CBSE Class 9 Science Practical Skills – Osmosis in Raisins

EXPERIMENT

AIM
To determine the mass percentage of water imbibed by raisins.

MATERIALS REQUIRED
A few raisins with stalk intact, blotting paper, weight box, physical balance, petri dish.

THEORY
Raisins are dehydrated grapes. These when kept in water swell up. This happens due to imbibition of water and endosmosis.
Imbibition is the adsorption of water or any other liquid by the solid particles of a substance causing it to increase in volume without forming a solution.
The solid particles which adsorb water or any other liquid are called imbibants. The liquid which is imbibed is called imbibate. The phenomenon of imbibition occurs due to presence of hydrophilic colloids. Proteins, starch and cellulose are some examples of hydrophilic substances. Different substances have different imbibing capacity. Proteins have very high imbibing capacity, starch has less imbibing capacity and cellulose is the weakest imbiber. Because of difference in imbibing capacities, proteinaceus pea seeds swell more on imbibition than starchy wheat seeds.

PROCEDURE

  1. Select a few raisins with intact stalks and weigh them.
  2. Now place these raisins in petri dish filled with water and cover it.
  3. Keep the set-up undisturbed overnight.
  4. Take out the swollen raisins from the petri dish and place them on a dry blotting paper to soak away the extra water.
  5. Weigh the swollen raisins and note their weight.
    CBSE Class 9 Science Practical Skills – Osmosis in Raisins 1

OBSERVATIONS AND CALCULATIONS

  1. Weight of dry raisins taken =x1 g
  2. Weight of swollen raisins =x2 g
  3. Weight of water imbibed by raisins = (x2 – x1 ) g = x g
  4. Percentage of water imbibed by raisins = \(\frac { X }{ { X }_{ 1 } }\) x 100 =………… %

RESULT
The percentage of water imbibed by raisins =……….. %

PRECAUTIONS

  1. Raisins should be with intact stalks.
  2. Proper care should be taken while weighing is done.
  3. Sufficient water should be taken in petri dish for soaking raisins.
  4. Before taking final weight, raisins should be dried with blotting paper to remove excess water.

INTERACTIVE SESSION

Examiner :
Define osmosis.
Examinee:
The movement of water from a low concentration solution to a high concentration solution through a semipermeable membrane is called osmosis.

Examiner :
What are the two types of osmosis?
Examinee:
Osmosis is of two types-

  1. Exosmosis
  2. Endosmosis.

Examiner :
What is exosmosis?
Examinee:
The outward movement of water from a cell through a semipermeable membrane is called exosmosis.

Examiner :
What would happen if grapes are placed in a concentrated sugar solution?
Examinee:
Exosmosis happens leading to shrinkage of grapes.

Examiner :
What is endosmosis?
Examinee:
The inward movement of water in a cell through a semipermeable membrane is called endosmosis.

Examiner :
Why do raisins swell up in water?
Examinee:
Raisins (e.g., a dried grape) swell because the water is imbibed in the raisins.

Examiner :
What is imbibition?
Examinee:
Imbibition is absorption of water by the solid particles of an adsorbent causing it to increase in volume without forming the solution.

Examiner :
State the factors that affect imbibition.
Examinee:
Factors include:

  1. Temperature
  2. Pressure
  3. pH
  4. Tonicity.

Examiner :
Name the constituents in plant cell that cause imbibition.
Examinee:
Cellulose, lignin, pectin, starch and proteins.

Examiner :
Name the first step hivolved in the germination of seeds.
Examinee:
Imbibition of water.

Examiner :
What is tonicity?
Examinee:
Tonicity is the phenomenon which deals with the external concentration.

Examiner :
Differentiate between hypertonic and hypotonic solution.
Examinee:
Hypertonic solution is an external solution whose concentration is more than the internal concentration of cell, whereas concentration is less than the internal concentration of cell of a Hypotonic solution.

