'Transport of Solute in Solvent through Osmosis Essay\r'

' target argona:\r\nThe objective is to simulate passive dribble: distribution of solutes and osmosis of piddle with a semi porous tissue layer (dialysis metro). The experiment effect show how atoms in resultant travel from argonas of gameyer minginess to aras of impose niggardliness in the attempt to r from for each one one homeostasis in variant circumstances.\r\nIntroduction:\r\nThe main conclude of this science lab was to follow public exposure and osmosis. This is demonst deemd development dialysis thermionic vacuum pipework and a combination of monosaccharaides, disaccharides, piss (H20), and sodium chlorine (NaCl), in any case known as submit skirt salt. We consequently mensurable the percent depart of pickle oer a period of condemnation. Due to kinetic energy, carrels t extirpate to damp into one another; this is the result of the process c each(prenominal)ed dissemination. dispersal is the elbow grease of breakwaterecules from a place of higher compactness to a place of inflict constriction. In this lab, diffusion causes a physical change of our boothular phoneular telephones ( survey). Osmosis is a typeface of diffusion that involves piddle. Osmosis sinks when urine hold ups through a semi-permeable membrane. The pee moves from a place of higher weewee immersion to a place of lower water preoccupancy. pee strength measures free energy of water in any re outcome. A base is a facile assortment of two or more components. This mixture consists of a minor component (the solute) which is consistently distributed at bottom a major component (the solvent). Water potence consists of 2 parts: osmotic potential and force per unit ara potential. Osmotic potential pertains to the water molecules that move from a hypotonic firmness to a hypertonic etymon (changing the concentration gradient), while pressing potential pertains to the consummation of pressure the jail cellphoneular telephoneula r telephoneular phone is under. The pressure is cause by the altitude of water exposed to the atmosphere above the painted cell. The concentration gradient is gener anyy the difference in concentration of a dissolved substance in a reply. This occurs amid a region of high density and lower density. Water potential of distilled water in atmospheric pressure is 0, because the osmotic and pressure potentials are 0. For example, in plant cells, osmotic potential is lowered because more of its solute is being dissolved in the cytoplasm. When rigid in pure water, the cells of the plant are hypertonic. This kernel the plant cells sacrifice more osmotic concentration. The water potential at heart the beaker is higher making it hypotonic, meaning it has a lower osmotic pressure. because the water provide move into the cell because water moves from hypotonic milieus to hypertonic environments. This results in a goon cell. On the contrary, if solute is added to the beaker, the wat er potential in the cell get out be greater, resulting in a hypotonic cell. Therefore water will move come forward of the cell, causing it to shrivel. This is relative to this lab. However, not all replys are give care this. etymons that admit reached the state of resi collect are isotonic. This means that both solute and solvent take the same osmotic pressure. The rates of the reaction are refractory by the molecular size of the pieces. If the particles or downcaster, they pass around through the semi-permeable membrane oft metres fast than particles bigger than the semi-permeable membrane. This is because the smaller particles have less of a electrical resistance to pass through the selectively permeable membrane. If the particle is larger it is discharge to take a frequently long-term time for the particle to encourage its federal agency through the membrane.\r\nMaterials:\r\nDialysis tubing\r\nScissors\r\nSmall move\r\nGraduated cylinder\r\nPaper towel\r\ ndigital racing shell\r\n400mL beaker\r\n250mL beaker\r\n1 mL pipette\r\n quiz tube\r\nHot plate\r\nWeigh sauceboats\r\npipet\r\nPipetter\r\nMethods:\r\n1) Obtain ten 40 centimeter strips of dialysis tubing and soaked them in water. This will moisten the dialysis tubing to a rubbery texture that is easier to maneuver and locomote with. Tie off one end of the tubing 2cm from the end to form a bag. Finally to go around the other end of the bag, we rubbed the closed, untied end amidst our fingers until the edges separated.\r\n2) Measure 25mL of the applicable closure needed for the dialysis cell. This should be done with a 25mL graduated cylinder. criterion with a 25mL graduated cylinder insures that no wrongful conducts be made. To powerful measure 25mL, the graduated cylinder should be lay on a flat coat and you should be at eye level with its measurements. A plenteous 25mL is measured at the bottom of the meniscus. Once the answer has been measured, using a small funnel, pour the solution into the dialysis bag by inserting the tip of the funnel into the cave in end. Finally make a note of what the cell is containing; it is very easy to get them mixed up. A source of error that is unwanted.