Science Practical Record

CLASS : X

Biology, Physics & Chemistry

Instruction to write Practical Record:
1.Preparing a report on each practical performed in the laboratory and maintaining a record of
the work done is an essential requirement. The report on each experiment should be such that
it informs all steps involved in performing the experiment and the result obtained. It is therefore
imperative that the report on an experiment should be presented under different headings so
that it is easily understood.
2.. On Lining page you have to write
Aim ,Material required, Theory and formula used, Procedure, Result, Precautions & Sources of
error.
3. On Plain Page you have to write:
Diagram, Observation table, Graph, Calculation graphical, Result graphical

4. Each experiment should be written in a new page

BIOLOGY
Experiment:1
Aim
To prepare a temporary mount of a leaf peel to show the stomata of a leaf.
Theory
Plants are the primary producers. They carry out physiological processes such as
photosynthesis and respiration which requires a gas exchange between the tissues of
plants and the atmosphere. This process is carried out through tiny openings located in
leaves, known as stomata.
Structure of stomata.
Stomata are small elliptical openings on leaves that contain chloroplasts. They are
girdled by two-kidney shaped cells known as guard cells on either side of the stomata.
The guard cells possess a thick inner wall and a thin outer covering which control the
closing and opening of the pores of stomata.
Closing and opening of the stomata.
Turgidity of the guard cells causes the stomata to open while the flaccid nature of the
guard cells causes the stomata to close.
Material Required
 Leaf
 Needles
 Forceps
 Watch glass
 Dropper
 Glass slides
 A brush
 Coverslips
  Blotting paper
 Safranin
 Glycerine
 Compound microscope

Procedure
 Pick a healthy leaf from the potted plant
 Fold the leaf to gently pull the peel apart to separate a peeled section from the lower surface
of the leaf. Use the forceps to perform this step. Allow the peel to remain in a watch glass
holding water for some time.
 In the watch glass, stain the sample by adding some drops of safranin through a dropper.
 Take the peel out after 2-3 minutes. Set it on a clear glass slide
 Add a drop of glycerin on the peel. Put a clear coverslip over it gently using a needle.
 Excess glycerin and stain can be removed using blotting paper
 Examine the slide first under a low-power and then under a high-power magnification of a
compound microscope.

Observation
 Visible epidermal cells. The cells in their outline are irregular with no intercellular spaces
 Small openings, stomata are scattered through the epidermal cells
 Guard cells are observed which have chloroplasts and nucleus
 Guard cells are observed having a thin outer covering and a thick inner boundary(concave)
 Guard cells control the closing and opening of the stomata.

Conclusions
Epidermal cells are found containing stomata on the lower surfaces of the leaf.
Precautions
 Avoid folding the leaf too much. The peel should be snipped to a proper size
 The peel should always be placed at the centre of the slide and the slides should be held
from the sides.
 The peel should neither be overstrained nor under strained
 A brush should be used to handle the peel, otherwise would damage cells.
 Coverslip needs to be placed in such a way that air bubbles are avoided
 Blotting paper can be used to remove excess stain
Experiment:2

Aim
To experimentally demonstrate that carbon dioxide is released during the process of
respiration.

Theory
The process of respiration is biochemically carried out wherein food, glucose to be
precise, is oxidized and energy is released. In this experiment, gram seeds (moistened)
are used. The purpose of using these seeds is that they release carbon dioxide and are
respiring actively. The released carbon dioxide is consumed by the solution of KOH.

Material Required
 Soaked gram seeds
 U-shaped delivery tube
 Conical flask
 Blotting paper (moist) /cotton wool
 Thread
 Water
 Beaker
 Test tube
 Rubber cork with a single hole
 Freshly prepared KOH solution (20%)

Procedure
 Germinate 10 seeds. This can be done by wrapping them in moist blotting paper
or cotton wool for around 3 to 4 days.
 Set up the germinated or sprouted seeds in the conical flask. Spray some water
into the flask to dampen the seeds.
 With the help of a thread, suspend the conical flask containing the test tube
having a freshly prepared 20% KOH solution.
 Use the rubber cork to seal the opening of the conical flask.
 One edge of the U-shaped glass delivery tube present in the conical flask should
be inserted through the hole in the rubber cork. The other edge should be placed
into a beaker that is saturated with water.
 All attachments of the set-up should be sealed. This can be done using vaseline
to create an air-tight environment.
 The initial water level present in the U-shaped delivery tube needs to be marked.
 Leave the experimental set-up uninterrupted for 1 to 2 hours. Observe the
fluctuations in the water level in the tube.
Diagram:

Observation
Careful observation after a certain period of time reveals that the water level in the U-
shaped delivery tube has risen in the beaker.

Conclusions
 The rise in level water indicates that carbon dioxide is released as a result of
germinating gram seeds during the process of respiration in the conical flask.
 The carbon dioxide that is released in the process is absorbed or consumed by
the KOH solution that is suspended in the test tube in the conical flask, creating a
vacuum or a void in the flask resulting in the upward water movement in the tube.
Hence, the water level in the tube changes.

