Category: School

 

S.No.		Contents	II	Page No.
I.		Theory		4
II.		Procedure		6
III.		Observation		7
IV.		Conclusion		8
V.		Bibliography		8

THEORY

Soap – Soap are the sodium or potassium salt of higher fatty acids. The fatty acid contains long chain of 16-18 carbon atoms.

Structure Of  Soap –

Soap contains two parts:

1. A long hydrocarbon chain, which is water repelling called a non polar tail.
2. Anionic part which is water attracting called hydrophobic. It is called polar tail.

Soap may be represented as :

CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂-CH₂– CH₂-CH₂-CH₂-CH₂-CH₂-COONa

Soap are also made from animal fats and vegetable oil. Fats and oils are esters of higher fatty acids are called Glycerides. When oils and fats are heated with a solution of NaOH, they break down to the sodium salt of respective fatty acid soap and glycerol. This process of making soap by hydrolysis of fats and oil with alkalis is called saponification. The soap is separated from the solution by an addition of common salt NaCl. Salt is added to the soap solution to decrease the solubility of soap due to which soap separates out from the solution in the form of solid and starts floating on the surface. The crust of soap thus formed is removed and put it in moulds to get soap cakes. The solution left behind contains glycerol and NaCl.

Limitation Of Soap –

Soap is not suitable for washing clothes with hard water because of the following reasons:

Hard water contains salt of Ca and Mg, when soap is added to hard water, Ca and Mg ions of hard water react with soap forming insoluble Ca and Mg salt of fatty acids.

2C₁₇H₃₅COONA + MgCl₂ – (C₁₇H₃₅COO)₂ Mg + 2NaCl

2C₁₇H₃₅COONA + MgCl₂ – (C₁₇H₃₅COO)₂ Ca + 2NaCl

Therefore a lot of soap is washed if water is hard.

When hard water is used, soap forms insoluble precipitates of Ca and Mg salt from which sticks of clothes being washed. Therefore it interfere with the cleansing ability of the soap and makes the cleansing process difficult.

PROCEDURE

1. Take 10 gm of quantity of each sample in which percentage of fatty material has to be determined.
2. Prepare the solution of each soap in water. Add 10 to 12 drops of HCl  in each solution and heat the solution for 5 to 10 min.
3. Fatty matter float on the soap solution surface by forming upper layer and how by filter paper are weighed for titration.
4. Now collect the fatty material from each solution by filtrate ion and again weigh the filter including filtrate (fatty material) are dissolved in the filtrate (fatty material) in ether for calculating oil materials.
5. Now take the solution in separating flask on the surface of solution and remove the solution except oily material.
6. Now, remaining solution is exposed in sunlight to evaporate ether from solution.
7. Now oily matter can be easily weighed by weighing machine.

OBSERVATION

S.NO.	NAME OF SOAP	WEIGHT OF FILTER PAPER (A)	WEIGHT OF FILTER FATTY MATERIAL (B)	FATTY MATERIAL (B-A)	PERCENTAGE [ % ]
1. 2. 3. 4.	Santoor Lux Lux Int. Lifebuoy	0.3856 0.2956 0.3203 0.4701	7.5856 7.5956 7.2701 7.2701	7.2 7.3 7.5 6.8	72 % 73 % 75 % 68 %

CONCLUSION

Soap contains alkali matter, which affects our skin and even skin may crack. To maintain the oily and moisture balance on our skin, fatty material required in soap. In general, the fatty matter in soap is approximately 70% to 80% fatty matter below 70% made our skin dry, rough and skin may crack whereas highest percentage [%] of fatty matter above 80% made the soap sticky and oily and washing become very difficult. From the table it is clear that the Lux international is the best soap for bathing purpose because it contains large amount of TFM or maximum percentage[%] of TFM.

BIBLIOGRAPHY

i     Introduction of Chemistry by Comprehensive.

ii     The complete reference Chemistry by S.Chand.

You can find other Chemistry Projects here.

 

 

S.No.		Contents	II	Page No.
I.		Theory		4
II.		Experiment 1		8
III.		Observation		9
IV.		Experiment 2		10
V.		Observation		11
VI.		Bibliography		12

Theory

Evaporation is the process whereby atoms or molecules in a liquid state (or solid state if the substance sublimes) gain sufficient energy to enter the gaseous state.

