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A tube has a length of 0.025 m and a cross-sectional area of 6.5 x 10-4 m2. The tube is filled with a solution of sucrose in water. The diffusion constant of sucrose in water is 5.0 x 10-10 m2/s. A difference in concentration of 5.2 x 10-3 kg/m3 is maintained between the ends of the tube. How much time is required for 5.7 x 10-13 kg of sucrose to be transported through the tube?

Respuesta :

Answer:

The time required for sucrose transportation through the tube is 8.4319 sec.

Explanation:

Given:

L = 0.025 m

A = 6.5×10^-4 m^2

D = 5×10^-10 m^2/s

ΔC = 5.2 x 10^-3 kg/m^3

m = 5.7×10^-13 kg

Solution:

t = m×L / D×A×ΔC

t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)

t = 8.4319 sec.

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