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Determination of Acetic Acid in Vinegar Investigatory Project

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Investigatory Project on Determination of Acetic Acid in Vinegar

 

Introduction

 

Acetic acid is a weak acid. It is widely used in industrial chemistry as glacial acetic acid. In food industry, it is mainly used as vinegar. Vinegar is a dilute solution of glacial acetic acid. Commercial vinegar is prepared by fermentation and it contains 4 - 5% of acetic acid.

 

In this project, the amount of acetic acid present in commercial vinegar is determined by a simple volumetric method. Here we titrate vinegar, a weak acid, with a standard solution of a strong base such as sodium hydroxide. Sodium hydroxide is not a primary standard. Therefore it should be standardised using a suitable primary standard such as oxalic acid using phenolphthalein indicator.

 

According to the theory of acid-base titrations, the end point in the titration of vinegar with sodium hydroxide will be observed between pH 8 and 10. Therefore, in this titration also, phenolphthalein is a siutable indicator. The reaction between vinegar and sodium hydroxide is given by,

 

CH3 - COOH + NaOH —> CH3COONa + H2O

 

By knowing the volume of vinegar solution reacting with a definite volume of NaOH solution, the strength of vinegar solution can be calculated.

 

Aim of the Investigatory Project

 

The aim of the project is to estimate the quantity of acetic acid present in commercial vinegar sample.

 

Apparatus and Chemicals Required

 

Burette and burette stand, Conical flasks, Pipette, Standard flask, Funnel, Weighing bottle, 0.1N NaOH solution (dissolving 4g NaOH in one litre of pure water), Phenolphthalein indicator (dissolving 0.4g phenolphthalein in 500 millilitre of ethanol and 500 millilitre of H2O by continuously stirring), Vinegar sample and Oxalic acid (AR)

 

Procedure

(i) Preparation of standard oxalic acid solution

 

Carefully weigh the bottle with around 0.63g of oxalic acid crystals taken in a chemical balance. After that transfer it into a funnel fixed over a 100 millilitre standard flask. Find out the weight of the bottle once more and check the accurate mass of oxalic acid moved by considering the variation amid the two weights. Then the oxalic acid must be washed down into the standard flask by some water (approximately 20 millilitre). Then rotate the flask till the oxalic acid is dissolved. The funnel and the stem are then washed down into the standard flask and then the funnel must be removed. Then solution should be make up to the mark and shake it to acquire a uniform concentration.

 

(ii) Standardisation of NaOH solution

 

Wash the burette with a few drops of oxalic acid and after that fill the acid until zero mark. Then pipette out 20 millilitre of standard NaOH solution in a 250 millilitre conical flask. Mix two or three drops of phenolphthalein indicator to produce a pink colour for the solution. Titrate the obtained solution against oxalic acid from the burette until the pink colour gone. Observe the burette reading and do again the titration to obtain concordant titre values.

 

(iii) Detemination of acetic acid in vinegar

 

Pipette out the 20 millilitre of comm. vinegar in a 100 millilitre standard flask. Fill the solution up to the mark of the flask with distilled water. Wash a fresh burette with some solution and fill it until zero mark with the same solution. Then pipette out 20 millilitre of NaOH solution in a 250 millilitre conical flask and put two or three drops of phenolphthalein indicator in the solution. The obtained solution is then titrated alongside vinegar solution from the burette until the pink colour gone. Observe the burette reading and do again the titration to obtain concordant titre values.

