Question

5. Consider sucrose diffusing along a 10 cm length of tube filled with water. The cross-sectional area of the tube is 6 cm². The diffusion coefficient is 5.0 x 10^–10 m²/s, and a total amount of sucrose of 8.1 x 10^–14 kg is transported in steady state along the tube in 15 s. What is the difference in the density levels of sucrose at the two ends of the tube? (Answer in grams per cubic meter, g/m³.)

Select one:

a. 4.0 gram per cubic metre

b. 18.0 gram per cubic metre

c. 27.0 gram per cubic metre

d. 1.8 gram per cubic metre

Answer 1

Here we have given that,

L = 10 cm length of tube filled with water

cross-sectional area of the tube A = 6 cm2

The diffusion coefficient D = 5.0 x 10^–10 m2/s, total amount of sucrose = 8.1 x 10–14 kg

Time t = 15 s.

So that the difference in the density levels of sucrose at the two ends of the tube will be given as

From the Ficks law we have a relationship as,

Delta C = RL/D×A

Here R = 8.1 × 10^-14/15 =5.4 ×10^-15

Now on inserting the values we will get

Delta C = (5.4 ×10^-15 ×0.1)/( 5 ×10^-10 × 6 ×10^-4)

Delta C = 1.8 ×10^-3 kg/m³

But answer demands in grams per cubic meter, g/m3.

So that the answer will be option d.) 1.8 gram per cubic metre

Delta C = 1.8 g/m³