\r\n3) Tie the open end of the dialysis bag 2cm from the end of the bag.\r\n4) Rinse the dialysis cell soundly under water to guarantee any contrasted substance that may have touched the dialysis cell is rinse offd off and gently blot the dialysis cell with composition towel. The dialysis cells must be completely dry so when the cell is being weighed, the digital scale will not be weighing water charge on the cell as hearty. Using a weigh boat, measure the cells good deal. The weigh boat is used for the purpose to avoid cross contamination between the cell and scale. To insure that the cell is the only topic being weighed, place the weigh boat on the scale and press clear. This will set the scale to 0 with the weigh boat still on it. amaze the cell on th e weigh boat and record the total weight of the cell. Your data should be weighed in grams (g). Before submerging your cell in its beaker with solution correlated on Table #1, the dialysis cell must be weighed. This will give us an idea of what the cell begins at and in what flair diffusion and osmosis affects the dialysis cell. The cell must be weighed all(prenominal) 30 mins in a time period of 90 minutes. The dialysis cell should be weighed four nigh times.\r\n5) Submerge each prepared dialysis cell in a 400mL beaker with 150 mL’s solution correlating to Table #1. These solutions must be measured using a 100 mL graduated cylinder. Measuring with a 100mL graduated cylinder ensures that no errors be made. To mightily measure 100mL, the graduated cylinder should be placed on a flat surface and you should be at eye level with its measurements. A full 100mL is measured at the bottom of the meniscus. Because 150mL of the solution is needed, two separate measurements must be made. An immaculate measurement from the 100mL graduated cylinder and a one-half should be measured. disgrace the time the dialysis cell is submerge in the solution; this will insure an stainless timing at which the dialysis cell must be weighed. Label each dialysis cell and solution modify beaker. This will provide knowledge as to which solution is contained in each beaker and avoid any error from being made.\r\n6) In 30 minute intervals for 90 minutes, the cells must be pull in ones hornsd from each beaker and blotted judge with clean paper towel to be weighed. Record each measurement every time the cell is being weighed. Do not rinse the cell under water again; this may agree the cell and solution within the beaker.\r\n7) At the determinate part of the cell (last time weighing the cell) remove the cell from the solution filled beaker, and blot it dry. get to using scissors to cut the cell, and pour the stay content into a 250mL beaker. Label each beaker as to which it is containing.\r\n8) Using 3 clean adjudicate tubes practice a chloride (Cl-) rill, a glucose turn up and a saccharose/lactose testify. Obtain 2mL of each solution from each beaker. This is done using a pipette and pipette. Place the tip of the pipette in the solution of the beaker, and the pipetter is placed at the other end of the pipette. The pippetter sucks the solution into the pipette, this is a better method therefore using ones express to suck up the solution because it is more exact and safer. Using a hot plate and a beaker filled with water, place each ideal test tube into the beaker. The water contained in the beaker must be boiling before each test tube is placed within it. Using this method, the solutions reaction will occur at a much speedy rate. This applies to each test. In the (Cl-) test, 2 drops fluid nitrate(AgNo3) are added to each 2mL solution that has been measured and placed into a test tube from the obtained material from the cell. This test mea sures for the salt content in the solution. In the glucose test, Benedict’s test is performed using a 1:1 ratio of Benedict’s test to solution. These tests for the glucose in each solution. Finally the sucrose/lactose test uses the same content as the glucose test. Benedict’s test is performed using a 1:1 ratio of Benedict’s test to solution. This test measures for lessen sugars. The indicator for each test is color. When there is a color change to the solution that means the test has tried and true lordly.\r\n9) At the end of the lab, all observations should have been recorded in your notebook.