Separation Processes Questions and Answers – Batch Crsytallization

«
»

This set of Separation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Batch Crsytallization”.

1. What type of size distribution occurs in batch crystallization?
a) Trimodal
b) Monomodal
c) Bimodal
d) No distribution occurs
View Answer

Answer: c
Explanation: Formation of tiny crystals by a secondary nucleation occurs simultaneously with growth of added seeds. The nuclei formed also grow alongside. This ultimately leads to a bimodal or even multimodal size distribution of the final product.
advertisement

2. Which of the following is not a common method in reducing the rate of nucleation?
a) To control supersaturation
b) To keep low level of turbulence of mixing
c) To keep the slurry density low
d) To keep high level of turbulence of mixing
View Answer

Answer: b
Explanation: A narrow crystal size distribution can be achieved if there is little or no nucleation, so that the final product is the result of growth of the seeds only. The common methods of reducing the rate of nucleation (a) to control the supersaturation level, (b) to keep a low level of turbulence or mixing, (c) to keep the slurry density low.

3. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 400 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.06lit/min
c) 0.07lit/min
d) 0.08lit/min
View Answer

Answer: a
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.05lit/min.

4. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 800 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.06lit/min
c) 0.07lit/min
d) 0.1lit/min
View Answer

Answer: d
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.1lit/min.

5. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 600 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.0625lit/min
c) 0.07lit/min
d) 0.08lit/min
View Answer

Answer: a
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.0625lit/min.
advertisement

6. Find the initial rate of evaporation if
Mass of seed=Ms= 0.924 kg,
Solubility C = 400 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.10lit/min
c) 0.15lit/min
d) 0.08lit/min
View Answer

Answer: c
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.15lit/min.

7. Find the initial rate of evaporation if
Mass of seed=Ms= 3.08 kg,
Solubility C = 400 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.5lit/min
c) 0.7lit/min
d) 0.8lit/min
View Answer

Answer: b
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.5lit/min.

8. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 200 kg/m3, G/Ls= 3.75 x 10-4
a) 0.0225lit/min
b) 0.06566lit/min
c) 0.0527lit/min
d) 0.5408lit/min
View Answer

Answer: a
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.0225lit/min.

9. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 80 kg/m3, G/Ls= 3.75 x 10-4
a) 0.01lit/min
b) 0.02lit/min
c) 0.03lit/min
d) 0.04lit/min
View Answer

Answer: a
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.01lit/min.
advertisement

10. Find the initial rate of evaporation if
Mass of seed=Ms= 0.308 kg,
Solubility C = 1600 kg/m3, G/Ls= 3.75 x 10-4
a) 0.05lit/min
b) 0.15lit/min
c) 0.20lit/min
d) 0.25lit/min
View Answer

Answer: c
Explanation: Since -dV/dt=3MsGc/Ls for initial rate of evaporaton, initial rate=0.20lit/min.

Sanfoundry Global Education & Learning Series – Separation Processes.

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

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!

advertisement
advertisement
advertisement
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