Separation Processes Questions and Answers – The MSMPR Crystallization Model

«
»

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.

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.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
advertisement
advertisement

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.

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.
advertisement

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.
advertisement

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.

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.

Sanfoundry Global Education & Learning Series – Separation Processes.

To practice all areas of Separation Processes for Aptitude test, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.