Precautions
 The seeds that are to be germinated need to be moistened
 Air-tight environment for all the connections in the experimental set-up
 The KOH solution that is used needs to be freshly prepared
 Care needs to be taken to ensure that one end of the delivery tube is placed in
the conical flask. The other edge is submerged in the water of the beaker
 The tube that contains the KOH solution needs to be suspended carefully
Experiment:3

Aim
To study about (a) Binary Fission in amoeba and (b) Budding in yeast with the help of
prepared slides

Principle/Theory
Budding and Binary fission are types of asexual reproduction observed in lower
organisms such as bacteria, unicellular protozoans and some other entities.

binary fission:
In this type of reproduction, the parent cell divides or is split into two daughter cells
through mitosis wherein each daughter cell develops into an adult. Amitosis is the
division of the nucleus.

Budding:
It is a kind of asexual reproduction wherein a new organism develops from a bud or an
outgrowth due to the process of cell division at a particular site.

Material Required
 Compound microscope
 Permanent slides of budding in yeast and binary fission in amoeba

Procedure
 Place the slide under a compound microscope
 Focus the slide, first under low power and later under high power of the compound
microscope
 Various stages of budding and binary fission can be carefully examined

Diagram
Observation
(a) Binary fission in Amoeba

 Initially, the pseudopodia are retrieved. The body of amoeba is coiled and becomes round
 Amitosis is observed, the division of the nucleus takes places which are followed by splitting
of cytoplasm
 At the point of fission in the body of the amoeba, a constriction starts to develop.
 The constriction or furrow turns deeper resulting in the formation of two daughter cells

(b) Budding in yeast

 Protuberance or a tiny outgrowth is observed on the parent cell

 Division of the nucleus is observed which is later seen in the bud
 Repetitive budding leads to the formation of a chain of cells

Conclusions
The prepared slides display asexual reproduction. One individual is involved to produce
a new offspring of its own kind.

Precautions
 Slides need to be aligned and focused accurately
 Sketch out your observation that is observed under a microscope
 The slides first need to be examined under a low-power magnification of the compound
microscope and then under high-power magnification.
Experiment:4

Aim
To identify the different parts of an embryo of a dicot seed

Theory
The process of fertilization in plants leads to the formation of fruits which forms the
ripened ovary. The seed can be one or many which form the mature ovule.

A seed consists of the following parts:

 Hilum – It is a scar that is located on the seed coat, associated with the stalk of the plant
 Seed coat – Forms the exterior covering of the plant, supplying with nourishment and
protection to the seed inside
 Endosperm – It is the tissue containing nutrients for the growth of the embryo
 Embryo – Several divisions of the zygote gives rise to this structure.

On the basis of cotyledons in the seed, angiosperms can be classified into the following:

 Monocots
 Dicots

Germination

is the process wherein the micropyle present in one part of the hilum takes up water
which under favorable conditions, such as suitable light, temperature, air, etc. it uses
up for the seedlings to develop into tiny, immature embryonic plants from the seed
coats.

Germination involves the following steps:

 Seeds swell, plumules develop into shoots

 From the radicle of the seeds, the roots arise
 Formation of cotyledons(one in monocots and two in dicots)

Material Required
 Seeds of red kidney bean/gram
 Forceps
 Magnifying glass
 Cloth
 Petri dish
 Water

Procedure
 Soak a few seeds overnight
 Next morning, drain the excess water out
 Now wrap the seeds in a clean and a moist cloth for a day, allow it to dry
 Next, carefully peel the seed coat
  With the help of forceps, dissect the seed so as to get two equal halves
 Examine with the help of a magnifying glass. Carefully identify and locate different parts of
the seed
 Sketch out the interior of the seed you examined labeling all the parts as shown in the
diagram.

Diagram

Observation
 The bean seed resembles the shape of a kidney. It has a convex and a concave side
 A scar known as the hilum is observed on the slightly darker side of the concave side
 A tiny pore known as the micropyle is located just adjacent to the hilum
 The seed is enclosed by a seed coat
 The embryo possesses two distinct and large cotyledons that resemble the shape of a
kidney and are white in color
 Lateral attachment of the cotyledons to the curved embryonal axis is observed
 Radicle is examined. It is the rod-shaped and lightly protrusive lower end of the embryonal
axis that is found placed towards the micropylar end.
 The upper end of the embryonal axis exhibits the plumule
 Hypocotyl is observed which is a section of the embryo axis found in between the radicle and
adjunct of cotyledon leaves
 The epicotyl is also observed which is the section of the embryo axis between the adjunct of
cotyledon leaves and plumule

Conclusion
Three principle parts of the embryo of dicot seeds are observed, they are:

 Cotyledons
 Plumule
 Radicle

Precautions
 Care needs to be taken while dissecting the seed as it may damage the seed
 The cloth that is used to wrap the seeds needs to be moist
 PHYSICS
Experiment:5

Aim:
To Determine Focal Length of a Given Concave Mirror
Materials Required
1. A concave mirror
2. A measuring scale
3. A screen holder
4. A mirror holder

Theory:
 A spherical mirror, whose reflecting surface is curved inwards, that is, faces towards the
centre of the sphere, is called a concave mirror.
 When parallel rays of light fall on a concave mirror along its axis, the rays meet at a point
in front of the mirror and the image formed of the object is real, inverted and very small in
size.
 As the image formed by concave mirror is real it can be obtained on a screen.
 The distance between the principal axis P of the concave mirror and the focus F is the
focal length of the concave mirror. It is denoted by letter ‘f’.

