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;
[tex]D_{AB}[/tex] = 0.252 × 10⁻⁹ m²/sec.
we know that; 1 L = 0.001 m³, 1mL = 0.001 L
[tex]C_{A1}[/tex] = 2.0 g sucrose/100 mL = 2.0 × 10⁻³ kg sucrose / 100 × 10 ⁻³ L
[tex]C_{A1}[/tex] = 2.0kg sucrose / 100 L
[tex]C_{A1}[/tex] = 2.0 kg sucrose / 100 × 10⁻³ m³
[tex]C_{A1}[/tex] = 20 kg sucrose / m³
[tex]C_{A2}[/tex] = 0.2 g sucrose/100 mL = 0.2 × 10⁻³ kg sucrose / 100 × 10⁻³ × 10⁻³ L
[tex]C_{A2}[/tex] = 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, [tex]N_{A}[/tex] = [tex]D_{AB}[/tex] /ß ( [tex]C_{A1}[/tex] - [tex]C_{A2}[/tex] )
we substitute
[tex]N_{A}[/tex] = (0.252 × 10⁻⁹ m²/sec / 5 × 10⁻³ m) ( 20 kg sucrose / m³ - 2 kg sucrose / m³ )
[tex]N_{A}[/tex] = (0.252 × 10⁻⁶ / 5) × 18 kg sucrose / m² sec
[tex]N_{A}[/tex] = 4.536 × 10⁻⁶ / 5 kg sucrose / m² sec
[tex]N_{A}[/tex] = 9.072 × 10⁻⁷ kg sucrose / m² sec
Therefore, the flux of sucrose is 9.072 × 10⁻⁷ kg sucrose / m² sec