# Calculating the reflection and solute permeability for a diffusion cell

Here is the question, at which I have some problem.

'The three characteristic parameters $$\sigma$$, $$L_p$$ and $$\omega$$ can be determined from two experiments; in the first experiment 4.6 ml of water has been permeated in 1 hour in a cell with a diameter of 7.5 cm whereas 10 bars of pressure has been applied. In the second experiment 1 g of sucrose (Mw: 342 g/mol) is dissolved in 100 ml of water. One compartment of a dialysis cell with a volume of 44 ml is filled with this solution whereas the other contains pure water. After two hour the liquid volume in the sucrose compartment has been increased 0.57 ml while the sucrose concentration has been decreased with 1.16%.'

My working

For Experiment 1, I was able to calculate the $$L_p$$ which is the permeability constant for water to be $$0.2892297237·10^{-5} \frac{cm}{bar ·s}$$ using the following equation.

$$\begin{equation} J_v = L_p(\Delta P-\sigma \Delta \pi) \end{equation}$$ Where I said that the osmotic pressure must be zero because the concentration on both sides is the same. So it was straight forward.

But I am having trouble with the second experiment to calculate the solute permeability $$\omega$$ and reflection coefficient $$\sigma$$.

I have calculated the following terms, $$\begin{equation} c= 1 \frac{g}{100 cm^3}=0.01 \frac{g}{cm^3} = \frac{1}{34200}\frac{mol}{cm^3} \end{equation}$$

The I calculated the diffusive fluxes for both volume and solute so that $$\begin{equation} J_v = 0.157 cm^3 / (7200s) = 0.2180555556·10^{-4} cm^3/s \end{equation}$$ $$\begin{equation} J_s = c · J_v = 6.375893439 · 10^{-10} mol/s \end{equation}$$

Apart from the above I also found the concentrations before and after. Since it is reducing by 1.6% therefore the new concentration is $$\begin{equation} c_2 = c_1 - \bigg(\frac{1.6}{100} · c \bigg)= 0.2877192982 · 10^{-4} · mol/cm^3 \end{equation}$$

And hence $$\Delta c = 4.6783626·10^{-7} mol/cm^3$$.

I also calculated the osmotic pressure to be 0.01143204685 bar.

Then using all these values and using the diffusion equations, $$\begin{equation} J_v = L_p(\Delta P-\sigma \Delta \pi) \end{equation}$$ $$\begin{equation} J_s = c · (1- \sigma)J_v + \omega \Delta \pi \end{equation}$$

But when I solve it get very wrong answers.

I can't understand what I wrong in the above. Any clue on how it should be calculated

P.S the correct answers are $$\omega = 6.5·10^{-6}$$ and $$\sigma = 0.86$$.

Any help will be highly appreciated.