Procedure
1. The distance between the selected distinct object should be more than 50 ft.
2. The concave mirror placed on the mirror stand and the distant object should be facing each
other.
3. The screen should be placed in front of the reflecting surface of the mirror. To obtain a clear,
sharp image the screen should be adjusted.
4. Using a metre scale the distance between the concave mirror and screen can be
determined. The distance is the same as the focal length of the given concave mirror.
5. Repeat the above procedure thrice to calculate the average focal length.
Experimental Setup
Observation Table
Sl.no Position of concave mirror (M) Position of screen (S) Focal length = (M-S)

cm

1 60 cm 40 cm 20 cm

2 50 cm 30 cm 20 cm

3 40 cm 20 cm 20 cm

Calculation:
mean value of the focal length of concave mirror:

f = 20+ 20+ 20 = 20 cm
3

Result
The focal length of the concave mirror is 20 cm

Precautions
1. To get a well illuminated and distinct image of the distinct object, the distant object should be
well illuminated.
2. A concave mirror should be always placed near an open window.
3. The polished surface of the concave mirror and the distinct object should be facing each
other.
4. There should not be any hurdle between the rays of light from the object and the concave
mirror.
5. The screen and the concave mirror stand should be parallel to the measuring scale.
6. The mirror holder along with the mirror should be perpendicular to the measuring scale
Experiment:6
Aim:
To determine focal length of a convex lens
Materials Required
1. A convex lens
2. A lens holder
3. A screen fixed to a stand
4. A measuring scale
Theory:
 Convex lens is is thicker in the middle and thinner at the edges. It is also called
converging lens because it converges a beam of light incident on it.

 When a parallel beam of light falls on a convex lens, the rays, after refraction converge at
a point on its other side.
 If the parallel beam of light comes from a distant object, a real, inverted image of very
small size is formed at the focus of the lens.
 The distance between the optical centre of lens ‘O’ and the principal focus ‘F’ of the lens
is called focal length of a lens, ‘f’ is the representation of focal length.
 Since the image formed by the lens is real, it can be obtained on a screen.

Procedure
1. Without disturbing the lens and screen, arrange both of them on the wooden bench.
2. Place the lens on the holder facing a distant object.
3. Place the holder with the screen on the bench.
4. The position of the screen should be such that the sharp image of the distant object is
obtained on it.
5. The difference between the position of the lens and the screen is equal to the focal length of
the given convex lens.
6. Now shift the focus to towards various other distant object and calculate the focal length of
the convex lens.

Experimental Setup
Observation Table

Sl.no Position of convex lens (L) Position of screen (S) Focal length = (L-S) cm

1 50 cm 35 cm f1 = 15cm

2 60 cm 45 cm f2 = 15cm

3 40 cm 25 cm f3 = 15cm

Calculation
mean value of the focal length of convex lens:
f = 15+ 15+ 15 = 15 cm
3

Result
focal length of the convex lens is 15 cm

Precautions
1. The placing of the convex lens should be vertical.
2. There should not be any hurdle between the rays of light from the object and the convex
lens.
3. To get a well illuminated and distinct image of the distinct object, the distant object should be
well illuminated.
4. The convex lens stand and the screen should be parallel to the measuring scale.
Experiment:7
Aim
To study the dependence of potential difference (V) across a resistor on the current (I)
passing through it and determine its resistance. Also, plot a graph between V and I.
Materials Required
1. A battery
2. An insulated copper wire
3. A key
4. An ammeter
5. A voltmeter
6. A rheostat
7. A resistor
8. A piece of sandpaper

Theory:
Ohm’s law
The potential difference V across the metallic wire is directly proportional to the current
flowing through it, provided its temperature is constant.
V ∝ I.
∴ V = IR, where R is the resistance, which is constant for a given metallic wire.
The factors affecting resistance-
Following are the factors affecting resistance:
 The nature of the resistor.
 With an increase in length, the resistance also increases. So resistance of a wire is directly
proportional to its length.
 With an increase in the cross-sectional area, the resistance decreases. So resistance is
inversely proportional to the cross-sectional area of the wire.

Circuit Diagram

Procedure
1. Arrange the devices as shown in the circuit diagram.
2. Connect the devices with the connecting wires keeping the key open.
3. The positive terminal of the battery should be connected to the positive terminal of the
ammeter.
4. Before connecting the voltmeter in the circuit, check for +ve and -ve terminals.
 5. Check for ammeter and voltmeter reading once the circuit is connected and also adjust the
slider of rheostat after inserting the key.
6. For current I and voltmeter V, record three different readings using a slider.
7. Record the observations in the observation table.
8. Using the formula R=V/I, calculate the resistance.
9. To plot the graph between V and I, take V on the x-axis and I on the y-axis.
10. For pure metals, resistance increases with an increase in temperature.

Observations:
Range of ammeters = 0A to 3 A
The least count of ammeter = 0.05 A
Range of voltmeter = 0 V to 5 V
The least count of voltmeter = 0.1 V

Observation Table:
Sl.no Current in Ampere (I) Potential difference in volts (V) Resistance in ohms
(ammeter reading) (voltmeter reading) R = V/I (Ω)

1 0.25 0.5 R1 = 2

2 0.5 1 R2=2

3 1.5 3 R3 = 2

4 2 4 R4 = 2

5 2.5 5 R5 = 2

Mean R = R1+R2+R3+R4+R5
5
= 2+2+2+2+2 = 2 Ω
5 I

Result: The resistance of the given wire is 2 Ω

V Graph
Conclusions
1. For all the Five readings, the R-value is the same and constant.
2. The ratio of potential difference V and current I is the resistance of a resistor.
3. With the help of the graph between V and I, Ohm’s law is verified as the plot is a straight line.

