Answer:
⇒D_AB= 1.21×10^(-9)
Explanation:
Wike chang equation is given as:
[tex]D_{AB}= \frac{117.3\times10^{-18}\times\(\phi\times M_B)^{0.5}\times T}{\mu\times\nu^{0.6}}[/tex]
Where
D_AB= diffusivity of chlorine in water
Φ= 2.26 for water as solvent
ν= 0.0484 for chlorine as solute
M_B = Molecular weight of water
τ= temperature=289 K
μ= viscosity = 1.1×10^{-3}
Now putting these values in the above equation we get
[tex]D_{AB}= \frac{117.3\times10^{-18}\times\(\2.26\times18)^{0.5}\times289}{\1.1\times10^{-3}\times\0.0484^{0.6}}[/tex]
⇒D_AB= 1.21×10^(-9)
Answer:
[tex]\large \boxed{2.8\times 10^{-5}\text{ cm$^{2}$/s}}[/tex]
Explanation:
The Wilke-Chang equation for the liquid diffusion coefficient is
[tex]D = 7.4 \times 10^{-8}\left (\dfrac{T\sqrt{xM}}{\eta V^{0.6}} \right)[/tex]
where
D = diffusion coefficient in square centimetres per second
T = kelvin temperature
x = an association parameter for the solvent
M = molar mass of solvent
η = viscosity of solvent in centipoises
V = molar volume of solvent at normal boiling point in cubic centimetres per mole
Data:
T = 289 K
x = 2.6
M = 18.02 g/mol
η = 0.890 cP
V = 18.9 cm³/mol
Calculation:
[tex]\begin{array}{rcl}D & = & 7.4 \times 10^{-8}\left (\dfrac{T\sqrt{xM}}{\eta V^{0.6}} \right)\\\\& = & 7.4 \times 10^{-8}\left (\dfrac{289\sqrt{2.6 \times 18.02}}{0.890 (18.9)^{0.6}} \right)\\\\& = & 7.4 \times 10^{-8}\left (\dfrac{289\sqrt{46.85}}{0.890\times 5.833} \right)\\\\& = & 7.4 \times 10^{-8}\left (\dfrac{289\times 6.845}{0.890\times 5.833} \right)\\\\& = & \mathbf{2.8\times 10^{-5}}\textbf{ cm$^{2}$/s}\\\end{array}[/tex]
[tex]\text{The liquid diffusion coefficient for chlorine in water is $\large \boxed{\mathbf{2.8\times 10^{-5}}\textbf{ cm$^{2}$/s}}$}[/tex]
The published value is 1.25 × 10⁻⁵ cm²/s.