The thermal motion of a molecule must be sufficient to overcome the surface tension of the liquid in order for it to evaporate, that is, its kinetic energy must exceed the work function of cohesion at the surface. Evaporation therefore proceeds more quickly at higher temperature and in liquids with lower surface tension. Since only a small proportion of the molecules are located near the surface and are moving in the proper direction to escape at any given instant, the rate of evaporation is limited. Also, as the faster-moving molecules escape, the remaining molecules have lower average kinetic energy, and the temperature of the liquid thus decreases.

If the evaporation takes place in a closed vessel, the escaping molecules accumulate as a vapour above the liquid. Many of the molecules return to the liquid, with returning molecules becoming more frequent as the density and pressure of the vapour increases. When the process of escape and return reaches equilibrium, the vapour is said to be “saturated,” and no further change in either vapour pressure and density or liquid temperature will occur.

Factors influencing rate of evaporation:-

1. Concentration of the substance evaporating in the air. If the air already has a high concentration of the substance evaporating, then the given substance will evaporate more slowly.

2. Concentration of other substances in the air. If the air is already saturated with other substances, it can have a lower capacity for the substance evaporating.

3. Temperature of the substance. If the substance is hotter, then evaporation will be faster.

4. Flow rate of air. This is in part related to the concentration points above. If fresh air is moving over the substance all the time, then the concentration of the substance in the air is less likely to go up with time, thus encouraging faster evaporation. In addition, molecules in motion have more energy than those at rest, and so the stronger the flow of air, the greater the evaporating power of the air molecules.

5. Inter-molecular forces. The stronger the forces keeping the molecules together in the liquid or solid state the more energy that must be input in order to evaporate them.

6. Surface area and temperature: –

Because molecules or atoms evaporate from a liquid’s surface, a larger surface area allows more molecules or atoms to leave the liquid, and evaporation occurs more quickly. For example, the same amount of water will evaporate faster if spilled on a table than if it is left in a cup.

Higher temperatures also increase the rate of evaporation. At higher temperatures, molecules or atoms have a higher average speed, and more particles are able to break free of the liquid’s surface. For example, a wet street will dry faster in the hot sun than in the shade.

Intermolecular forces: –

Most liquids are made up of molecules, and the levels of mutual attraction among different molecules help explain why some liquids evaporate faster than others. Attractions between molecules arise because molecules typically have regions that carry a slight negative charge, and other regions that carry a slight positive charge. These regions of electric charge are created because some atoms in the molecule are often more electronegative (electron-attracting) than others. The oxygen atom in a water (H₂O) molecule is more electronegative than the hydrogen atoms, for example, enabling the oxygen atom to pull electrons away from both hydrogen atoms. As a result, the oxygen atom in the water molecule carries a partial negative charge, while the hydrogen atoms carry a partial positive charge. Water molecules share a mutual attraction—positively charged hydrogen atoms in one water molecule attract negatively charged oxygen atoms in nearby water molecules.

Intermolecular attractions affect the rate of evaporation of a liquid because strong intermolecular attractions hold the molecules in a liquid together more tightly. As a result, liquids with strong intermolecular attractions evaporate more slowly than liquids with weak intermolecular attractions. For example, because water molecules have stronger mutual attractions than gasoline molecules (the electric charges are more evenly distributed in gasoline molecules), gasoline evaporates more quickly than water.

Experiment no.1

Aim:

To compare the rate of evaporation of water, acetone and diethyl ether.

Materials required:

China dish, Pipette, Beaker, Weighing balance Measuring flask, Acetone, Distilled water, Diethyl ether, Watch

PROCEDURE:

1. Take three china dishes.

2. Pipette out 10 ml of each sample.

3. Dish A-Acetone

Dish B-Water

Dish C-Diethyl ether

4. Record the weights before beginning the experiment.

5. Leave the three dishes undisturbed for ½ an hr and  then wait patiently.

6. Record the weights of the samples after the given time.

7. Compare the prior and present observations.

OBSERVATION:

	Water (gm)	Acetone (gm)	Diethyl Ether (gm)
Weight of dish	50	50	50
Weight of (dish + substance) before evaporation	60	57.85	57
Weight of (dish + substance) after evaporation	59.8	55.55	54.33
Weight of substance evaporated	0.2	2.30	2.67

Inference and conclusion: –

The rate of evaporation of the given three liquids is in order :-

Diethyl Ether>Acetone>Water

Reason: –

Water has extensive hydrogen bonding in between oxygen atom of one molecule and hydrogen atom of another molecule. But this is absent in the case of acetone.