 

Observations

 

Mass of bottle + Oxalic acid crystals = m1 g

Mass of bottle after the transfer of salt = m2 g

Then, Mass of Oxalic acid transferred = m1 – m2 = m g

 

Oxalic acid Vs Sodium hydroxide solution – Phenolphthalein indicator


No

Volume of NaOH (mL)

Initial burette reading (mL)

Final burette reading (mL)

Volume of Oxalic acid

1

20

 

 

 

2

20

 

 

 

3

20

 

 

 

 

Vinegar solution Vs Sodium hydroxide solution – Phenolphthalein indicator


No

Volume of NaOH (mL)

Initial burette reading (mL)

Final burette reading (mL)

Volume of Vinegar Soln

1

20

 

 

 

2

20

 

 

 

3

20

 

 

 

 

Calculations

 

(i) Estimation of strength of NaOH soln

 

Equivalent Mass of oxalic acid = 126/2 = 63

Mass of of oxalic acid put in the 100 mL soln = m g

Normality of oxalic acid soln, N1 = (m x 1000) / (63 x 100) = _______ N

Volume of Sodium Hydroxide soln (V2) = 20 mL

Volume of oxalic acid needed to react with 20 mL Sodium Hydroxide soln = V1 mL

We know that, Normality of Sodium Hydroxide soln is N2

Therefore, N1V1 = N2V2

i.e., N2 = _______ N

 

(ii) Determination of strength of Vinegar soln

 

Normality of Sodium Hydroxide soln, N2 = _______ N

Volume of Sodium Hydroxide soln, V2 = 20 mL

Volume of vinegar soln reacting with 20 mL Sodium Hydroxide soln = V3 mL

We know that normality of vinegar soln, N3,

N2V2 = N3V3

N3 = ______ N

Here 20 mL of vinegar is diluted to 100 mL, so the normality of comm. vinegar = N3 x (100/20) = ______ N

Equivalent mass of acetic acid = 60

i.e., Strength of comm. vinegar = 5N3 x 60 = ______ gL-1

 

Result

 

Normality of vinegar soln = ______

Normality of comm. vinegar = _____

Strength of comm. Vinegar = _____

 

Viva Questions and Answers

 

1. What is vinegar?

Ans: Vinegar is dilute solution of acetic acid

 

2. How much acetic acid is normally present in vinegar?

Ans: Vinegar contains 4 - 5% acetic acid.

 

3. Why is acetic acid a weak acid ?

Ans: It is incompletely ionised in solution. It has low Ka (1.8 x 10-5).

 

4. What is a standard solution ? What is the standard solution used in this experiment ?

Ans: A standard solution is one whose normality is known. In this experiment crystalline oxalic acid is used to make standard solution.

 

5. What is the amount of crystalline oxalic acid used to make 100 millilitre of 0.1N oxalic acid solution? 

Ans: Equivalent mass of oxalic acid is 63. Therefore 0.63 g oxalic acid in 100 millilitre solution forms 0.1N solution.

 

6. How will you find the equivalent mass of oxalic acid ?

Ans: Molecular mass of crystalline oxalic acid (H2C2O4 . 2H20) is 126. Its basicity is two.

Therefore equivalent mass = molar mass/basicity = 126/2 = 63

 

7. Indicate the reaction between oxalic acid and NaOH ?

Ans: Neutralisation.

H2C2O4 + 2NaOH è Na2C2O4 + 2H2O

 

8. Why do you choose phenolphthalein as indicator as the titration of vinegar against NaOH ?

Ans: It is a weak acid - strong base reaction for which the end point appears between pH 8 and 10. So phenolphthalein is suitable.

 

9. Why can't you use methyl orange as indicator in this experiment?

Ans: Methyl orange is suitable for weak base-strong acid titration. The pH range of methyl orange is from 3 to 4.

 

10. Why can't you prepare a standard solution of NaOH directly to titrate against vinegar solution?

Ans: Sodium hydroxide is not a primary standard. It is not available in the pure state. So to standardise it we need a primary standard (oxalic acid).

Investigatory Project on Determination of pH Value

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INVESTIGATORY PROJECT ON STUDY OF THE CHANGE IN pH VALUES OF ACETIC ACID AND AMMONIUM HYDROXIDE BY THE ADDITION OF SALTS HAVING COMMON IONS

 

Introduction

pH value is a convenient method for expressing the hydronium ion concentration (H3O+) of a solution. The negative logarithm of hydronium ion concentration in terms of moles per litre is called as pH of a solution.


pH = - log [H3O+]


In general, at room temperature all neutral solutions would have a pH equal to 7, all acidic solutions would have a pH less than 7 and all basic solutions would have a pH more than 7.