Precautions
1. Thick copper wires should be used as connecting wires and using sandpaper, their insulation
should be removed.
2. To avoid external resistance, the connections should be tight.
3. The current should enter from the positive terminal and exit from the negative terminal of the
ammeter and it should be connected in series with the resistor.
4. Resistor and voltmeter should be connected in parallel.
5. The least count of ammeter and voltmeter should be recorded properly.
6. When there is no current flow, the pointers of ammeter and voltmeter should be at zero.
7. To avoid unnecessary heating in the circuit, the current should be passed for a short time.
Experiment:8

Aim
To determine the equivalent resistance of two resistors when connected in series.

Theory
The resistance can be increased or decreased depending on the combination and
connections in a circuit. The difference between the series and parallel circuit is based
on the arrangement of the resistors. Resistors are said to be connected in series if their
ends are joined. The potential difference across each resistor would be different, but the
current would be the same.

If two resistors are connected in series, then;

Resistance, R = R1+R2

Current, I = constant

Potential difference, V = V1+V2

On applying Ohm’s law, we get,

V1 = IR1

V2 +IR2

V = V1+V2

V = I(R1+R2)

∴ R = R1+R2

Materials Required
 Two resistors of different values
 A battery of 6 volts
 Ammeter
 Plug key
 Connecting wires
 A piece of sandpaper
 Voltmeter
 Rheostat
Circuit Diagram

2Ω 4Ω

Procedure
1. With the help of a circuit diagram, make the connections.
2. Do not switch on the key.
3. The ammeter should be connected in series, the voltmeter in parallel, and the rheostat in
series.
4. Make the connections as shown in the experimental setup and check of +ve and -ve
terminals of the battery.
5. By inserting the key, record the ammeter and voltmeter readings.
6. Note three readings by adjusting the rheostat.
7. Note down the readings of a voltmeter by connecting it to each resistor.
8. Measure the potential difference, V1 across the first resistor by plugging in the key.
9. Measure the potential difference, V2 across the second resistor by plugging in the key.
10. Calculate the relationship between V, V1, and V2.

Observations:
Range of ammeters = 0A to 3 A
The least count of ammeter = 0.05 A
Range of voltmeter = 0 V to 5 V
The least count of voltmeter = 0.1 V
Observation Table:
Resistor used No.of Voltmeter Ammeter R = V/I Mean value of
observations reading in reading in (in resistance
Volts (V) Ampere (I) ohm) (ohm)

R1 (first resistor) a 1 0.25 4

4Ω
b 2 0.5 4

c 4 1 4

R2 (second resistor) a 1 0.5 2

2Ω
b 2 1 2

c 4 2 2

Rs=R1+R2 (series a 1.5 0.25 6

combination)
b 3.0 0.5 6 6Ω

c 4.5 1.5 6

Result
The calculated value of Rs Rs=R1+R2= 6 Ω

The experimental value of Rs 6Ω

Hence, it is verified that Rs=R1+R2.

Precautions
1. Voltmeter and resistor should always be in parallel.
2. The least count of voltmeter and ammeter should be calculated properly.
3. Connections should be as per the experimental setup.
4. When no current flows through the ammeter and voltmeter, the pointers should be at zero.
5. The connecting wires that are used should be thick copper wire and using sandpaper the
insulation at the end of the wires should be removed.
Experiment:9
Aim
To determine the equivalent resistance of two resistors when connected in parallel.

Theory
A number of resistors are said to be in a parallel connection if one end of each
resistance is connected to one point and the other is connected to another point. The
potential difference across each resistor is the same and is equal to the applied
potential difference between the two points. I = I1+I2+….
Therefore from Ohm's Law
I1 = V/R1
and
I2 = V/R2
Substituting the values derived for I1 and I2 gives us
I = V/R1 + V/R2
I= (1/R1+1/R2) x V
I/V=1/R1+1/R2
1/R = 1/R1+1/R2
Materials Required
1. A battery
2. A plug key
3. Connecting wires
4. An ammeter
5. A voltmeter
6. Rheostat
7. A piece of sandpaper
8. Two resistors of different values
Procedure
1. Make all the connections as shown in the experimental setup I by keeping the key off.
2. Insert the key when the circuit is connected appropriately.
3. For resistors R1 and R2, note three readings of ammeter and voltmeter.
4. The resistors are connected in parallel and voltmeter is also connected in
parallel.
5. Use the rheostat and record three different readings of ammeter and voltmeter.
6. Remove the key.
7. With the help of the observation table, do the calculation.
Observations:
Range of ammeters = 0A to 3 A
The least count of ammeter = 0.05 A
Range of voltmeter = 0 V to 5 V
The least count of voltmeter = 0.1 V

Observation Table
Resistor used No.of Voltmeter Ammeter R=V/I Mean value of
observations reading in reading in (in resistance
R1 (first resistor) a 1
Volts (V) 0.25
Ampere (I) Ohm) R 1 = 1 Ohm
(Ohm)
b 2 0.5

c 4 1

R2 (second a 1 0.5 R2 = 2 Ohm

resistor)
b 2 1

c 4 2

1/Rp=(1/R1)+(1/R2) a 1 0.76 1.32 Rp= 1.3 Ohm

Parallel b 1.5 1.2 1.25
b
c 1.5
2 1.16
1.5 1.29
1.33
combination
c 2 1.5 1.33

Result
The calculated value of RP Rs= 1.3 Ω

The experimental value of RP 1.3 Ω

Precautions
1. The connections should be tight to avoid introducing external resistance.
2. To make connections, the circuit diagram should be referred to.
3. To make the current entry from the positive terminal and exit from the negative terminal, the
ammeter should be connected in series.
4. Resistor and voltmeter should be connected in parallel.
5. The least count of ammeter and voltmeter should be calculated properly.
Experiment: 10

Aim
To trace the path of a ray of light passing through a rectangular glass slab for different
angles of incidence. Measure the angle of incidence, angle of refraction, and angle of
emergence, and interpret the result.