Experiment no.2

Aim:-To study the effect of surface area on the rate of evaporation of Diethyl ether.

Requirements

Three Petri dishes of diameter 2.5 cm,5 cm, and 10 cm with covers ,10 ml pipette and stopwatch.

Procedure

1. Clean and dry the Petri dishes and mark them as A,B,C.

2. Pipette out 10 ml of Diethyl ether in each of the Petri dishes a, band C cover them immediately.

3. Uncover all the three Petri dishes simultaneously

and start the stopwatch.

4. Note the time when diethyl ether evaporates completely from each Petri dish.

Observation

Petri dish Mark	Diameter of Petri dish	Time taken for complete evaporation
A	2.5 cm	11min 45sec
B	5.0 cm	8min 45sec
C	7.5 cm	6min 30sec

Result

It will be observed that maximum evaporation occurs in Petri dish with largest diameter followed  by smaller and the smallest Petri dish. It is, therefore , concluded that rate of evaporation increases with increase in surface area.

BIBLIOGRAPHY

• NCERT CHEMISTRY XII
• ENCARTA ENCYCLOPEDIA 2009

You can find other Chemistry Projects here.

Study the electrolysis of products of Potassium Iodide (KI)

 

S.No.		Contents	II	Page No.
I.		Theory		4
II.		Procedure		8
III.		Observation		9
IV.		Precaution		10
V.		Conclusion		10
VI.		Bibliography		11

THEORY

Electrolysis-

It is defined as a process of decomposition of an electrolyte by the passage of electricity through its aqueous solution  or molten (fused) state.

Mechanism  Of  Electrolysis-

Whenever an electrolyte is dissolved in water or is taken in the molten state, the electrolyte dissociates to produce Positively and Negatively charged ions. On passing electric current, the positively charged ions move towards the cathode and hence are called cations, whereas the negatively charged ions move towards the anode and hence are called anions. On reaching their respective electrodes, ions lose their charge and become neutral. The cations accept electrons from the cathode to become neutral species. Thus, oxidation occurs at the anode while reduction takes place at the anode. The conversion of ions into neutral species at their respective electrodes is called Primary change. The product formed as a result of primary change may be collected as such or it may go under a Secondary change to form the final products.

Quantitative Aspects Of Electrolysis-

Michael Faraday was the first scientist who described the quantitative aspects of electrolysis.

Faraday’s Laws Of Electrolysis-

First Law:-

The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).

Second Law :-

The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (atomic mass of metal – number of electrons required to reduce the cations).

Products Of Electrolysis –

Products of electrolysis depend on the nature of material being and the type of electrodes being used .If the electrode is inert, it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert electrodes. The products of electrolysis depend on the different oxidizing and reducing species present in the electrolytic cell and their standard electrode potentials. Moreover, some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these do not seem to take place and extra potential (called overvoltage) has to be applied, which makes such processes more difficult to occur.

Reactions involved:-

In the electrolysis of an aqueous solution of KI, I ions are oxidized at the anode preferentially to water molecules. Possible reactions at anode are as follows:-

2 I^– (aq) I₂ (g) + 2 e^– …………(1)

2 H₂O (l) 4 H⁺ (aq) + O₂ + 4e^– ………….(2)

Reaction (1) occurs in preference to reaction (2) due to standard electrode potential value of the following reaction.

I₂ (g) + 2 e^– 2 I^– (aq)                           …………(3)

E^o/volt = + 0.53V

4 H⁺ (aq) + O₂ (g) + 2e^– 2 H₂O (l)                …………(4)

E^o/volt = + 1.53V

Possible cathode reactions are:

K⁺ (aq) + e^– K (s)                                                 …………..(5)

E^o/volt = – 2.92V

2 H₂O (l) + 2e^– H₂ (g) + 2 OH^– (aq)                      …………..(6)

E^o/volt = – 0.83V

E^o value of reduction reaction (5) is much smaller than that of reaction (6). Thus, reaction (6) occurs competitively over reaction (5) at cathode .Thus, violet colour of anode is due to formation of iodine and its subsequent reaction with starch  Pink colour at cathode is due to formation of OH^– ions which also render the solution alkaline. OH^–ions give pink colour with phenolphthalein.