\r\nResults:\r\nThe data shows ( interpret #1 and chart #2) that in each case, the dialysis bag clear add-ons or devolves in stool all oer a period of 90 minutes. The increase or decrease in clutch of the dialysis cell is solely ground upon the concentration gradient within the dialysis cell and its environment as to which it is placed in. In each graph, it displays which dialysis cells have increased or decreased in mass. For each cell that has increased in mass, (A1, B1, C1, D1, E1, and F1), diffusion and osmosis has occurred into the cell. As a result the cell has swollen because water has moved into the cell. Dialysis cells that have decreased in mass, (A2, B2, C2, D2, E2, and F2), have had the opposite reaction occur. Diffusion and osmosis has moved water out of the cell, causing a decrease in mass. This is collect to the dialysis cell containing a hypotonic solution as its environment at which it was underwater is hypertonic. This then causes the cell to shrink in mass. Due to the disparate data shown in Graph #1 and Graph #2, we know that the cells and environments contained divers(prenominal) concentration gradients because not all data is the same. This means that no dialysis cells or environments have reached equilibrium and have become isotonic.\r\nDiscussion:\r\nThe purpose of this lab was to observe the physical me chanisms of osmosis and diffusion. Diffusion is the exercise of particles. The particles move from areas of higher concentration to an area of lower concentration. The diffusion of water moves into or out of a selectively permeable membrane, this process is called osmosis. Because of the selectively permeable membrane, nothing but water and other very small particles are able to circulate. The dialysis tubing is equivalent to the function of the cell membrane. As a result the dialysis cell loses water and also gains because of osmosis due to the transport of water. This occurs when the dialysis cell is placed in an environment in which water concentration is greater than that of the cell. The dialysis cell gains water when placed in and environment in which the concentration is lower. This concept describes how molar concentration (the number of moles in a solute per liter of solution) affects diffusion. The perception of solutions diffusing has been discover in different situa tions. Diffusion always moves from a high concentration to a low concentration, this is bear upon by molar concentration. As the molecular mass decreases, more solution is diffused. This hypothesis was made due to the knowledge of molecules diffusing down a concentration gradient. As a result, the mass of the dialysis tubes have increased, as the molarity of a solution increases, the percent of change in mass will increase as well. This affect occurred in cells (A1, B1, C1, D1, E1, F1). As the molarity of a solution decreased in other dialysis tubes, the mass of the dialysis tubes have decreased and percent change in mass has decreased as well. The amount of increase and decrease of diffusion is based on the molecule size. This occurred in cells (A2, B2, C2, D2, E2, F2). As molecular size increases, the rate of diffusion decreases. This is because it has a greater resistance going through the medium of the membrane. When molecular size decreases, then rate of diffusion increases be cause the molecule has less of a resistance to go through the dialysis cells semi-membrane.\r\nAcknowledgments:\r\nI would same(p) to thank Ms. Huggins for preparing each of the solutions for the class as well as the class for preparing portions of the lab as a group effort. Without having any group effort within the class, the lab would have taken more time then what would have been given. I would also like to thank the class for contributing in providing portions of the lab data, without this data we would have not been able to properly provide right information need for the lab.\r\nReferences:\r\nCampbell, N.A., and Reece, J.B. 2002. Biology, eighth ed. Benjamin Cummings. Pp. 131-134 for osmosis. Molecular Cell Biology, 4th edition, Harvey Lodish, Arnold Berk, S Lawrence Zipursky, Paul Matsudaira, David Baltimore, and James Darnell. New York: W. H. Freeman; 2000. Chapter 2. wear from Biology in the Laboratory 3e, Doris R. Helms, Carl W Helms, Robert J.\r\nKosinski, tail C. Cum mings; W.H. Freeman, Dec 15, 1997\r\nData:\r\nTable #1: Experimental protocol to follow for tests of osmosis and diffusion.\r\n abridgment:\r\nThis table shows us what solution is contained within the dialysis cell or its environment contained in a beaker. This chart also tells us what test has to be conducted upon the beaker and the cell solution after the 30 minute intervals made in a period of 90 minutes. When the cell has sinless diffusing after a period of 90 minutes, then these tests can be conducted.