Question

Calculate the amount of diffusion of acetic acid (A) in 2 hours across a film of non-diffusing water (B) solution 1 mm thick at 17 °C when the concentrations on the opposite side of the film are 9 and 3 weight percent acid respectively. The diffusivity of the acetic acid in solution is 0.95 x 10⁻⁹ m²/s. At 17 °C, the following additional data are available:

Density of 3% and 9% solution are 1003 and 1012 kg/m³ respectively. Molecular weight of A and B are 60 kg/kmol and 18 kg/kmol respectively.

Answer 1

Na = 9.25 x 10⁻⁷

Step-by-step explanation:

ρ1 = 1003 kg/m³

ρ2 = 1012 kg/m³

Dab = 0.95 x 10⁻⁹ m²/s

Z = 10⁻³

`x_(a1)=(9/60)/(9/60+91/18) =0.0288\\`

xa2 = (3/18) / (3/18 + 97/60) = 0.0092

Mavg1 = (9/100 * 60) + (91/100 * 18) = 21.78

Mavg2 = (3/100 * 18) + (97/100 * 60) = 19.26

ρavg = (ρ1 + ρ2) / 2 = (1003 + 1012) / 2 =1007.5 kg/m³

(ρ/M)avg = (ρavg/Mavg1 + ρavg/Mavg2) / 2 = (46.25 + 52.31) / 2 = 49.28

Na = -(ρ/M)avg * Dab / Z * ln [(1-xa1) / (1 - xa2)]

= -49.28 * 0.95 x 10⁻⁹ / 10⁻³ * ln [(1-0.0288) / (1 - 0.0092)] = 9.25 x 10⁻⁷