Procedure

Prepare 0.1M solution of potassium iodide. Fix a U- shaped tube in a stand and insert two graphite electrodes into both ends of the U- tube through the corks. Assemble the apparatus as shown in the figure. Take about 30ml of 0.1M solution of potassium iodide in a 100ml beaker add five or six drops of phenolphthalein solution and five to six drops of freshly prepared starch solution. Stir the solution and transfer it into an electrolysis – tube fitted with graphite electrodes. Pass electric current through the electrolyte and observe the appearance of colour. A pink colour appears at the cathode and a violet colour appears at the anode. Bubble formation also occurs on the surface of the cathode.

Observations

TEST  SOLUTIONS	OBSERVATIONS	INFERENCE
Aqueous solution of potassium iodide with five drops of phenolphthalein and five drops of starch solution.	At the anode, violet colour. At the cathode: (i )Pink colour (ii)Formation of bubbles	Free iodine is evolved. (i)OH– ion is formed (ii)Hydrogen is evolved

Precautions

1)    Both the electrodes should be loosely fixed into the U- tube so as to allow the escape of evolved gasses.

2)    Electrodes should be cleaned before use.

Conclusion

In the electrolysis of an aqueous solution of potassium iodide, I^– ions are oxidized at the anode preferentially to water molecules. Violet colour at anode is due to iodine. Pink colour at cathode is due to formation of OH^– ions which renders the solution alkaline. OH^– ions give pink colour with phenolphthalein.

BIBLIOGRAPHY

• NCERT CHEMISTRY XII
• ENCARTA ENCYCLOPEDIA 2009

You can find other Chemistry Projects here.

 

Study of the Effect of Metal Coupling on the Rusting of Iron

 

S.No.		Contents	II	Page No.
I.		Introduction		4
II.		Aim Of The Project		6
III.		Requirement		7
IV.		Procedure		8
V.		Observation		9
VI.		Conclusion		9
VII.		Bibliography		10

Introduction:

Metals and alloys undergo rusting and corrosion. The process by which some metals when exposed to atmospheric condition i.e., moist air, carbon dioxide form undesirable compounds on the surface is known as corrosion, The compounds formed are usually oxides . Rusting is also a type of corrosion but the term is restricted to iron or products made from it. Iron is easily prone to rusting making its surface rough. Chemically, rust is a hydrated ferric oxide

Titanic‘s bow exhibiting microbial corrosion damage in the form of ‘rusticles’

Rusting an Electrochemical Mechanism ;

Rusting may be explained by an electrochemical mechanism. In the presence of moist air containing dissolved oxygen or carbon dioxide, the commercial iron behave as if composed of small electrical cells. At anode of cell, iron passes into solution as ferrous ions. The electron moves towards the cathode and form hydroxyl ions. Under the influence of dissolved oxygen the ferrous ions and hydroxyl ions interact to form rust, i.e., hydrated ferric oxide.

Methods of Prevention of Corrosion and Rusting

Some of the methods used to prevent corrosion and rusting are discussed here :

1) Barrier Protection : In this method , a barrier film is introduced between Iron surface and atmospheric air. The film is obtained by painting, varnishing etc.

2) Galvanization ; The metallic iron is covered by   a layer of more reactive metal such as zinc. The active metal losses electrons in preference of iron. Thus, protecting from rusting and corrosion.

Galvanized Metals

Aim of the project

In this project the aim is to investigate effect of the metals coupling on the rusting of iron. Metal coupling affects the rusting of iron . If the nail is coupled with a more electro-positive metal like zinc, magnesium or aluminium rusting is prevented but if on the other hand , it is coupled with less electro – positive metals like copper , the rusting is facilitated.

Requirement :

1)Two Petri dishes

2) Four test – tube

3) Four iron nails

4) Beaker

5) Sand paper

6)Wire gauge

7) Gelatine

8) Copper, zinc & magnesium strips

9)Potassium ferrocyanide solution

10)Phenolphthalein

Procedure

1)At first we have to clean the surface of iron nails with the help of sand paper.

2) After that we have to wind zinc strip around one nail, a clean copper wire around the second & clean magnesium strip around the third nail. Then to put all these three and a fourth nail in Petri dishes so that they are not in contact with each other.

3) Then to fill the Petri dishes with hot agar solution in such a way that only lower half of the nails are covered with the liquids .Covered Petri dishes for one day or so.

4) The liquids set to a gel on cooling. Two types of patches are observed around the rusted nail, one is blue and the other pink. Blue patch is due to the formation of potassium ferro-ferricyanide where pink patch is due to the formation of hydroxyl ions which turns colourless phenolphthalein to pink.