Examiner :
Name the type of solution whose concentration is same as compared to the internal composition of cell.
Examinee:
Isotonic solution.

Examiner :
What do you mean by differentially permeable membrane?
Examinee:
Differentially permeable membrane allows selective passage of solutes through them.

Examiner :
What is diffusion?
Examinee:
The movement of molecules, atoms of gases, liquids and solids without any semipermeable membrane from high concentration to low concentration is diffusion.

Examiner :
How is diffusion different from osmosis?
Examinee:
In diffusion, the movement of a substance is from the place of its high concentration to an area of its low concentration that does not require a semi-permeable membrane, whereas osmosis is the movement of water from its high concentration to its low concentration that requires a semi-permeable membrane.

Examiner :
Define plasmolysis.
Examinee:
Shrinkage of protoplasm of a cell from its cell wall under the influence of a hypertonic solution is called plasmolysis.

Examiner :
What will happen to weeds if excess salts are added to weeds in lawn tennis ground?
Examinee:
The cells of weeds undergo plasmolysis and consequent death of weeds.

Examiner :
What is deplasmolysis?
Examinee:
The swelling up of a plasmolysed protoplast under the influence of hypotonic solution is called deplasmolysis.

NCERT LAB MANUAL QUESTIONS

Question 1:
Will a piece of iron also swell when it is kept in water? Justify your answer.
Answer:
No, iron does not swell but rather undergoes rusting in water.

Question 2:
Have you experienced difficulty in closing wooden doors or windows during rainy seasons? Give a suitable explanation.
Answer:
During rainy season, imbibition of water from atmosphere makes the doors/windows swell up.
Question 3:
Suggest an experiment by which the swollen raisins can be shrunk again.
Answer:
Place the swollen raisins in 10% salt solution.
Question 4:
In some plants seed coats are very hard and thick. How do they break before seeds germinate?
Answer:
The process of imbibition leads to breakdown of seed coats when soaked in water.

Question 5:
What will happen to the shape of a grape when it is placed in a viscous sugar solution?
Answer:
Grape will shrink.

Question 6:
What is the effect of temperature on the rate of imbibition?
Answer:
Imbibition increases.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
A student weighed 108 raisins and designated the weight as A. She then soaked them in 50 ml distilled water in a beaker. After 2 hours, she removed the raisins wiped them dry from outside and weighed again and called that weight as B. For determining the percentage of water imbibed by raisins, she should calculate as follows
(a)\(\frac { B-A }{ A }\)
(b)\(\frac { B-A }{ B }\) x 100
(c)\(\frac { B-A }{ A } x\frac { 1 }{ 100 }\)
(d)(B-A)x100

Question 2:
Raisins swell up after being placed them in a beaker containing water for sometime because i
(а) the concentration of water in the cell sap is higher than the water in the beaker.
(b) the concentration of water in the cell sap is lower than the water in the beaker.
(c) the concentration of water in the cell sap is the same as that of water in the beaker.
(d) water inside the raisins passes out of them when placed in a beaker of water.

Question 3:
Raisins selected for the experiment should
(a) have intact stalks.
(6) be swollen raisins.
(c) be without stalks
(d) none of these.

Question 4:
Hypertonic solution as compared to hypotonic solution will have
(a) less solute.
(6) same solute.
(c) more solute.
(d) equal solvent.

Question 5:
Which of the following component facilitates imbibition in a plant cell?
(a) Cellulose
(b) Pectin
(c) Lignin
(d) All of these

Question 6:
Rate of osmosis is affected by rise in temperature as
(a) osmosis increases with increase in temperature.
(b) osmosis decreases with increase in temperature.
(c) rate of osmosis remains same.
(d) none of the above.

Question 7:
The water imbibed by raisins is calculated as
(a) weight of wet raisin – weight of dry raisin.
(b) weight of dry raisin – weight of wet raisin.
(c) weight of wet raisin – weight of wet raisin.
(d) weight of wet raisin + weight of dry raisin.