Theory
Laws of refraction:
Following are the laws of refraction:

 The incident ray, the normal at the point of incidence, and the refracted ray lie in the same
plane.
 Snell’s law states that the ratio of the sine of the angle of incidence to the sine of the angle of
refraction is constant.

Refraction of light:
The refraction of light is a property of light due to which it changes its path when it
passes from one medium to the other.
Lateral displacement:
Lateral displacement is defined as the perpendicular shift in the path of light when it
emerges out from the refracting medium.
Materials Required
1. A drawing board
2. 4-6 all pins
3. White sheet of paper
4. Rectangular glass slab
5. A protractor
6. A scale
7. A pencil
8. Thumb pins

Procedure
1. Fix a white sheet on the soft drawing board using thumb pins.
2. Place the glass slab at the centre of the white paper and draw its outline boundary using a
sharp pencil.
3. Let ABCD be the rectangular figure obtained by drawing.
4. Mark a point E on AB and draw a perpendicular EN and label it as a normal ray.
5. Draw one angle of 30° with the help of protractor with EN. Fix pins at P and Q at 4-5 cm on
the ray that is obtained by the angle.
6. Place the glass slab on the rectangular figure ABCD.
7. To fix R and S, see through the glass slab from side CD, such that when seen through the
glass slab, all the pins P, Q, R, and S should lie in a straight line.
8. Draw small circles around the pins P, Q, R and S and remove the pins.
 9. Remove the glass slab.
10. Join points R and S such that it meets CD at point F. Draw a perpendicular N’M’ to CD at
point F.
11. Using a pencil, join the points E and F.
12. Measure the angles formed at AB and CD, i.e, the incident angle, refracted angle, and
emergent angle.
13. The lateral displacement is obtained by extending the ray PQ in a dotted line which is parallel
to ray FRS.
14. Measure the lateral displacement.
15. Repeat the same procedure for angles 45° and 60°.

Diagram:

Observation Table
Sl.no Angle of incidence Angle of refraction Angle of emergence ∠i – ∠e
∠i = ∠PEN ∠r = ∠MEF ∠e = ∠SFM’ ∠PEN – ∠SFM’

1 30° 28° 30° 0°

2 45° 43° 44.8° 0.2°

3 60° 56° 59.8° 0.2°

Conclusion
1. The angle of incidence and the angle of emergence are almost equal.
2. As the light is traveling from rarer to denser optical medium, the angle of refraction will be
lesser than the angle of incidence.
3. For different angles of incidence, the lateral displacement will remain the same.
4. The light will bend towards the normal when it travels from an optically rarer medium to an
optically denser medium.

Precautions
1. The rectangular glass slab used should have perfectly smooth faces.
2. The drawing board should be soft so that pins can be easily fixed on it.
3. The angle of incidence should lie between 30° and 60°.
4. All pins base should be in a straight line.
5. The distance between the pins P and Q or the pins R and S, about 5 cm gap should be
maintained.
6. Using a sharp pencil, draw thin lines.
7. The perpendiculars should be drawn with care.
 CHEMISTRY

Experiment:11

Aim:
To determine the pH of the given samples using pH paper or universal indicator. The
samples whose pH has to be determined are-
1. Dilute CH3COOH
2. Dilute NaOH
3. Salt NaCl
4. Dilute NaHCO3
5. Water
6. Lemon juice

Materials required:
1. Six test tubes
2. Test tube stand
3. Dilute acid CH3COOH
4. Dilute base NaOH
5. Salt NaCl
6. Water
7. Lemon juice
8. Dilute NaHCO3
9. Glass rod
10. Standard pH colour chart
11. pH paper
12. Glass rod
13. Dropper
14. Universal indicator

Theory:
pH is a measure of hydrogen ion concentration to determine the alkalinity or acidity of a
solution.
 If the pH value of a solution is less than 7 it is an acidic solution
 If the pH value of a solution is greater than 7 it is a basic solution
 If the pH value of a solution is equal to 7 it is a neutral solution

pH scale:
The pH scale consists of values which range from 0 (very acidic) to 14 (very alkaline).
The numbers on the scale help to determine the hydrogen ion concentration.

pH paper:
pH paper can help us to know if a solution is basic, acidic or neutral. When the pH
paper is dipped into a solution whose pH has to be determined, a colour will be
developed. This colour is compared with the standard pH colour chart. Instead of pH
paper, we can also use universal indicator paper or universal indicator solution.
Universal indicator:
A universal indicator is a mix of pH indicator solutions that are designed to determine
the pH of solutions over a wide range of values. Put a drop of solution on the universal
pH indicator paper. The colour developed on the paper is matched with the standard pH
colour chart.

