The evaporation rate of chloroform (CHCl3) in air has been experimentally measured in an Arnold diffusion cell at 25°C and 1.00 atm. Over a period of 10 hr, the chloroform dropped 0.44 cm from the initial surface 7.40 cm from the top of the tube. The density of liquid chloroform is 1.485 g/cm3 and its vapor pressure at 25°C is 200 mm Hg. 1. Assuming air is an ideal gas, determine the molar concentration of air in mol/cm3. 2. Determine the mole fraction of chloroform in air at the liquid-air interface. 3. Determine the flux of chloroform through the air column in mol / (cm2 s) 4. Determine the diffusion coefficient of chloroform in air in cm2 / s.

1. C = 0.041 mol/L, 2. Mole Fraction = 0.2632, 3. N (A) = 9.322 x 10⁻⁵ mol/cm². s, 4. D (AB) = 5.688 x 10⁻⁵ cm²/s

Explanation:

Use the ideal gas equation

PV = nRT

Can be re4arranged as

P = nRT/V, where n/V = C

Hence

P = CRT

1. Concentration of air

C = P/RT

C = 101325 / (8.314 x 298)

C = 40.89mol/m³ = 0.041 mol/L

2. Mole Fraction of chloroform in air at the liquid - air interface

Mole fraction = vapor pressure (chloroform) / vapor pressure (total)

Mole Fraction = 200/760 = 0.2632

3. Flux of chloroform

N (A) = (D (AB) / /Z). (P (T) / RT). (P (A1) - P (A2)) / P (BM)

P (BM) = (760 - 560) / ln (760/560) = 655 mm Hg

D (AB) = RTP (BM) (Z²₁ - Z²₀) / {2PM (A) (P (A1) - P (A2)) }

D (AB) = 8.314 x 298 x 655 (0.0758² - 0.074²) / (2 x 101325 x 0.1195 x 200 x 36000)

D (AB) = 5.688 x 10⁴ = 5.688 x 10⁻⁵cm²/s (multiplying by 10⁴ to convert into cm2)

N (A) = (5.688 x 10⁻⁵) / (7.62) x (101325/8.314 x 298) x (200/655)

N (A) = 9.322 x 10⁻⁵ mol/cm². s

4. D (AB) has already been calculated in the solution of 3

D (AB) = 5.688 x 10⁻⁵ cm²/s