[AP Bio 10]: Diffusion + Osmosis Lab ~RN

% change in mass formula

(finalmass - initialmass)/initialmass x 100

How do you make a 120 ml of 0.2 M sucrose? 0.4 M sucrose? 0.6 M sucrose? 0.8 M sucrose? 1.0 M sucrose? 0.0 M sucrose?

- 24 ml of sucrose (2% of 120) + 96 ml of distilled water (120 - sucrose) - 48 ml sucrose + 72 ml dH2O - 72 ml sucrose + 48 ml dH2O - 96 ml sucrose + 24 ml dH2O - 120 ml sucrose - 120 ml dH2O

lab B results

0.0 M - isotonic, so mass stays same (least % mass change) 0.2 M - bag hypertonic to dH2O, so water flows into bag, increasing mass 0.4 M - bag hypertonic to dH2O, so water flows into bag, increasing mass 0.6 M - bag hypertonic to dH2O, so water flows into bag, increasing mass 0.8 M - bag hypertonic to dH2O, so water flows into bag, increasing mass 1.0 M - bag hypertonic to dH2O, so water flows into bag, increasing mass (most % mass change)

lab C results

0.0 M - potato cells hypertonic to solution, so water flows in, mass increases (highest % mass change) 0.2 M 0.4 M 0.6 M 0.8 M 1.0 M - solution hypertonic to cells, so water flows out, mass decreases (lowest % mass change) molar concentration of vegetable is at isotonic, where the % mass change is 0

lab B summary

1) Each person prepared a solution: distilled water, 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M sucrose 2) Tie one end of dialysis bag 3) Pour in 15 ml of solution into bag 4) Remove air from bags by drawing bags with two fingers and tie off the other end, leaving space for expansion of bag contents (solution should only fill about one-third of bag) 5) Rinse each bag with distilled water to remove any spilled sucrose during filling 6) Hammock each bag and mass them in grams 7) Place each bag in empty cups with labels to indicate molarity of solution in dialysis bag 8) Fill each cup two-thirds with distilled water, completely submerging each bag 9) Let stand for 30 minutes 10) Remove bags from water, hammock again, and mass

lab C summary

1) put remaining solution (after taking 15 ml out for lab B) into a cup 2) take four cubes of potato and mass them 3) put the four cubes in the sucrose solution 4) let it stand overnight

relationship between the the molarity of sucrose in the bags and percent change in mass

As the amount of sucrose in the bag increased in molarity, the percent change in mass also increased. This is a directly proportional relationship and would form a linear graph

If a plant cell has a lower water potential than its surrounding environment, and if pressure is equal to zero, is the cell hypertonic or hypotonic to its environment? Will the cell gain water or lose water? Explain your response.

If the plant cell has lower water potential, that means the water will come into the cell, the cell is hypertonic to its environment. This cell will gain water because water follows its concentration gradient.

A dialysis bag is filled with distilled water then placed in a sucrose solution. The bag's initial mass is 20 g and its final mass is 18 g. Calculate the percent change of mass, showing your calculations here.

Initial- 20g Final- 18g (18g-20g)/20g x 100 Percent change of mass= -10%. 18g(final mass) - 20g(initial mass)=-2/20g(initial mass) x 100, which gives you a -10% change of mass.

potato errors

Some mistakes that could have taken place are mathematical miscalculations while finding the initial and final masses. A piece of potato skin could have been left in the beakers along with the potato. This causes problems in the data tables. Another possible source of error could be that the students did not pat dry the potato sample well enough and increased the masses of the cores. Numerous may have occurred while using the electronic balance.

what would happen to the mass of each bag in the experiment if they were placed in a 0.4M sucrose solution instead of distilled water

The higher the solute concentration, the higher the rate of osmosis and the smaller the solute potential. Therefore any bag with a sucrose molarity of less than .4 would lose mass and would gain mass if it was greater than a .4 sucrose molarity. A bag with .4 sucrose molarity would result isotonic, meaning the concentrations stayed the same. -, 0.4 is same, + ----- the solutions would reach equilibrium somewhere between the different concentrations, they become inversely proportional because whenever the sucrose molarity inside the bag is more concentrated, it will become more dilute and vice versa ----- If each bag were placed in .4M sucrose, bags with molarities of sucrose less than .4 would lose mass while those with more would gain mass. This is due to a desire for equilibriuman equal distribution of sucrose molecules on either side of the membrane.

If a potato is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why?

The water potential of the potato would decrease because water moves from a high water potential region to a low potential region, and a dehydrated potato cell is hypertonic in comparison with the environment, forcing water to come into the cell. The moving in part shows that the potato cell had a low water potential.

A biology student could have forgotten to rinse the dialysis tube after tying it closed.

This would have left some sucrose solution on the outside of the tube, and decreased the difference in tonicity from the water in the cup to the solution in the tube. Less water would have moved into the tube, and the % change in mass would have been less.

