Question

Diffusion of Sucrose in Gelatin. A layer of gelatin in water 5 mm thick and con-taining 5.1 wt % gelatin at 293 K separates two solutions of sucrose. The concentra-tion of sucrose in the solution at one surface of the gelatin is constant at 2.0 g sucrose/100 mL solution, and 0.2 g/100 mL at the other surface. Calculate the flux of sucrose in kg sucrose/s

Answer 1

the flux of sucrose is 9.072 × 10⁻⁷ kg sucrose / m² sec

Step-by-step explanation:

Given the data in the question;

water-gelatin solution separating two different concentration solutions of sucrose; 5.1 wt % gelatin at 293 K; at this conditions diffusivity of sucrose in water gelatin solution is;

`D_(AB)` = 0.252 × 10⁻⁹ m²/sec.

we know that; 1 L = 0.001 m³, 1mL = 0.001 L

`C_(A1)` = 2.0 g sucrose/100 mL = 2.0 × 10⁻³ kg sucrose / 100 × 10 ⁻³ L

`C_(A1)` = 2.0kg sucrose / 100 L

`C_(A1)` = 2.0 kg sucrose / 100 × 10⁻³ m³

`C_(A1)` = 20 kg sucrose / m³

`C_(A2)` = 0.2 g sucrose/100 mL = 0.2 × 10⁻³ kg sucrose / 100 × 10⁻³ × 10⁻³ L

`C_(A2)` = 2 kg sucrose / m³

Thickness ß = 5 mm = 5 × 10⁻³ m

Now, flux of sucrose in kg sucrose / m³sec will be;

using the formula,
`N_(A)` =
`D_(AB)` /ß (
`C_(A1)` -
`C_(A2)` )

we substitute

`N_(A)` = (0.252 × 10⁻⁹ m²/sec / 5 × 10⁻³ m) ( 20 kg sucrose / m³ - 2 kg sucrose / m³ )

`N_(A)` = (0.252 × 10⁻⁶ / 5) × 18 kg sucrose / m² sec

`N_(A)` = 4.536 × 10⁻⁶ / 5 kg sucrose / m² sec

`N_(A)` = 9.072 × 10⁻⁷ kg sucrose / m² sec

Therefore, the flux of sucrose is 9.072 × 10⁻⁷ kg sucrose / m² sec