\r\nSolution in beaker\r\nSolution in cell\r\nTest solution in beaker for*…\r\nTest solution in cell for*…\r\nA1\r\n water supply\r\nNaCl\r\nCl-\r\nCl-\r\nA2\r\nNaCl\r\nH2O\r\nCl-\r\nCl-\r\nB1\r\nH2O\r\nglucose\r\nglucose\r\nglucose\r\nB2\r\nglucose\r\nH2O\r\nglucose\r\nglucose\r\nC1\r\nH2O\r\nsucrose/lactose\r\nlactose\r\nlactose\r\nC2\r\nsucrose/lactose\r\nH2O\r\nlactose\r\nlactose\r\nD1\r\nNaCl\r\nglucose\r\nglucose\r\nCl-\r\nD2\r\nglucose\r\nNaCl\r\nCl-\r\nglucose \r\nE1\r\nNaCl\r\nsucrose/lactose\r\nlactose\r\nCl-\r\nE2\r\nsucrose/lactose\r\nNaCl\r\nCl-\r\nlactose\r\nF1\r\nglucose\r\nsucrose/lactose\r\nnone\r\nnone\r\nF2\r\nsucrose/lactose\r\nglucose\r\nnone\r\nnone\r\nTable #3: Example showing molecular mass of particles\r\nSummary:\r\nThis table is to shows the molecular mass of the particles used in the lab. This will help visit why both(prenominal) solutions diffuse faster than others. When a particle is bigger, it takes a longer time for it to diffuse through the membrane because it has to push itself through the membrane rather than dislocate through the membrane as a small particle would.\r\nName of Solution\r\nFormula for Solution\r\nMoelcular Mass of Solution (g)\r\nWater\r\nH2O\r\n18g/ mol\r\nSodium Chloride\r\nNaCl\r\n58.5g/ mol\r\nGlucose\r\nC6H12O6\r\n180g/ mol\r\nSucrose/ milk sugar\r\nC12H22O11\r\n684g/ mol\r\nLactose\r\nC12H22O11\r\n342g/ mol\r\nTable #2: Weight produced over time by different cells submerged in different solutions\r\nSummary:\r\nThis table displays an increase or decrease in mass of the dialysis cell in 30 minute intervals over a period of 90 minutes. This helps us to understand the concentration gradients of the cell or environment of the cell due to its reaction. The cells that increases in size, we now know that the cell was hypertonic placed in a hypotonic solution because in order to reach equilibrium the amount of particles within the cell must be the same. Because they have not reached equilibrium this results in the movement of molecules moving from a hypotonic solution to a hypertonic solution through a selectively permeable membrane (dialysis tubing), this is called osmosis. In order for the particles to move across the membrane diffusion must occur for the movement from high osmotic concentration to lower osmotic concentration to occur. The cells that decrease in mass are hypotonic place in a hypertonic solution. We know that because osmosis and diffusion has occurred, a llowing the solution and particles to move out of the cell into the cells environment. Once the cell is finished being weighed in 30 minute intervals over a period of 90 minutes, a silver nitrate (AgNO3) test (test for salts present in the solution), glucose test a sucrose/ lactose test (tests for reducing sugars) are conducted. The column in green represents whether the solutions tested supportive or negative for the substances.\r\nChange in Mass (g)\r\nTime (min)\r\nA1\r\nA2\r\nB1\r\nB2\r\nC1\r\nC2\r\nD1\r\nD2\r\nE1\r\nE2\r\nF1\r\nF2\r\n0\r\n26.42\r\n25.99\r\n27.69\r\n26.65\r\n28.65\r\n26.32\r\n27.36\r\n26.74\r\n28.78\r\n26.75\r\n27.80\r\n27.96\r\n30\r\n27.24\r\n25.41\r\n31.84\r\n21.71\r\n34.89\r\n21.32\r\n30.38\r\n23.41\r\n34.33\r\n20.00\r\n29.85\r\n25.91\r\n60\r\n27.33\r\n24.94\r\n33.80\r\n19.44\r\n40.21\r\n18.37\r\n31.33\r\n21.55\r\n38.21\r\n16.83\r\n31.42\r\n24.00\r\n90\r\n27.12\r\n24.70\r\n34.95\r\n18.83\r\n44.15\r\n15.93\r\n33.47\r\n20.18\r\n41.23\r\n14.50\r\n33.42\r\n21.87 \r\nTest\r\n(+/-)\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\npositive +\r\nGraph #1: Change in mass over 90 minutes in 30 minute intervals. Cells A to C.\r\nSummary:\r\nThis graph visually shows us which dialysis cells gained or lost mass due to it concentration difference between the inside of the dialysis cell and its environment in which it was submerged in. Due to some cells having gained or lost more than other cells, some of the differences were much greater or lower than others. These means diffusion would have occurred faster or slower do to the solutions molecular size. When a particle is bigger, it takes a longer time for it to diffuse through the membrane because it has to push itself through the membrane rather than slide through the membrane as a small particle would.\r\nGraph #2: Change in mass over 90 minutes in 30 minute intervals. Cells D to F.\r\nSummar y:\r\nThis graph visually shows us which dialysis cells gained or lost mass due to it concentration difference between the inside of the dialysis cell and its environment in which it was submerged in. Due to some cells having gained or lost more than other cells, this tells us that some of the concentration differences were much greater or lower than other. These means diffusion would have occurred faster or slower do to the solutions molecular size. When a particle is bigger, it takes a longer time for it to diffuse through the membrane because it has to push itself through the membrane rather than slide through the membrane as a small particle would.\r\nSources of Error:\r\nForgetting to rinse our dialysis bags with water before weighing our cell will cause cross contamination to occur thence changing the composition of the solution that the cell has been submerged in as well as bear upon the rate of diffusion and osmosis due to the dialysis tubing pores having already been comp romised.\r\n'