Common ion effect


Weak acids and weak bases are ionised only to small extent in their aqueous solutions. The degree of ionisation of a weak electrolyte (may be a weak acid or a weak base) is more suppressed if a strong electrolyte which can provide several ion common with the ions provided by weak electrolyte, is added to its solution. This effect is called common ion effect.


For example, degree of ionisation of acetic acid (a weak acid) is suppressed by the addition of sodium acetate (a strong electrolyte). The ionisation of acetic acid and sodium acetate in solution is given below.


CH3COOH CH3COO- + H+

CH3COONa  CH3COO- + Na+


Due to the presence of excess acetate ions furnished by sodium acetate, the equilibrium for the ionisation of acetic acid shifts in the backward direction. i.e., more acetate ions combine with hydrogen ions to form unionised acetic acid. This reduces the concentration of hydrogen ion (or more correctly hydronium ion) in solution. Therefore the pH of the solution increases. Similarly, the degree of ionisation of NH4OH (a weak base) is suppressed by the addition of NH4Cl (a strong electrolyte).


NH4OH NH4+ + OH-

NH4CI NH4+ + Cl-


So the concentration of OH- decreases in solution or the concentration of H3O+ increases. Thus the pH of the solution is lowered.


Determination of pH


pH paper can be used to determine the approximate pH value of any solution. A pH paper is a narrow piece of paper that is prepared by dipping the piece in the solutions of various indicators and then drying it. It is available commercially.


Aim of the project


The aim of the project is


(a) to study the change in pH of acetic acid by the addition of sodium acetate, and

(b) to study the change in pH of ammonium hydroxide by the addition of ammonium chloride.


Apparatus and chemicals


1. Test tubes and test tube stand

2. Measuring cylinders

3. Glass tubes

4. pH paper

5. Dilute acetic acid solution

6. Sodium acetate

7. Ammonium hydroxide solution

8. Ammonium chloride


Procedure


(a) Take 10 mL. of dilute acetic acid in a clean test tube Place a drop of this solution on a strip of pH paper and note its colour. Compare the colour with the colour on chart paper and note the pH of the acid. Now weigh 1 g of solid sodium acetate and dissolve it in the acetic acid taken. Verify the pH of the solution using pH paper. Weigh again 1g of sodium acetate and dissolve it in the same solution. Determine the pH again. Record the observations.


(b) Take 10 mL of dilute ammonium hydroxide solution in a clean test tube. Place a drop of this solution on a strip of pH paper and determine its pH. Now weigh 1g of solid ammonium chloride and dissolve it in the ammonium hydroxide solution. Verify the pH of the solution using pH paper. Weigh again 1g of ammonium chloride and dissolve it in the same solution, Determine the pH of the solution. Record the observations.


OBSERVATIONS

(a)

Exp No:

Solution

Shade of colour

pH Value

1

Acetic acid

 

 

2

Acetic acid + 1g sodium acetate

 

 

3

Acetic acid+ 2g sodium acetate

 

 

 

(b)

Exp No:

Solution

Shade of colour

pH Value

1

Ammonium hydroxide

 

 

2

Ammonium hydroxide + 1g ammonium chloride

 

 

3

Ammonium hydroxide + 2g ammonium chloride

 

 

 

Result:


1. pH value of acetic acid goes on increasing on adding more and more of sodium acetate to it.

2. pH value of ammonium hydroxide goes on decreasing on adding more and more ammonium chloride to it.

 

Viva Questions and Answers

 

1. What is pH?

Ans: See the 'introduction' of this experiment.


2. Give the pH of pure water at a temperature of 25°C ?

Ans: 7


3. The pH of a solution is 3. What is its hydrogen ion concentration? Is the solution acidic or basic ?

Ans: Its hydrogen ion concentration is 10-3 molL-1. The solution is acidic.


4. The pH of two solutions A and B are 4 and 5 respectively. Which is more acidic ?

Ans: Solution A is more acidic.