Observation

S.No.	Metal Pair	Colour of the patch	Nails rusts or not
1	Iron- Zinc
2	Iron -Magnesium
3	Iron- Copper
4	Iron – Nail

Conclusion

It is clear from the observation that coupling of iron with more electropositive metals such as zinc and magnesium resists corrosion and rusting of iron. Coupling of iron with less electropositive metals such as copper increases rusting.

BIBLIOGRAPHY

• NCERT CHEMISTRY XII
• ENCARTA ENCYCLOPEDIA 2009

You can find other Chemistry Projects here.

Introduction

When substances are brought in contact with each other, they intermingle with each other. This phenomenon is known as diffusion. Diffusion takes place very rapidly in case of gases, to a lesser extent in case of liquids, and not at all in the case of solids.

However, diffusion of solids in liquids does take place, albeit at a very slow rate. If a solid is kept in contact with excess of solvent in which it is soluble, some portion of the solid gets dissolved. This process is known as dissolution of a solid in liquid, and it takes place due to the diffusion of solid particles into liquid medium.

Molecules of solute are in constant random motion due to the collision between molecules of solute and that of the solvent. It is this physical interaction between solute-solvent particles that leads to diffusion.

Objective

To demonstrate that rate of diffusion depends upon the following factors:

 Temperature: As temperature increases, the kinetic energy of the particles increases. Thus, the speed of particles also increases, which in turn increases the rate of diffusion.

 Size of the particle: As the size of particle increases, rate of diffusion decreases. This is because the particles become less  mobile in the solvent.

 Mass of the particle: As the mass of the particle increases, the rate of diffusion decreases; as the particle becomes less mobile.

Experiment 1

To study diffusion when copper sulphate is brought in contact with water (liquid).

Requirements

Copper sulphate (CuSO₄) crystals, 100 mL beaker

Procedure

 Take about two grams of copper sulphate crystals in 100 mL beaker.

 Add about 50 mL of water and allow it to stand for few minutes.

 Note the development of blue colour in water.

 Allow to stand further till it is observed that all copper sulphate disappears.

 Note the blue colour change in water.

Conclusion

When solids such as copper sulphate are brought in contact with liquids such as water, intermingling of substances, i.e., diffusion takes place.

Experiment 2

To study the effect of temperature on the rate of diffusion of solids in liquids.

Requirements

Copper sulphate (CuSO₄) crystals, three 100 mL beakers, watch glass, wire gauge, burner, tripod stand, thermometer, stop watch.

Procedure

 Take five gram of copper sulphate each in three beakers.

 Pour 100 mL of distilled water slowly in one of the beakers.

 Cover this beaker with a watch glass.

 Pour 100 mL of cold water in a second beaker slowly.

 Place a third beaker containing 100 mL of water on a tripod stand for heating.

 Observe the diffusion process which begins in all the beakers.

 Record of copper sulphate the time taken for the dissolution of copper sulphate in all the three cases.

Observations

S. No.	Crystal Size	Time Taken to Diffuse
1	Big	19 minutes
2	Medium	13 minutes
3	Small	5 minutes

Conclusion

The rate of diffusion of copper sulphate in water is in the order as given below:

Beaker 3 > Beaker 2 > Beaker 1

Thus, the rate of diffusion varies directly with temperature.

Experiment 3

To study the effect of size of particles on the rate of diffusion of solids in liquids.

Requirements

Graduated 100 mL measuring cylinders, copper sulphate (CuSO₄) crystals of different sizes, stop watch.

Procedure

 Add 50 mL of water to each of the three cylinders.

 Take five gram each of big size, medium size, small size crystals of copper sulphate, and add them separately in three cylinders.

 Allow them to stand for sometime.

 Note the time taken for blue colour to reach any fixed mark in each of the cylinders and note the observations.

Observations

 

S. No.	Crystal Size	Time Taken to Diffuse
1	Big	19 minutes
2	Medium	13 minutes
3	Small	5 minutes

Conclusion

The rate of diffusion of copper sulphate in water is in the order as given below:

Beaker 3 > Beaker 2 > Beaker 1

Thus, smaller particles undergo diffusion more quickly than bigger particles.

Result

 When solids such as copper sulphate are brought in contact with liquids such as water, intermingling of substances, i.e., diffusion takes place.

 The rate of diffusion varies directly with temperature.

 Small particles undergo diffusion more quickly than bigger particles.

