilibrium stages
b) Theoretical stages
c) Equilibrium or theoretical stages
d) None of the mentioned

Explanation: The extractor stages are maintained in equilibrium to get the effect mass transfer.

2. Extraction includes mixing alone.
a) True
b) False

Explanation: Extraction needs mixing and separation.

3. Every stage of extractor act as mixer and settler in the extraction process.
a) True
b) False

Explanation: Every extracting process needs mixing and separating unit.
Note: Join free Sanfoundry classes at Telegram or Youtube

4. If the pure solvent is used find the value of “ys” by analysing the below single stage extractor. Where, F,S are feed and solvent
a) 0
b) 1
c) 0-1
d) None of the mentioned

Explanation: Pure solvent in the sense no solute present initially so ys= 0.

5. Find the rate of extract in mol/hr if

Rate(mol/hr) Composition
Feed 100 0.5
Solvent 50 0
Residue 25 0.25
Take Mass Transfer Mock Tests - Chapterwise!
Start the Test Now: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

a) 100
b) 125
c) 150
d) 50

Explanation: Feed + solvent = Extract + residue
100+50-25 = 125 mol/hr.

6. Is the extraction is possible if a gas solvent is used for extracting solute from liquid mixture.
a) True
b) False

Explanation: The extraction is possible only if liquid-liquid equilibrium occurs. If gas solvent is used such equilibrium cannot be achieve.

7. Find the process. a) Single stage batch process
b) Single stage continuous process
c) Multistage continuous process
d) None of the mentioned

Explanation: Here stage used is one and there is a continuous input and output so the process is a single stage continuous process.

8. Find the sum of R and E if the sum of F and S is 80 kmol. a) 20 kmol
b) 40 kmol
c) 60 kmol
d) 80 kmol

Explanation: M= F+M= R+S =80.

9. Find the pure solvent rate if the feed ( F= 100 kmol/hr) composition is 0.65 and mixture solute composition is 0.5.
a) 10 kmol/hr
b) 20 kmol/hr
c) 30 kmol/hr
d) 40 kmol/hr

Explanation: S/F = 0.65- 0.5/0.5 -0
= 0.3
= 100 x 0.3 = 30 kmol/hr.

10. Find the mixture composition of feed and pure solvent if the ratio of solvent to feed rate is 0.5 and the feed composition is 0.5.
a) 0.5
b) 0.25
c) 0.33
d) 0.4

Explanation: S/F= 0.5-x/x-0
0.5x+x = 0.5 then x= 0.33.

Sanfoundry Global Education & Learning Series – Mass Transfer.

To practice advanced questions on all areas of Mass Transfer, here is complete set of 1000+ Multiple Choice Questions and Answers 