5. Explain why the addition of NH4Cl decreases the pH of a solution of NH4OH.

Ans: See the 'introduction' of this experiment.


6. What is common ion effect ?

Ans: See the 'introduction' of this experiment.


7. What is an acid base indicator ? Give one example

Ans: An acid base indicator is an organic compound which changes its colour within a certain pH range in solution. Methyl orange is an example. It is golden yellow in alkaline medium and rose red in acid medium.


8. What do you mean by universal indicator ?

Ans: It is an indicator made by mixing a number of acid base indicators. It exhibits various colour changes over a wide range of pH. Each colour corresponds to a particular value of pH. Universal indicator is used to determine the pH of a solution.


9. What is the connection between the pH and pOH of an aqueous solution ?

Ans: pH + pOH = 14 (at 298K)


10. What is a buffer solution ?

Ans: A buffer solution is one which resists changes in its pH value even when small quantity of acid or base is added to it.

Relative Reactivity of Metals (Practical)

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Investigatory Project on Study of the relative reactivity of metals

 

Introduction

 

The relative reactivity of various metals can be conveniently predicted from the electrochemical series. In the electrochemical series different metals are arranged in the increasing order of their standard reduction potentials. Hence the metals occupying higher position in the series are more reactive than those occupying lower positions. Any metal in the electrochemical series can displace all other metals lying below it from the solution of their salts. This confirms that a more reactive metal can displace a less reactive metal from its salt solution. For example, zinc is more reactive than copper. Therefore zinc can displace copper from copper sulphate solution. Thus, by studying the interaction of metals with the salt solution of other metals it is possible to compare the reactivities of different metals.

Aim of the project

 

The aim of the project is to arrange magnesium, iron, zinc and copper in the decreasing order of their reactivities by studying their interaction with salt solutions.

 

Apparatus and Chemicals

 

1. Test tubes and test tube stand

2. Small measuring cylinders

3. 0.25M solution of zinc sulphate, copper sulphate, ferrous sulphate and magnesium sulphate.

 

Procedure

 

(1) Measure out 5 ml. of zinc sulphate, ferrous sulphate and copper sulphate solutions separately into three test tubes.

(ii) Put a piece of clean magnesium wire in each solution.

(iii) After about half an hour, examine the metal surface and note the change in colour of the solution, if any, in all the test tubes. Record the observations.

(iv) Clean the test tubes and perform the experiment with the following combination of metal and salt solutions.

(a) Zinc metal and solutions of magnesium sulphate, ferrous sulphate and copper sulphate.

(b) Iron metal and solutions of magnesium sulphate, zinc sulphate and copper sulphate.

(c) Copper metal and solutions of magnesium sulphate, zinc sulphate and ferrous sulphate.


Metal Added

Salt Solution

Colour of Solution

Appearance of metal surface

Inference

Initial

Final

Mg

Mg

Mg

ZnSO4

FeSO4

CuSO4

 

 

 

 

Zn

Zn

Zn

MgSO4

FeSO4

CuSO4

 

 

 

 

Fe

Fe

Fe

MgSO4

ZnSO4

CuSO4

 

 

 

 

Cu

Cu

Cu

MgSO4

ZnSO4

FeSO4

 

 

 

 

 

Result

 

1. Since Mg displaces all the other three metals from their salt solutions, it is tile most reactive among the four metals.

2. Since Cu does not displace any of these metals from their salt solutions, it is the least reactive among the four.

3. Zn is more reactive than Fe and Cu but less reactive than Mg.

4. Fe is more reactive than Cu but less reactive than Mg and Zn.

5. The order of reactivity of these metals is Mg > Zn > Fe > Cu.

 

Viva Questions and Answers (Lab)

 

1. What is electrochemical series?

Ans: The arrangement of elements according to the increasing value of their standard reduction potentials is termed as electrochemical series otherwise activity series.

 

2. What do you mean by electrode potential?

Ans: Electrode potential is the measure of the tendency of an electrode to lose or gain electrons if in contact with solution of its own ions.

 

3. How does the activity of metals vary in electrochemical series?

Ans: Activity of metals decreases on going down the electrochemical series.

 

4. What is the action of zinc on copper sulphate solution?

Ans: Since Zn is more reactive than Cu, zinc displaces copper from copper sulphate solution. Zn gets oxidised to Zn2+ while Cu2+ gets reduced to Cu,

Zn + CuSO4 ---> ZnSO4 + Cu

 

5. Can we store copper sulphate solution in an aluminium vessel ?

Ans: No. Aluminium, being more reactive than copper, will displace copper horn copper sulphate solution. Aluminium gets oxidised to Al3+ ions and Cu2+ ions get reduced to Cu. So the aluminium container gets corroded.

 

6. Name two metals that will not displace hydrogen from dilute sulphuric acid?

Ans: Copper and silver

 

7. Why copper does not displace hydrogen from dilute sulphuric acid and dilute hydrochloric acid?

Ans: As copper comes below hydrogen in the electrochemical series, it cannot displace hydrogen from dilute sulphuric acid and dilute hydrochloric acid.

Effect of Herbicides on lawn weeds

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Investigatory Project on Effect of Herbicides on lawn weeds

 

STUDY OF THE EFFECT OF A HERBICIDE ON LAWN WEEDS (To compare the effect of a herbicide on monocotyledonous and dicotyledonous plants)

 

Introduction

 

Weeds are unwanted plants which grow in cultivated fields and compete with the useful plants for the resources. They compete for the light, space, nutrients, water and reduce the crop yield. Weeds acts as alternate host for certain diseases and these help in spreading the diseases. Many weeds cause allergy and other skin diseases in human beings. Weeds can be removed from the fields, play grounds and lawns either by mechanical or chemical method. In chemical method, certain chemicals are used to kill the weeds. These chemicals are called weedicide. The synthetic auxins like 2, 4-D (2, 4 Dichlorophenoxy acetic acid) interferes with the translocation of carbohydrate and thus kills the plants. Dicot and monocot plants respond differently to different concentrations of 2, 4-D. Specific concentrations of 2, 4 - D can destroy dicot plants, while monocot plants remain unaffected. Therefore, 2, 4 - D is commonly used as selective herbicide.


Materials and Methods

 

2, 4 - D, sprayer, beakers, measuring cylinder, cord, nails etc were collected. Then prepared ten gram 2, 4-D was dissolved in 3 ml of ethanol. This solution was mixed with 100 ml of distilled water. A weed growing lawn was selected for the study. A square area was masked out by using nails and cord. 2, 4 - D solution was spread over the selected area once in a week.

 

Observation and Results

 

After a few days the marked and treated area was observed. The broad leaved dicot plants showed wilting after a day or two and gradually died. But the monocot plant such as grass remain unaffected.

 

Discussions and Conclusions

 

2, 4 - D had killing effect on broad leaved dicot plants and not on monocot plants.

 

Model Viva Questions and Answers

 

1. What are weeds?

Weeds are unwanted plants which compete with economically useful plants for the resources and reduce the yield.

 

2. Name a few common weeds.

Amaranthus, Alternanthera, Convolvulus, crotton sps. etc.

 

3. What are herbicides?

Herbicides are chemical substances which are used to kill herbs.

 

4. Why 2, 4 - D is used as a common herbicides for lawns?

Because 2, 4 - D kills broad leaved dicot plants without affecting narrow leaved monocot plant or the grass of the lawn.

 

5. Name a few commonly used herbicides.

2, 4 - D, Atrazine, Simazine etc.


References

 

1. G. Ray Noggle and George J. Fritz. Introductory Plant Physiology. Prentice, Hall of India Pt. Limited. New Delhi.

2. William. G. Hopkins, Introduction to Plant Physiology. John Wiley and Sons Inc. NewYork.

Zener Diode Experiment Viva Questions

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Zener Diode Experiment Viva Questions and Answers

 

1. What is a Zener diode?

 

A junction diode which can operate in the reverse breakdown voltage region is called a Zener diode.

 

2. Is Ohm's law obeyed in a semiconductor or not?

 

In a semiconductor, Ohm's law is valid only for low electric fields.

 

3. What is meant by a junction diode?

 

When a p-type semiconductor is fixed with n-type semiconductor, it forms a junction diode.

 

4. What is meant by forbidden gap?

 

The band separating the valance band and the conduction band is called forbidden gap.

 

5. What is meant by reverse current?

 

The reverse current is the current in the p-n junction diode circuit, when the junction is reverse biased.

 

6. What is meant by reverse breakdown?

 

Sudden increase in reverse current of a p-n junction diode when a definite reverse voltage is applied is called reverse breakdown.

 

7. What is the cause of reverse breakdown?

 

The cause of reverse breakdown will rupture a very large number of covalent bonds due to high reverse voltage applied.

 

8. What is zener current?

 

The value of reverse current after the zener breakdown is called zener current.


9. What is meant by the break down voltage of a diode


Materials are known as Conductors, Semiconductors, and Insulators, based on their electrical features. Materials that can easily conduct electricity are conductors. In comparison, insulators are categorised as materials that do not conduct any electricity. Among conductors and insulators lie the features of semiconductor materials. Researchers have observed during their work with insulators that when a certain amount of electricity is applied to them, insulator material can be made to act as a conductor. Breakdown was named for this phenomenon, and the minimum voltage at which this happens is known as Breakdown Voltage. For various materials, these voltage levels are distinct and often depend on their physical properties.


The main use of zener diodes in electronic circuits is they can be used as basic building blocks. In order to provide a reference voltage to the electronic circuits, zener diodes are used. Zener diodes works in the breakdown regions of the diode. 


Generally zener diodes are huge diodes, so that they can work in the reverse biased regions.  Due to the zener effect, the breakdown takes place. When the electric field of the reverse-biased P-N diode is increased in the Zener effect, tunnelling of the valence electrons into the conduction band occurs. This results in an improvement in the reverse current of the minority charge carriers. The Zener effect is known as this phenomenon, and the minimum voltage at which this phenomenon begins is known as the Zener voltage breakdown.


10. What are the differences between an ordinary diode and zener diode?


The main differences between ordinary diode and zener diodes are:


(i) The operation of diode is uni-directional, while the zener diode operates bi-directionally in both forward and reverse biased directions.


(ii) For an ordinary diode and a zener diode, the doping characteristics is also different. The conventional diode is moderately doped, while the zener diode is sharply doped.


(iii) In normal P-N Junction diodes, the breakdown voltage is high, but in case os zener diodes, the breakdown voltage is sharp.


(iv) It is not possible to operate a conventional diode in reverse biased mode, but it is possible to operate a zener diode in reverse biased mode.


(v) The main use of conventional diodes are they can be used in clipper, clamper, rectifiers, etc. The main use of zener diode is in voltage regulator circuits.


11. Distinguish between zener break down and avalanche break down?


Avalanche Breakdown:


(i) The condition of occurrence of Avalanche breakdown is when both sides of the PN junction are lightly doped and high depletion layer.


(ii) In Avalanche breakdown, it has a weak electric field across the depletion region.


(iii) Covalent bonds are broken due to the collision with valence electrons and electron-holes are created.


(iv) In Avalanche breakdown, from the applied potential,  the charge carriers acquires energy and produces  more carriers. This process is termed as Avalanche multiplication and the breakdown is termed as the avalanche breakdown.


Zener Breakdown


(i) The depletion layer is narrow and zener breakdown occurs when both sides of the PN junction are strongly doped.


(ii) Zener breakdown produces a a strong electric field by applying a small reverse bias voltage.


(iii) In Zener Breakdown, the covalent bonds are braked by the field, hence producing large number of electrons and holes.


(iv) The electrons and holes yields to the reverse saturation current (Zener current). This zener current and the applied voltage is independent.


12. What is the use of a zener diode


The main uses of zener diodes are:

(a) Zener diode can be used as constant - voltage diodes

(b) It can be used as voltage detection devices.#

(c) It provides voltage clipping


13. Give the Symbol of Zener Diode

14. Aim of Zener Diode Experiment

 

(i) To draw I—V characteristic of a zener diode under reverse biased condition.

(ii) To determine the zener breakdown voltage from the graph.

 

15. Apparatus of Zener Diode Experiment

 

A zener diode (ex: FZ 6.2 A), milliammeter, multimeter or dc voltmeter, dc regulated supply which can be varied from 0 to 10 V in steps of IV, 100 Ω resistor, etc.

 

16. Procedure of Zener Diode Experiment

Connections are made as shown in the figure. The type of zener diode is noted. Its breakdown voltage Vz, maximum current rating and maximum power rating are noted from the data book. (This helps us to fix the value of current limiting resistor R and the range of voltmeter or multimeter).


The regulated supply is switched on. The voltage V across the zener diode is measured between the points C and D by the multimeter and the zener current I is noted in the milliammeter. The zener voltage V is varied from zero in steps of, say, IV by varying the input voltage Vi of the regulated till the zener current I is about 50% of the maximum rated value of the zener current. In each step the zener current I is noted. The input voltage Vi is also measured in each step across the points A and B by the multimeter. Once the break down occurs zener voltage remains almost constant.

A graph is plotted between I and V in the third quadrant of the graph paper This gives the I—V characteristic of the zener diode. From the graph the breakdown voltage Vz of the zener diode is noted at any point on the steep portion of the graph. The dynamic resistance of the diode is also calculated at the point.

Pollen Germination Experiment for Class 12

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STUDY OF RATES OF POLLEN GERMINATION OF VARIOUS SPECIES IN SUGAR AND BORON SOLUTION

Introduction

 

Pollen grains or microspores are the male gametophyte of angiosperms. The process of transfer of pollen grains from anther to the stigma of a flower is called pollination. Pollen grains germinate after reaching on the stigma. A pollen grain has usually two cells, generative cells and vegetative cell at the time of pollination. A mature pollen grain has 2 layers outer thick exine and inner thin intine. Here and there in the exine small thin areas are present, they are called germ pores. During pollen germination a pollen tube grows out from the pollen grain through a germ pore. The generative cell move down to the pollen tube and divide into two male gametes. The pollen tube enters the embryosac and releases the male gametes to ensure fertilisation. The rate of pollen germination differs in different plant species.

 

Materials and Methods

 

Flowers of different plants, cavity slides, sucrose, boric acid, distilled water, beaker, micro-scopes etc were collected.

 

A nutrient solution was prepared by dissolving 1 gm of sucrose in 100 ml distilled water. Five clean cavity slides were taken and a few drops of nutrient solution was put in the cavity of each slides. In this nutrient solution the pollen grains from different flowers were dusted. Similarly another nutrient medium was prepared by dissolving 1 gm of boric acid in 100ml of distilled water and the experiment was repeated with the pollen grains of flowers of different plants.

 

Observation and Results

 

Observe each slide containing sucrose solution one by one under five different microscopes after 5 minutes and after one minute each. Similarly observe each slide containing boric acid solution one by one under five different microscopes. Record the observations in the tabular column given below.

 

Rates of pollen germination of different species of plants in sucrose solution


Rates of pollen germination of different species of plants in boric acid solution


Discussions and Conclusions

 

The viable pollen grains germinate in both the nutrient solutions. But the germination percentage varies in boric acid solution and sucrose solution. This also depends upon the species variation and time factor.

 

References

 

Give a list of books and other reading materials you have referred for the present project work.

 

Viva Questions and Answers

 

1. What are viable pollen grains?


The pollen grains which are able to germinate and produce male gametes are called viable pollen grains.

 

2. Why a nutrient solution is required for pollen germination?


Because nutrients are essential for the growth of pollen tube.

 

3. Where do pollen grains germinate?


Pollen grains germinate on the stigma.