Water potential and molarity mixup

Claim

After a series of experiments our group claims that the colored solutions have the following concentrations of sugar: yellow 1.0M red .8M purple .6M blue.4M orange .2M and green 0M. I claim that the isomolar point of the parsnip we tested is 1.79M.

Evidence

The evidence we have for this was collected from an experiment we conducted that involved filling dialysis tubing with different solutions. For instance, we put a dialysis tube filled with the purple solution into a cup filled with yellow solution. We mixed and matched the bags twice to get these pairings purple in yellow, yellow in green, orange in blue, green in orange , red in purple, blue in red, yellow in blue, red in yellow, blue in  green, green in red. The change in mass of the dialysis bags is as follows respectively:  from 2.99g to 2.7g, from 1.99g to 3.9g, from 3.2g to 3.2g, from 2g to 1.7g, from  2.73g to 3g, from 3.3g to 3.5g, from 4.9 to 5.85, from 4.7g to 6.3g, from 4.2 to 4.1g. From this data we determined the molar concentrations of the solutions. For the parsnip we weighed pieces of parsnip each of them in a colored solution overnight and reweighed them these are the results: purple from 1.3g to 1.5g, blue from 2.3g to 2.7g, red from 2.2g to 2.5g, yellow from 2.4g to 2.5g,green from 2.1g to 2.5g, orange from 1.5g to 1.7g. I then constructed this graph of the percent change in mass for the parsnip from 0-1M:

Reasoning

The experiment was based on a principal like such, if a is greater than b and b is greater than c then a must also be greater than c. A solution of higher molarity has a lower concentration of water and the opposite is true for a solution of lower molarity. Water will travel from a solution of lower molarity to a solution of higher molarity (through osmosis).  We could figure out if a solution in a dialysis bag was of a higher or lower molarity than the solution it was in, by observing how water traveled through the selectively permeable membrane. If the bag gained mass when we reweighed it then we knew water entered the bag and if it lost mass then we knew water left the bag. In this way, we did not need to figure out exactly what molarity each solution was. In every case, the yellow solution gained water molecules whether it was in the bag if it had a bag placed in it. Using the aforementioned logic one could deduce that yellow is of the highest molarity and the highest molarity possible was 1m so we determined yellow’s concentration was 1m. I also reasoned that the isomolar point of the parsnip is 1.79M  by calculating the percent change in mass and making the above graph. I then attained the equation of the trend line and set y equal to 0.

Then I used the formula for solute potential to get the water potential of the parsnip (I could do this because the parsnip was in an open container meaning there was a pressure potential of 0).