Procedure:
1. Wash six test tubes with distilled water and put them on test tube stand and label them A, B,
C, D, E, F.
2. Add 2ml of CH3COOH in test tube A, Add 2ml of NaOH in test tube B, Add 2ml of NaCl in
test tube C, Add 2ml of NaHCO3 in test tube D, Add 2ml of Water in test tube E, Add 2ml of
Lemon juice in test tube F.
3. Take white tile, place 6 pH paper and label them A, B, C, D, E, F.
4. Use a dropper or glass rod to put the respective sample solutions on the labelled pH paper
placed on the white tile.
5. Observe the colour change.

Observation:

Test tube Solution pH colour paper pH Nature

Sample A CH3COOH Orange 3 Weak acid

Sample B NaOH Dark blue 14 Strong base

Sample C NaCl Green 7 Neutral

Sample D NaHCO3 Light blue 9 Weak base

Sample E Water Green 7 Neutral

Sample F Lemon juice Pink 2 Weak acid

Precautions:
 Use freshly prepared test sample for the experiment.
 The fruit juice sample should also be fresh to get the proper pH values.
 Glass rod or dropper used for one sample should be washed thoroughly before using it for
the other samples.
Experiment:12
Aim:
To study the following properties of acetic acid (ethanoic acid)-
1. Odour (Smell)
2. Solubility in water
3. Effect on litmus
4. Reaction with sodium bicarbonate

Materials required:
1. Test tube
2. Litmus paper
3. Dropper
4. Cork fit
5. Test tube stand
6. Water
7. Beaker
8. Acetic acid
9. Sodium bicarbonate
10. Distilled water
11. Lime water (freshly prepared)

Theory:
The chemical name of acetic acid is ethanoic acid and has the chemical formula
CH3COOH. The COOH group is called the carboxylic group which is responsible for the
properties of ethanoic acid. This acid freezes at 16.6° C hence called glacial acetic acid.
It smells like vinegar and dissolves in water. It is a weak acid as it dissociates
particularly in water. It reacts with alcohol to obtain ester. It reacts with sodium
bicarbonate to give carbon dioxide gas.
Procedure:
To determine the odour:

 Take a test tube of 20 mL.

 Add 5 mL of ethanoic acid into it
 Bring the test tube near your nose and smell it by wafting.

To check the solubility in water:

 Take a test tube of 20 mL.
 Add 2 mL of ethanoic acid to it
 Pour 10-15 mL of distilled water and mix it.

To determine the effect on litmus:

 Take a clean dropper

 Take a blue litmus paper
 Pour 2-3 drops of ethanoic acid on the litmus paper.
Reaction with sodium bicarbonate:

 Take a test tube of 20 mL.

 Add a pinch of sodium bicarbonate to it
 Pour 1 mL of dilute ethanoic acid into the test tube.
 To the mouth of the test tube fix a cork with a bent delivery tube

 The other end of the delivery tube is required to be dipped in lime water.

Observation:
Properties Observation

Determination of odour Pungent/vinegar smell

Checking its solubility in water Dissolves in water

Determining the effect on litmus Blue litmus paper turns red

Reaction with sodium A colourless gas is produced which turns the lime water
bicarbonate milky

Results:
 Ethanoic acid or acetic acid or glacial acetic acid smells like vinegar.
 Glacial acetic acid is water-soluble.
 Ethanoic acid turns blue litmus paper red.
 When acetic acid reacts with sodium bicarbonate, carbon dioxide gas is
liberated.

Precautions:
 Handle the solution with care.
 Add a small amount of sodium bicarbonate.
 Do not inhale the vapours of the chemicals.
 Lime water should be freshly prepared.
Experiment:14

Aim:
1. Arranging the metals Zinc, Copper, Iron, and Aluminium in the decreasing order
of reactivity based on the results obtained from the bellow
2. Observing the action of zinc, iron, copper and aluminium metals for the following
salt solutions.
a. Zinc sulphate (ZnSO4)
b. Copper sulphate (CuSO4)
c. Ferrous sulphate (FeSO4)
d. Aluminium sulphate (Al 2(SO4)3)

Materials required:
1. Aluminium foil
2. Zinc granules
3. Copper turnings
4. Iron filings
5. Copper sulphate solution
6. Aluminium sulphate solution
7. Zinc sulphate solution
8. Four 50ml beakers
9. Test tube
10. Test tube stand

Theory:

Reactivity series:
Different metals have different reactivities with chemical reagents. The metals form
positive ions by losing electrons more readily. A more reactive metal displaces a less
reactive metal from salt solution. Such reactions are called displacement
reactions. Consider an example, if a small amount of zinc is dipped in copper sulphate
solution, zinc is displaced from copper sulphate. The blue colour of copper sulphate
fades and colourless zinc sulphate solution is obtained. Therefore, zinc is more reactive
than copper.

 Aluminium has the ability to displace zinc, iron, copper from their respective
salt solutions and therefore is more reactive than them.
Reactions:

2Al(s) + 3ZnSO4(aq) –→ Al2(SO4)3 + 3Zn(s)

2Al(s) + 3FeSO4(aq) –→ Al2(SO4)3 + 3Fe(s)

2Al(s) + 3CuSO4(aq) –→ Al2(SO4)3 + 3Cu(s)

 Zinc has the ability to displace iron, copper from their respective salt solutions
and therefore is more reactive than them.
Reactions:
Zn(s) + CuSO4(aq) –→ ZnSO4 (aq) + Cu(s)
Zn(s) + FeSO4(aq) –→ ZnSO4 (aq) + Fe(s)
Zn(s) + Al2(SO4)3(aq) –→ no reaction
Zn(s) + ZnSO4(aq) –→ no reaction

 Iron has the ability to displace copper from its salt solution and therefore is
more reactive when compared to Cu.
Reactions:

Fe(s) + CuSO4(aq) –→ FeSO4 (aq) + Cu(s)

Fe(s) + FeSO4(aq) –→ no reaction

Fe(s) + Al2(SO4)3(aq) –→ no reaction

Fe(s) + ZnSO4(aq) –→ no reaction

No reaction occurs when any of the metals from zinc, iron, aluminium, and
copper are placed in aqueous solution of Al 2(SO4)3.

