Separation Processes Questions and Answers – The MSMPR Crystallization Model

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This set of Separation Processes Questions and Answers for Aptitude test focuses on “The MSMPR Crystallization Model”.

1. In crystallization, solvent should dissolve large amount of solute at:
a) Room temperature
b) Boiling point
c) Freezing point
d) Slip melting point
View Answer

Answer: b
Explanation: Solvent should dissolve large amount of solute at boiling point of solvent.
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2. Insoluble impurities during crystallization are removed by:
a) Decantation
b) Filtration
c) Distillation
d) Fractional crystallization
View Answer

Answer: b
Explanation: Since the impurities are insoluble they can be easily removed by filtration and no other methods are required.

3. The crystals obtained in crystallization are dried by:
a) Dryer
b) Oven
c) Filter paper
d) Exposing to sun
View Answer

Answer: c
Explanation: The moisture is sucked with the help of a filter paper and the crystals are dried.

4. Which of the following is not an assumption of MSMPR model?
a) Perfect mixing of the magma
b) No crystal breakage
c) Continuous, steady-flow, steady-state operation
d) Crystals are of various sizes.
View Answer

Answer: d
Explanation: The crystals are assumed to be of same size in the MSMPR model.

5. What is the number of crystals per unit mass of crystals if
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 5
a) 0.01
b) 0.7
c) 0.6
d) 0.8
View Answer

Answer: a
Explanation: Number of crystals per unit mass of crystals= 4.5/fvρL3, hence N=0.01.
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6. What is the number of crystals per unit mass of crystals if
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.05
a) 0.01
b) 0.7
c) 0.6
d) 100000
View Answer

Answer: d
Explanation: Number of crystals per unit mass of crystals= 4.5/fvρL3, hence N=1000000.

7. What is the number of crystals per unit mass of crystals if
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.78
a) 1.56
b) 2.34
c) 2.56
d) 0.8
View Answer

Answer: c
Explanation: Number of crystals per unit mass of crystals= 4.5/fvρL3, hence N=2.56.

8. What is the number of crystals per unit mass of crystals if
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.56
a) 5.67
b) 6.78
c) 7.11
d) 9.56
View Answer

Answer: c
Explanation: Number of crystals per unit mass of crystals= 4.5/fvρL3, hence N=7.11.

9. What is the number of crystals per unit mass of crystals if
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.67
a) 1.56
b) 7.89
c) 4.22
d) 4.15
View Answer

Answer: d
Explanation: Number of crystals per unit mass of crystals= 4.5/fvρL3, hence N=4.15.
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10. What is the nucleation rate if
Mass of production of crystals C=0.5kg
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.5
volume of mother liquor in magma V= 1m3
a) 0.5
b) 1.5
c) 1.8
d) 2.7
View Answer

Answer: c
Explanation: The nucleation rate=4.5C/fvρL3V, hence rate= 1.8.

11. What is the nucleation rate if
Mass of production of crystals C=0.5kg
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.5
volume of mother liquor in magma V= 0.5m3
a) 5.4
b) 3.6
c) 1.8
d) 2.7
View Answer

Answer: b
Explanation: The nucleation rate=4.5C/fvρL3V, hence rate= 3.6.

12. What is the nucleation rate if
Mass of production of crystals C=0.5kg
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.6
volume of mother liquor in magma V= 1m3
a) 1.45
b) 1.5
c) 1.07
d) 1.04
View Answer

Answer: d
Explanation: The nucleation rate=4.5C/fvρL3V, hence rate= 1.04.

13. What is the nucleation rate if
Mass of production of crystals C=0.5kg
Volume shape factor fv= 0.5
Density of particle= 0.5kg/m3
Characteristic size of crystal L= 0.5
volume of mother liquor in magma V= 1m3
a) 0.5
b) 15
c) 1.8
d) 18
View Answer

Answer: d
Explanation: The nucleation rate=4.5C/fvρL3V, hence rate= 18.
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14. What is the nucleation rate if
Mass of production of crystals C=10kg
Volume shape factor fv= 0.5
Density of particle= 5kg/m3
Characteristic size of crystal L= 0.5
volume of mother liquor in magma V= 1m3
a) 9
b) 18
c) 180
d) 36
View Answer

Answer: c
Explanation: The nucleation rate=4.5C/fvρL3V, hence rate= 180.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn