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Experiment To determine the rate of osmosis using different concentration of solutes
Introduction
Plants live due to nutrients, mineral salts and water that they obtain from the ground. These materials are obtained to the plant through osmosis, diffusion and active transport. The mineral salts are obtained through active transport. Water gets into the plant by means of the roots by a process known as osmosis. The water moves from a region of high concentration to low concentration. They pass through cell membrane which regulates the amount of water that is needed at each time. The plant need to have enough salt and water to for its survival. The excess water and needs to be eliminated and if the salt is less, it needs to be added. The addition of water is by osmosis (process by which water moves from a region of high concentration to low concentration across a semi-permeable membrane). When there is excess water outside the cell vacuole, the plant gains water but when there is less it loses to balance the outside environment.
Objectives:
a) To determine the rate of osmosis using different concentration of solutes.
b) To find out the rates of osmosis using similar but with different concentration of the same e substance (0.5M lactose and 1M lactose).
Methods
Parameters
Water
Different Solutions with different concentrations
The following are the test solutions that were used.
TEST SOLUTION CHEMICAL FORMULA MOLECULAR WEIGHT CONCENTRATION
(M)
Distilled water H2O 18 1
Sodium chloride
salt chloride NaCl 58 0.5
Glucose reducing sugar C6H12O6 180 1.0
Lactose reducing sugar C12 H22 O11 342 0.5
Lactose reducing sugar C12 H22 O11 342 1.0
Starch C6 H12 O6 232.5 250 ml of distilled water
Tray
Test tubes
Dialysis membrane
Procedure A
a) Place water in a tray
b) Weigh the Dialysis membrane without the solution. Record this x
c) Place solutions in Dialysis membrane and record the value to be y
d) Place the Dialysis membrane and the solution in tray containing 250ml of distilled water.
Wait for 15 minutes
Record the new weight after 15 minutes. Repeat the procedure for all the solutions and then tabulate them as shown below. Table1.
Find the change in the mass and tabulate them as well.
The change is gotten by (y-x) g.
The rate of osmosis is measured by determining the rate of change in the weight of the contents of the dialysis tube.
The change is taken and the time as well.
Rate is change in mass divided by change in time taken.
RESULTS:
The results obtained are below. Table two.
.
Time/test solution 0 5 15 30 45 60 Average change in mass
Water 13.7 14 13.9 14 14 14 Tube Cumulative Change 0 +0.3 +0.2 +0.3 +0.3 +0.3 0.5M NaCl 15.3 15.5 15.6 15.6 15.7 16 Wt Cumulative Change 0 +0.2 +0.3 +0.3 +0.4 +0.7 1.0M GLUCOSE 15.6 16.1 16.5 16.8 17.7 18.8 Wt Cumulative Change 0 +0.5 +0.9 +1.2 +2.1 +3.2 0.5M LACTOSE 17.0 17.7 18.2 18.5 20.2 22.1 Wt Cumulative Change 0 +0.7 +1.2 +1.5 +3.2 +5.1 1.0M LACTOSE 17.1 19.3 20.2 20.9 20.9 26.4 Wt Cumulative Change 0 +2.2 +3.1 +3.8 +5.8 +9.3 Starch suspension 15.7 15.8 15.8 15.8 15.8 16.1 Wt Cumulative Change 0 +0.1 +0.1 +0.1 +0.1 +0.4 Chemical test
Substance Procedure Results Deductions Average change in each test tube
Test for Chloride (Cl-) Add 2 drops of silver nitrate (AgNO3) the test tubes A whitish-grey precipitate (Cl-) present Test for Glucose Add about 1ml of Benedict’s reagent to the test tubes and place the tubes in a boiling water bath for 1 minute. color change from blue to yellow Presence of reducing sugar Test for Lactose Add about 1ml of Benedict’s reagent to the test tubes and place the tubes in a boiling water bath for 1 minute. color change from blue to greenish-yellow Presence of reducing sugar Test for starch Add several drops of potassium iodide-iodine (IKI) solution to the tubes blue-black color Starch present DISCUSSION
From the result obtained it indicates that there is a gradual increase in the masses and this can be the indication o f osmosis. Water moved from the tray through Dialysis membrane and into the solution. This increased the mass. The possible errors can be with the accuracy of the measurement instrument. It can only measure to one decimal place and this does not indicate the best results. The further improvement should be done on the measurement. A more sensitive machine should be used.
The vacuole of plant cells contains a variety of substances dissolved in water (eg. salts, pigments, amino acids) but they do not move out of the cell vacuole due to their sizes. They are large but the pores are too small to permit them to pass. This ensures that the plants do not lose them and become hypotonic. The mineral salt and Cholesterol as an example enter the vacuole of the cell by the means of active transport. Energy is needed and this is carried out by carriers which move them across the membrane. The energy is provided by the mitochondria. This indicates their importance in the cell.
When animal or plant cells are put into an environment that has higher concentration of water in the cell in relation to that outside, the cell will lose water through osmosis and it becomes plasmolysed. If the cells are placed in more concentrated environment than its contents, then it will gain water and bust (for animal cells alone) and the condition called hymolysis. Plant cells have rigid cell wall that prevents them from busting. This is essential for water plant where the cell contents are always hypertonic to the water outside. These two are dangerous to the affected cells. Some cells are in surroundings where water movement in and out is equivalent. If such occurs then we say the cells are in “osmotic equilibrium” and the cell experiences no dangerous conditions. In the balance of salt and water in the cells, the conditions outside and inside are considered by the cell. If the condition outside has more water than inside the cell, then the cell gains water. In this we say the cell is hypertonic and if the condition outside has less water, then the cell loses water and the condition of the cell compared to the environment is called hypotonic. If the two have equal percentages of water, then the cell neither loses nor gain water and the condition is referred to as isotonic.
When the solutions with higher concentrations were placed, the result obtained indicated different rates of osmosis. The more concentrated ones showed a greater rater than the less concentrated ones. This indicates that osmosis depends on the concentration of the solutions
The control in this experiment was the tube containing distilled water only. No net change in the weight of the tube would be expected because water molecules would be entering and leaving the tube at the same rate. IF any weight change did occur it must be due to some experimental error.
Report
There is uniform trend of rise across all excerpts with water. The masses increases gradually and then stops after some time. All the data obtained shows a positive increment. This can also be obtained from the graph.
The following test confirmed that it was only water that moved. It confirmed the presence of the salts.
Discussions
From the result it is evident that 0.5M lactose tube gained water faster than 0.5M NaCl tube. This can be because of the differences in the particle. Though the two have similar concentrations but it can be assumed that the cause is the differences in mass in the tubes. Lactose has 17.1 but Sodium Chloride has 15.6.
The tube that contained 1M lactose gained water faster than the tube containing 0.5M lactose. The rate of gain or loss of water depends on the concentration of the solute. 0.5M lactose had a lower concentration compared to 1M lactose. The higher concentration in 1M lactose facilitated the faster osmosis.
All the tubes increased in weight. The tube containing water shored unexpected results it rose from 13.7 to 14 then fell back to 3.9. This can be an experimental error. When placed plastic tray containing water, they were hypertonic to then water. In the balance of the two environments, the water moved from the dish to the tubes and this caused an increase in mass.
Part b
The result is graphed and the following figure obtained.
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