Therefore, from the above, we can conclude the reactivity of elements in the following
order –

 Aluminium is more reactive than zinc.

 Zinc is more reactive than Fe.
 Fe is more reactive than Cu.
Hence, Aluminium is the most reactive and Copper being the least reactive metal.

The below table gives you an understanding of the colours exhibited by the metals and
their salts:

Procedure:
1. Wash four beakers with distilled water, dry them and label them A, B, C, D.
2. Prepare 50 mL solution with 5% concentration by volume of ferrous sulphate,
aluminium sulphate, copper sulphate, zinc sulphate.
3. Pour ZnSO4 in beaker A, FeSO4 in beaker B, CuSO4 in beaker C, and add
Al2(SO4)3 in beaker D.
4. Wash four test tubes with distilled water and dry them.
 5. Put them in a test tube stand and label them as A, B, C and D as shown in the
figure above.
6. Take 10 mL of the saturated ferrous sulphate (FeSO 4), copper sulphate (CuSO4),
zinc sulphate (ZnSO4), and aluminium sulphate (Al 2(SO4)3 ) in the test tube as
shown in the figure.
7. Take metal strips of aluminium (Al), iron (Fe), copper (Cu), and zinc (Zn). Clean
them with sandpaper before using them in the experiment.
8. Now dip one of the four metal strips in all the four test tubes and observe if any
colour change occurs in all four test tubes.
9. Continue step 8 with other metal strips by dipping in fresh aqueous solutions of
metals and see the displacement reaction.

Metal Dipped Observation

in

Al ZnSO4 No change in the solution. Zinc metal which is greyish accumulates on

Aluminium metal.

Al CuSO4 Blue solution becomes colourless. Copper metal which appears reddish
brown gets accumulated.

Al Al2(SO4)3 No reaction.

Al FeSO4 Green solution becomes colourless. Iron metal which appears greyish
black gets accumulated.

Fe ZnSO4 No reaction.

Fe CuSO4 Solution becomes light green. Copper metal which appears reddish brown
gets accumulated.

Fe Al2(SO4)3 No reaction.

Fe FeSO4 No reaction.

Zn ZnSO4 No reaction.

Zn CuSO4 Blue solution turns colourless. Reddish brown coppers accumulates on

zinc metal.

Zn Al2(SO4)3 No reaction.

Zn FeSO4 Light green solution turns colourless. Greyish black iron accumulates on
zinc metal.

Cu ZnSO4 No reaction.

Cu CuSO4 No reaction.

Cu Al2(SO4)3 No reaction.

Cu FeSO4 No reaction.
Conclusion:
Metals Inference

Aluminium Aluminium displaces zinc, copper, and iron from their solutions. Aluminium is the
most reactive metal.

Zinc Zinc displaces copper, and iron from their solutions. Zinc is more reactive than Cu
and Fe metal.

Iron Iron displaces copper from its solution. Therefore, copper is less reactive than
iron.

Copper Copper does not displace any metal therefore it is the least reactive.

Result :
The decreasing order of reactivity of metals is as follows: Al, Zn, Cu, Fe

Precautions:
 Use cleanly washed and dried apparatus for the experiment.
 Clean the metals before dipping them in the solution. You can clean it with
sandpaper.
 Use small quantities of aluminium sulphate, iron sulphate, copper sulphate and
copper sulphate solutions for the experiment.
 Handle the chemicals with care and do not allow them to come in contact with
your skin.
 Make sure you wash the test tubes after every observation.
Experiment:15

Aim:
Types of Chemical Reactions -

Performing and observing the actions of-

1. Water on quicklime
2. Heat on ferrous sulphate crystals
3. Iron nails dipped in copper sulphate solution
4. Reaction between sodium sulphate and barium chloride solution
On the following reactions-

1. Combination reaction
2. Decomposition reaction
3. Displacement reaction
4. Double displacement reaction

Materials Required
1. Quick lime (Calcium oxide)
2. Ferrous sulphate crystals
3. Copper sulphate solution
4. Iron Nail
5. Barium chloride solution
6. Sodium sulphate solution
7. Burner
8. Borosil beaker
9. Glass rod
10. Distilled water
11. Dropper
12. Test tubes

Theory:
The types of chemical reaction depend upon the reactivity of the chemicals. The types
of reaction are:
1. Combination reaction (Synthesis reaction experiment)
2. Decomposition reaction
3. Single displacement reaction experiment (Single replacement reaction)
4. Combustion reaction
 1. Combination Reaction-

When calcium oxide is mixed in water it dissolves and forms calcium hydroxide (basic in
nature). During this reaction, a lot of heat is liberated and therefore, it is called an
exothermic reaction. The equation is as follows:

CaO(s) + H2O(l) → Ca(OH)2 (aq) + Heat ………(1)

In the reaction (1) slaked lime is formed by combining two products viz, quick lime and
water. Therefore, when two or more than two substances combine to give a single
product it is termed as combination reaction. The reactions accompanied by the
evolution of heat are called exothermic reactions.