Question 8:
Entry of water in a cell depends on
(a) solute concentration.
(b) tonicity.
(c) temperature.
(d) all of these

Question 9:
For the experiment, “To show imbibition of water”, raisins should be immersed into water
(a) partially.
(b) completely.
(c) both of the above.
(d) none of these.

Question 10:
While performing the experiment with raisins to determine the percentage of water imbibed by them, a student recorded the following data
Weight of water in the beaker = 50 g
Weight of raisins before soaking = 20 g
Weight of raisins after soaking for one hour = 30 g
Weight of water left in the beaker after the experiment was over = 40 g
The percentage of water imbibed by raisins is:
(a) 10
(b) 20
(c) 45
(d) 50

Question 11:
If the water imbibed by soaked raisins is 50% then the weight of raisins before soaking in water (W1) and the weight of raisins after soaking in water (W2) might have been
(а) W1= 30 g and W2= 45 g
(B)W1= 30 g and W2 = 50 g
(c)W1= 50 g and W2= 100 g
(d)W1= 0.5 g and W2= 25 g

Question 12:
A student dissolved 5 g of sugar in 100 mL of distilled water in beaker A. Then she dissolved 100 g of sugar in 100 mL of distilled water in beaker B. After that she dropped a few raisins of equal weight in each beaker. After two hours she found the raisins in A swollen and those in B shrunken. The inference drawn is that
(a) sugar concentration of raisins is lower than that of solution A and higher than that of solution B.
(b) sugar concentration of raisins is higher than that of solution A and lower than that of solution B.
(c) in B the cell membranes of raisins were damaged resulting in leaching.
(d) in A the permeability of water of the cell membranes of raisins was enhanced.

Question 13:
Each of the three beakers ‘A’, ‘B’ and ‘C’ contained 50 ml of distilled water. A student placed five raisins in each beaker. The raisins for each beaker weighed the same. The beakers were kept at room temperature. The raisins were removed from beaker. ‘A’ after 10 minutes, from beaker ‘B’ after 20 minutes and from beaker ‘C’ after one hour. On calculating the percentage of water absorbed by the raisins, it was found that
(a) maximum absorption of water by raisins was in beaker ‘C
(b) maximum absorption of water by raisins was in beaker ‘B
(c) minimum absorption of water was by raisins in beaker ‘C’.
(d) absorption of water was equal in raisins of all the three beakers.

Question 14:
A student soaked 5 raisins each of equal weight in three beakers A, B and C each containing 100 ml of distilled water and left them at room temperature. After 10 minutes he removed the raisins from beaker A, after 20 minutes from beaker B and after one hour from beaker C. He calculated the percentage absorption of water in each case. If PA, PB and Pc denote per cent
absorption, then which of the following is correct?
(o) PA = PB = Pc
(b) PA < PB > Pc
(c) PA > PB < Pc
(d) PA < PB < Pc

Question 15:
A student soaked 5 g of raisins in beaker (A) containing 25 mL of ice-chilled water and another 5 g of raisins in beaker (B) containing 25 mL of tap water at room tempe-rature. After one hour the student observed that
(a) water absorbed by raisins in beaker (A) was more than that absorbed by raisins of beaker (B).
(b) water absorbed by raisins in beaker (B) was more than that absorbed by raisins of beaker (A).
(c) the amount of water absorbed by the raisins of both beakers (A) and (B) was equal.
(d) no water was absorbed by raisins in either of the beakers (A) and (B).

ANSWER KEY
Multiple Choice Questions/VSA

  1. (a)
  2. (b)
  3. (a)
  4. (c)
  5. (d)
  6. (a)
  7. (a)
  8. (d)
  9. (b)
  10. (d)
  11. (a)
  12. (b)
  13. (a)
  14. (d)
  15. (b)

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