Bibliography

 Chemistry (Part I) – Textbook for Class XII; National Council of Educational Research and Training

 Concepts of Physics 2 by H C Verma; Bharti Bhawan (Publishers & Distributors)

Bibliography

NCERT Chemistry for Class XII

You can find other Chemistry Projects here.

 

S.No.		Contents	II	Page No.
I.		Introduction		4
II.		Objective		5
III.		Experiment 1		6
IV.		Experiment 2		7
V.		Experiment 3		9
VI.		Result		11
VII.		Bibliography		12

INTRODUCTION

When substances are brought in contact with each other they intermix, this property is known as Diffusion. This property of diffusion takes place very rapidly in case of gases and to a lesser extent in case of liquids, whereas solids do not show this process of diffusion with each other. But what we can observe in case of solids is that the diffusion of solids in liquids takes place at a very slow rate.

If a solid is kept in contact with an excess of solvent in which it is soluble, some portion of the solid gets dissolved. We know that this process is known as dissolution of a solid in liquid and this process has taken place due to the diffusion of solid particles into liquid.

Molecules of solute are in constant random motion due to the collision between molecules of solute and that of the solvent.

OBJECTIVE

Rate of diffusion depends upon:-

Temperature: As temperature increases, the kinetic energy of the particles increases so the speed of particles also increases which thus increases the rate of diffusion.

Size of the particle: As the size of particle increases, rate of diffusion decreases.

Mass of the particle: As the mass of the particle increases the rate of diffusion decreases.

EXPERIMENT 1

To study diffusion when copper sulphate is brought in contact with water (liquid)

REQUIREMENTS:

Copper sulphate crystals, 100ml beaker.

PROCEDURE:

• Ø Take about 2g of copper sulphate crystals in 100ml beaker.
• Ø Add about 50ml of water and allow it to stand for few minutes.
• Ø Note the development of blue colour in water.
• Ø Allow to stand further till it is observed that all copper sulphate disappears.
• Ø Note the blue colour change in water.

CONCLUSION:

When solids such as copper sulphate, potassium permanganate are brought in contact with liquids such as water, intermixing of substances, i.e. diffusion takes place.

EXPERIMENT 2

To study the effect of temperature on the rate of diffusion of solids in liquids

REQUIREMENTS:

Copper sulphate crystals, 200ml beaker, watch glass, wire gauge, burner, tripod stand, thermometer and stop watch.

PROCEDURE:

• Ø Take 5g of copper sulphate each in three beakers.
• Ø Pour 100ml of distilled water slowly in one of the beakers.
• Ø Cover this beaker with a watch glass.
• Ø Pour 100ml of cold water in a second beaker slowly.
• Ø Place a third beaker containing 100ml of water on a tripod stand for heating.
• Ø Observe the diffusion process which begins in all the beakers.
• Ø Record the time taken for the dissolution of copper sulphate in all the three cases.

OBSERVATIONS:

S.No.	Temperature of water	Time Taken in Minutes
1.	25 0C	15 Min.
2.	10 0C	20 Min.
3.	70 0C	10 Min.

CONCLUSION:

The Rate of diffusion of copper sulphate in water is in the order of Beaker 3 > Beaker 1 > Beaker 2. Thus, the rate of diffusion varies directly with temperature.

EXPERIMENT 3

To study the effect of size of particles on the rate of diffusion of solids in liquids

REQUIREMENTS:

Graduated 100ml measuring cylinders, copper sulphate crystals of different sizes, stop watch

PROCEDURE:

• Ø Add 50ml of water to each of the three cylinders.
• Ø Take 5g each of big size, medium size, small size crystals of copper sulphate and add them separately in three cylinders.
• Ø Allow to stand for sometime.
• Ø Note the time taken for blue colour to reach any fixed mark in each of the cylinders and note the observations.

OBSERVATION:

S.No.	Crystal size	Time Taken in Minutes
1.	Big	20 Min.
2.	Medium	15 Min.
3.	Small	10 Min.

CONCLUSION:

Small particles undergo diffusion more quickly than bigger particles.

RESULT

• Ø When solids such as copper sulphate, potassium permanganate are brought in contact with liquid such as water, intermixing of the substances, i.e. diffusion takes place.
• Ø The rate of diffusion varies directly with temperature.
• Ø Small particles undergo diffusion more quickly than bigger particles.

BIBLIOGRAPHY

• NCERT CHEMISTRY XII
• ENCARTA ENCYCLOPEDIA 2009

You can find other Chemistry Projects here.

