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

Answer:

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

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