Procedure
1. Wash a borosil beaker with distilled water and dry it.
2. Take a small amount of calcium oxide (quick lime) and slowly add water to it.
3. Wash and take a clean glass rod to stir the mixture of quick lime and water.
4. Touch the beaker carefully from outside.
5. Observe the change.
6. With the help of dropper take a few drops of the mixture from the beaker and
place it on red and blue litmus paper strips.
7. Wait and observe.

Observation:
A hissing sound is heard during the reaction when water is added to the beaker
containing quick lime. Due to the evolution of heat during the reaction the temperature
increases and makes the solution hot.

2. Decomposition reaction:

Ferrous sulphate crystals are ferrous sulphate heptahydrate with a chemical

formula FeSO4.7H2O and are green in colour.

On heating the ferrous sulphate heptahydrate it loses seven water molecules to form
anhydrous ferrous sulphate (FeSO4) and is white in colour.

In the reaction one substance FeSO4 (Ferrous sulphate) splits into three substances
ferric oxide (Fe2O3), sulphur dioxide (SO2), and sulphur trioxide (SO3) due to heat.
Therefore, this reaction is called decomposition reaction or also known as a thermal
decomposition reactionThe reaction is as follows:

FeSO4.7H2O → FeSO4(s) + 7H2O ………….(1)

(Green colour) (white colour)

Ferrous sulphate when heated is decomposed to ferric oxide, sulphur trioxide, and
sulphur dioxide. The reaction is as follows:

FeSO4(s) → Fe2O3(s) + SO2(g) + SO3(g) ………….(2)

(White colour) (brown colour) (colourless) (colourless)

We can combine reaction (1) and (2) and write it as follows:

2FeSO4.7H2O(s) → Fe2O3(s) + SO2(g) + SO3(g) + 14H2O(g)

Procedure
1. Wash a boiling tube with distilled water and dry it.
2. Take 2 grams of ferrous sulphate crystals in the tube.
3. Make a note of the colour of the crystals.
4. Use a test tube holder to hold the boiling tube.
5. Heat the boiling tube on the bunsen burner as shown in the figure.
6. Observe the colour of the residue got and smell the odour of the gases evolved.
7. Tiny colourless water droplets are seen near the neck of the tube.
8. Gently turn it towards your nose and smell for any gas evolved.
9. Classify the type of reaction.

Observation:
Observation

 Colour of Ferrous sulphate crystals changes from green to white and later brown.
 The gas evolved smells like burning sulphur.

3. Single displacement Reaction

As per the reactivity series, the more reactive metals displace the less reactive metals.
When iron is compared with copper, it is placed above copper in activity series.
Therefore the metals placed above are more reactive whereas the metals placed below
are less reactive.

When iron nails are placed in CuSO4 iron displaces copper from copper sulphate to form
ferrous sulphate. The iron nails get deposited with a brownish red substance of the
copper metal. The reaction is as follows:

Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)

Metallic iron displaces copper ion (Cu2+) from its salt is an example of a chemical
displacement reaction. Metallic iron is converted to ferrous iron, the cupric ion is
converted to metallic copper. The reaction is as follows:

Fe(s) + Cu2+(aq) → Fe2+(aq) +Cu(s)

Procedure
1. Take Copper sulphate Solution in a test tube and put one iron nail into it by cleaning it with
sand paper wait for 10 minutes .
2. Observe the intensity of the blue colour of CuSO4 before and after the experiment performed
in test tube.
3. Record your observation.

Observation
Experiment Before Reaction After Reaction

Colour of CuSo4 Blue Green

Colour of iron nail Silvery grey Brownish red

4. Double displacementReaction:

Reactions occurring in the solution by exchanging ionic compounds to form new

compounds are called double displacement reactions. The ionic compounds considered
as reactants are water soluble. One of the products is formed as a precipitate or as a
gas which is water soluble.

When two solutions viz sodium sulphate and barium chloride are mixed, double
displacement reaction as below occurs.

Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(s)

Sulphate ions from the solution of sodium sulphate are displaced by chloride ions and
the chloride ions from the solution of barium chloride are displaced by sulphate ions.

Procedure
1. Take two test tubes, wash them with distilled water and dry them.
2. Label the test tube as P and Q.
3. Pour 5mL of barium chloride in the test tube P and observe the colour.
4. Pour 5mL of sodium sulphate in the test tube Q and observe the colour.
5. Take a conical flask and pour the solutions from both the test tube into it.
6. Stir the mixture added to the conical flask with a glass rod.
 7. Keep it undisturbed for some time.
8. Observe the change in colour of the solution.
9. Record your results in the below-given table.

Observation
Experiment Observation

Colour of test tube P and test tube Q Colourless

Mixture of solution in conical flask Precipitation is formed

Precautions:
 Do not keep the mouth of the boiling tube towards yourself.
 Do not smell the gas by directly getting it under your nose but gently turn it towards your
nose and blow it with your hand.
 Wearing safety glasses is important while you are performing this experiment.
 Thoroughly wash the boiling tube with distilled water and dry it before using it.
 Use good quality boiling tube while heating the ferrous sulphate crystals.
 Take care to keep the sulphur dioxide and sulphur trioxide gas coming in contact with your
eyes as they cause irritation to eyes.