# Mass Transfer Questions and Answers – Molecular Diffusion in Liquids

This set of Mass Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Molecular Diffusion in Liquids”.

1. Before collision, the distance between the two particles is mean free path.
a) True
b) False

Explanation: The mean free path can be found after collision.

2. Diffusion co-efficient in molecular diffusion is estimated by _________
a) Daltons law
b) Diffusion law
c) Ficks law
d) None of the mentioned

Explanation: Fick’s law is the ratio of J-flux to the concentration gradient.

3. Find the total flux for a particle A and B in a steady state if flux of A and B is 2.44 and 4.44 mol/sq.m sec.
a) 2
b) 4.88
c) 6.88
d) 8

Explanation: Total flux= flux A+ flux B= 6.88.

4. Find the flux of A ( xA= 0.2) if Total flux is 5 mol/sq.m sec and J flux of A is 2 mol/sq.m sec.
a) 2
b) 2.5
c) 3
d) 3.5

Explanation: N flux of A = J flux of A + Total flux N * xA
= 2+ 0.5*2 = 3.

5. Find J flux of A ( xA = 0.5) if the total N flux is 6 mol/sq.m sec and N flux of A is 3 mol/sq.m sec.
a) 0
b) 1
c) 2
d) 3

Explanation: N flux of A = J flux of A + Total flux N * xA
3= J flux of A + 6 * 0.5
J flux = 0.
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6. Diffusivity of the liquids can be determined by
a) Wilke- lee equation
b) Wilke- chan equation
c) Lee and chan equation
d) None of the mentioned

Explanation: Wilke- Chan derived an equation for diffusivity of liquids.

7. For the molecular diffusion of liquids, if the diffusing molecules having criteria of steady state diffusion of B over non diffusing A then the N flux of A is
a) 1
b) 0
c) Negative
d) Infinity

Explanation: For non-diffusing A then N flux of B is zero.

8. For liquid molecular diffusion of A and B, steady state equimolar counter diffusion the N flux of A is negative of N flux of B.
a) True
b) False

Explanation: For equimolar counter diffusion, the one component flux is negative of other.

9. Find the change in concentration for a steady state equimolar counter diffusion if D(AB)= 6 sq.m/sec, the change in distance is 2 m and the N flux of A is 5 mol/sq.m sec.
a) 0.67
b) 1.67
c) 2.67
d) None of the mentioned

Explanation: N flux of A = D(AB)/z * (Concentration difference)
Concentration difference = 5/3 =1.67.

10. Find the diffusivity of AB in sq.m/sec if N flux of A 5 mol/sq.m sec, concentration difference is 2mol/cu.m and distance is 3 m.
a) 2.5
b) 5.5
c) 7.5
d) 10.5

Explanation: N flux of A= D(AB)/z * concentration difference
D(AB) = 7.5 sq.m/sec.

Sanfoundry Global Education & Learning Series – Mass Transfer.

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