 

 

S.No.		Contents	II	Page No.
I.		Introduction		4
II.		Apparatus		6
III.		Procedure		7
IV.		Precaution		9
V.		Bibliography		9

Introduction

Cellulose is nature’s own giant molecule. It is the fibrous material that every plant from seaweed to the sequoia makes by baking glucose molecules in long chains; the chains are bound together in the fibres that give plants their shape and strength. Wood has now become the main source of cellulose. Since it contains only 40% to 50% cellulose, the substance must be extracted by ‘pulping’. The logs are flaked, and then simmered in chemicals that dissolve the tarry lignin, resins and minerals. The remaining pulp, about 93% cellulose, is dried and rolled into sheets-raw material for paper, rayon and other products.

It can be obtained in 2 ways:

1. Viscose Process: Cellulose is soaked in 30% caustic soda solution for about 3 hrs. The alkali solution is removed and the product is treated with CSi. This gives cellulose xanthate, which is dissolved in NaOH solution to give viscous solution. This is filtered and forced through a spinneret into a dilute H₂SO₄ solution, both of which harden the gum like thread into rayon fibres. The process of making viscose was discovered by C.F. Cross and EJ. Bevan in 1891.
2. Cuprammonium Rayon: Cuprammonium rayon is obtained by dissolving pieces of filter paper in a deep blue solution containing tetra-ammine cupric hydroxide. The latter is obtained from a solution of copper sulphate. To it, NH₄OH solution is added to precipitate cupric hydroxide, which is then dissolved in excess of NH₃.

Reactions:

CUSO₄+ 2NH₄OH —> Cu(OH)₂+ (NH₄)₂S0₄

Pale blue ppt

Cu(OH) ₂ + 4NH₄OH —> [Cu(NH₃)₄](0H) ₂ + 4H₂O

[Cu(NH3) ₄](OH) ₂ + pieces of filter paper left for 10-15 days give a viscous solution called VISCOSE.

Apparatus Required

a) Conical flask (preferably 250 ml)

b) Funnel

c) Glass rod

d) Beaker (preferably 250 ml)

e)Water bath

f) Filter paper

Chemicals Required

a) CuSO₄

b) NaOH solution

c) Liquor ammonia solution

d) Dilute H₂SO₄

e) Whitman Paper

f) Distilled H₂O

Procedure

A. Preparation of Schweitzer’s Solution:

a)     Weight of CuSO₄.5H₂0.

b)     Transfer this to a beaker having 100ml distilled water and add 15ml of dilute H₂SO₄ to prevent hydrolysis of CuSO₄.

c)      Stir it with a glass rod till a clear solution is obtained. Add 11ml of liquor ammonia drop by drop with slow stirring. The precipitate of cupric hydroxide is separated out.

d)    Filter the solution containing cupric hydroxide through a funnel with filter paper.

e)     Wash the precipitate of cupric hydroxide with water until the filtrate fails to give a positive test for sulphate ions with barium chloride solution.

f)       Transfer the precipitate to a beaker that contains 50ml of liquor ammonia or wash it down the funnel. The precipitate when dissolved in liquor ammonia gives a deep blue solution of tetra-ammine cupric hydroxide. This is known as SCHWEITZER’S SOLUTION.

B. Preparation of Cellulose material

a)     After weighing 2g of filter paper divide it into very fine pieces and then transfer these pieces to the tetra-ammine cupric hydroxide solution in the beaker.

b)     Seal the flask and keep for 10 to 15 days, during this period the filter paper is dissolved completely.

C. Formation of Ravon Thread

a)      Take 50ml of distilled water in a glass container. To this add 20ml of conc H₂SO₄ drop by drop. Cool the solution under tap water. In a big glass container pour some of the solution.

b)     Fill the syringe with cellulose solution prepared before.

c)      Place the big glass container containing H₂SO₄ solution produced before in ice (the reaction being spontaneous results in excess release of energy in the form of heat which makes the fibres weak and breaks them).

d)     Immerse the tip of the syringe in the solution and press gently. Notice the fibres getting formed in the acid bath. Continue to move your hand and keep pressing the syringe to extrude more fibres into the bath.

e)      Leave the fibres in solution till they decolorize and become strong enough.

f)       Filter and wash with distilled water.

Precautions

a)     Addition of excess NH₃ should be avoided.

b)     Before taking the viscose in the syringe make sure that it does not contain any particles of paper, otherwise, it would clog the needle of the syringe.

c)      Addition of NH₃ should be done in a fume cupboard and with extreme care. The fumes if inhaled may cause giddiness.

d)    Use a thick needle otherwise the fibres won’t come out.