A student could have forgotten to leave room in the dialysis bag for water to flow in before tying the second knot.

This would result in a lower percent change in mass because less water will be able to flow into the bag.

Why did you calculate the % change in mass rather than simply using the change in mass?

When using percentages of masses, it allows the mass numbers to be compared accurately and results in universal scale that can disregard the initial mass of the beaker, letting the lab in general easier to analyze. ----- each group began with different amounts of solution for their initial mass meaning the real results are less accurate so the percent gives the exact difference and includes the quantities of solution ----- Since the initial mass of each bag was not the same, differences alone would not reflect any sort of pattern. ----- Because this is the only way you can compare different potato chips in the osmosis experiment. Absolute increase will be meaningless for comparison. An extreme example - a 10 gram chip increases by 2 gram; a 5 gram chip increase by 1 gram. If you plotted actual increase in weight there is a difference between 1 and 2 gram increase. This is not valid since the 2 gram increase was twice as haevy to start with. In fact the % increase is the same for both - 1/5 & 2/10 = 20%. Also by taking % increases you remove the error in the different chips in the different solutions not being exactly the same weight. ----- Percents are a useful way to compare values that are not equal. Absolute changes are only relevant when comparing two samples of equal mass, using percent change removes the requirement that they be the same. ----- it is the only way to compare different dialysis bag in osmosis and the percent increase remove the error in different bag in different solutions of not being exactly the same weight. Absolute increase would be meaningless for comparison (graphing). ----- Because if something weighs 1g and changes by 1g, that's really important. If it ways 1kg and changes by 1g, that's much less important. As such, percentage mass is always used to show how big the change was, relative to the existing mass. ----- Percent change in mass was calculated rather than using change in mass because the differences in mass do not deal with the proportional aspect of the solutions. The percent is calculated to give the exact differences ----- The differences in mass don't deal with the proportional aspect of the solutions, making the real results less accurate. The percent was calculated to give the exact difference, along with considering the quantities of solution. ----- In the data collected, the percent change in mass was calculated to show how greatly the mass increased or decreased. The difference in mass is not enough to go by because the initial masses of the dialysis bags were not all the same. ----- Each group began with different amounts of solution for their initial mass. Therefore, results cannot be based on those numbers. The differences in mass don't deal with the proportional aspect of the solutions, making the real results less accurate. The percent was calculated to give the exact difference, along with considering the quantities of solution. ----- different volumes

Before the distillation bag was put into the solution, the individual drying the bag could have hammock-ed the bag too much TERRIBLE

caused the mass of the solution to decrease which would later lower the percent change of the mass of the bag

lab B graph

direct linear relationship, use 2SEM title: Percent Change in Mass of Dialysis Bags of Varying Sucrose Concentrations in Distilled Water x-axis: sucrose concentration (M) y-axis: % change in mass of dialysis bag positive correlation, meaning that the higher the molarity of sucrose the higher the percent change in mass within the dialysis bags

lab C graph

direct linear relationship, use 2SEM title: Percent Change in Mass of Potatoes in Varying Sucrose Concentrations x-axis: sucrose concentration (M) y-axis: % change in mass of potato cores positive correlation, meaning that the higher the molarity of sucrose the higher the percent change in mass within the dialysis bags

hypertonic

higher solute concentration than another solution

hypotonic

lower solute concentration than another solution

forgetten to hammock the dialysis bags before massing them before the experiment

make the intial mass of the dialysis bag too large and would lower the % gain in mass

The dialysis bag while being transferred from the pool of water it came from to the cup for the osmosis test could have been left out for a longer period of time than the other dialysis bags,

meaning that the tube dried out an lost more mass that it should have.

diffusion

movement from area of high concentration to low

osmosis

special case of diffusion w/ 2 characteristics: 1) diffusion of water 2) across a semi-permeable membrane - lower solute concentration to a higher solute concentration

solvent

substance doing the dissolving (water)

solute

substance to be dissolved (sugar/sucrose)

student neglected to wash hands before handling the dialysis tubes

the oil on their hands may have obstructed the small holes on the dialysis bag, preventing the water to be able to properly osmose, and the end mass would be smaller than it should be ----- This would mean that the oil or other substances on their fingers would be on their fingers when they handle the dialysis tubes and clog it. This would make it harder for water to travel across the dialysis tube, resulting in a lower percentage gain in mass.

isotonic

two solutions equal in solute concentration

water potential

water moves from an area of high water potential (higher free energy; more water molecules) to an area of lower water potential (fewer water molecules) affected by: - solute dissolved in solution (lowers water potential) - pressure on solution (raises water potential)

failed to completely submerge the dialysis bag, leaving it floating with air bubbles

would decrease the amount of surface area for water to diffuse across ---> making the final mass not as large as it should have been, and thus, decreasing the % mass change of the dialysis bag