# Separation Processes Questions and Answers – Rate Based Model for Leaching

This set of Separation Processes Multiple Choice Questions & Answers (MCQs) focuses on “Rate Based Model for Leaching”.

1. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.6h
c) 7.8h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=3.79h.

2. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.6M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.56h
c) 7.87h
d) 9.4h

Explanation: T= ρr2/6DeMbbCab, hence t=4.56h.

3. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 4.2h
b) 4.3h
c) 4.76h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=4.76h.

4. If the density of CuO in ore ρ = 0.08
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 6.08h
b) 6.32h
c) 6.35h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=6.08h.

5. If the density of CuO in ore ρ = 0.06
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 10.2h
b) 10.42h
c) 10.54h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=10.42h.
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6. If the density of CuO in ore ρ = 0.03
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 2.1h
b) 2.28h
c) 7.8h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=2.28h.

7. If the density of CuO in ore ρ = 0.04
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 3.8h
b) 4.75h
c) 7.8h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=4.75h.

8. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 6*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 6.12h
b) 6.22h
c) 6.62h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=6.62h.

9. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 4.44*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 5.13h
b) 5.3h
c) 5.65h
d) 6.4h

Explanation: T= ρr2/6DeMbbCab, hence t=5.13h.

10. If the density of CuO in ore ρ = 0.05
It is leached with H2SO4 = 0.5M
Effective diffusivity De= 9*10-6
Molecular weight Mb=79.6
B=0.5
Cab=0.01
Calculate the time required for total leaching.
a) 2.53h
b) 3.4h
c) 5.56h
d) 6.78h

Explanation: T= ρr2/6DeMbbCab, hence t=2.53h.

Sanfoundry Global Education & Learning Series – Separation Processes.

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