# Separation Processes Questions and Answers – Maloney-Schubart Graphical Equilibrium Stage Method

This set of Separation Processes Quiz focuses on “Maloney-Schubart Graphical Equilibrium Stage Method”.

1. How a sharp separating point is occurs for solute and carrier in type 2 systems?
a) Extract reflux
b) Extract influx
c) Extract preflux
d) Detract influx

Explanation: With the help of extract relux, a sharp separating point is occurs for solute and carrier in type 2 systems.

2. For total reflux, difference points lie at:
a) Y=0
b) Y=-100 and Y=100
c) Y=0 and Y=infinity
d) Y=infinity and Y= -infinity

Explanation: Because F=B=D=0, for total reflux, the difference points lie at Y=infinity and Y= -infinity.

3. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.5 (YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 250

Explanation: Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 100*(0.5+0.5)-0.7/0.7-0.5=150.

4. Y which is ratio of mass of solvent to mass of solvent in liquid free phase: What does the value of Y depend on?
a) Viscosity of solvent
b) Amount of solvent
c) Mass fraction of solvent
d) Carrier solubility of solvent

Explanation: Y which is ratio of mass of solvent to mass of solvent in liquid free phase depends on carrier solubility of solvent.

5. Following the feed-stage location, the stripping stages are _______ until desired raffinate concentration is achieved.
a) Not done
b) Stepped on
c) Stepped off
d) Increased

Explanation: Following the feed-stage location, the stripping stages are stepped off until desired raffinate concentration is achieved.
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6. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.5 (YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 200kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 300

Explanation: Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR, LR= 200*(0.5+0.5)-0.7/0.7-0.5=300.

7. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.4 (YD)
For extract leaving stage N is 0.6 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 40 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 250

Explanation: Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 100*(0.4+0.4)-0.6/0.6-0.4=100.

8. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.3(YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 125
c) 200
d) 250

Explanation: Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 250*(0.3+0.5)-0.7/0.7-0.5=125.

9. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.5 (YD)
For extract leaving stage N is 0.7 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 400
b) 500
c) 600
d) 700

Explanation: Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR= 400*(0.5+0.5)-0.7/0.7-0.5=600.

10. If Y which is ratio of mass of solvent to mass of solvent in liquid free phase:
For extract is 0.2 (YD)
For extract leaving stage N is 0.4 (YVN)
Amount of extract is 100kg (D)
Amount of solvent in cascade is 50 (SD)
What is the extract reflux?
a) 100
b) 150
c) 200
d) 250

Explanation: Since LR/D = (YD)+SD/D)-( YVN)/YVN-YD, LR 100*(0.2+0.5)-0.4/0.4-0.2=150.

Sanfoundry Global Education & Learning Series – Separation Processes.

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