+1 vote
in Mass Transfer by (37.1k points)
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

This question was addressed to me by my college professor while I was bunking the class.

My question is based upon Molecular Diffusion in Liquids topic in portion Diffusion in Solids and Molecular Diffusion in Fluids of Mass Transfer

1 Answer

0 votes
by (39.8k points)
The correct option is (b) 1.67

For explanation I would say: N flux of A              = D(AB)/z * (Concentration difference)

             Concentration difference = 5/3 =1.67.

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