Bibliography

Chemistry (Part I) –  Textbook for Class XII

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S.No.		Contents	II	Page No.
I.		Introduction		4
II.		Material Required		5
III.		Theory		6
IV.		Procedure		8
V.		Conclusion		10
VI.		Bibliography		10

Introduction

An alloy is a homogeneous mixture of two or more metals or a metal and non-metal.

They are generally harder than their components with reduced malleability and ductility. Alloys are prepared to enhance certain characteristics of the constituent metals, as per requirement.

In this project, we shall qualitatively analyze the chemical composition of two alloys:

MATERIALS REQUIRED

1) BRASS AND BRONZE PIECES

2) CHINA DISHES

3) FILTRATION APPARATUS

4) NITRIC ACID

5) HYDROGEN SULPHIDE GAS

7) AMMONIUM CHLORIDE

8)POTASSIUM FERROCYANIIDE

9) AMMONIUM SULPHIDE

10) DIL HYDROCHLORIC ACID

Theory

Brass

Brass contains Cu and Zn . Both dissolve in nitric acid.

4Zn+ 10 HNO₃ –> 4Zn(NO)₂+ N₂O + 5HO

3Cu + 8 HNO₃ –> 3Cu(NO₃)₂ + 4HO+2NO

Further analysis is carried out for respective ions.

Cu dissolves in H₂S to give black ppt. of CuS. It is filtered to get the soln of Zinc Sulphide. It precipitates out in the form of ZnCl₂ in an ammonical soln. of Ammonium chloride. The precipitate is dissolved in dilute HCl and then treated with Potassium ferrocyanide to get a bluish-white ppt. of Zn₂[Fe(CN)₆].

Bronze

Bronze contains Cu and Sn. Their nitrates are obtained by dissolving the sample in conc. Nitric acid. The nitrates are precipitated as sulphides by passing H₂S through their solution in dil. HCl.

The CuS is insoluble in yellow ammonium sulphide, while SnS is soluble. The ppt. is separated by filtration.

The ppt. is dissolved in cone HNO₃ and then Ammonium hydroxide solution is passed through it. Blue colouration confirms the presence of Cu.

The filtrate is treated with conc. HCl followed by Zinc dust to obtain SnCl₂ . Then HgCl₂ solution is added. Formation of slate-coloured ppt. indicates the presence of Sn.

Procedure

Brass:

1. A small piece of brass was placed in a china dish and dissolved in minimum quantity of 50% conc. H₂SO₄.

2. The soln. was heated to obtain a dry residue. The residue was dissolved in Dilute HCl gas was passed and a black.ppt. was observed. The soln. was filtered and the ppt. was dissolved in NH₄OH soln. A blue coloration observed indicates the presence of Cu.

4.   The filtrate was tested for presence of Zn. Ammonium hydroxide and chloride solutions were added and then H₂S gas was passed. A dull grey ppt. was separated and dissolved in dil. HCl. followed by addition of Potassium ferrocyanide. A bluish white ppt. confirms the presence of Zn.

Bronze:

1. The sample was dissolved in 50% HNO₃ and then heated to obtain nitrates.

2. The nitrates were dissolved in dil. HCl and then precipitated as sulphides by passing H₂S gas.

3. The precipitates were treated with yellow amm.sulphide.

4. The ppt. was tested for Cu as in the case of brass.

5. The filtrate was treated with conc. HCl followed by Fe dust.

6. Then HgCl₂ soln. was added. Formation of a slate-coloured ppt. confirmed the presence of Sn.

Conclusion

Brass contains Copper and

Bronze contains Copper and Tin.

Bibliography

1. Comprehensive practical Chemistry- Class 12.

2. www.alloyanalyzer.niit.edu

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S.No.		Contents	II	Page No.
I.		Objective		4
II.		Introduction		5
III.		Theory		7
IV.		Material required		8
V.		Procedure		9
VI.		Observation		11
VII.		Bibliography		11

OBJECTIVE

The purpose of the experiment is – to compare the rate of fermentation of the given samples of wheat flour, gram flour, rice flour and potatoes.

I became interested in this idea when I saw some experiments on fermentation and wanted to find out some scientific facts about fermentation. The primary benefit of fermentation is the conversion of sugars and other carbohydrates ,e.g., converting juice into wine, grains into beer, carbohydrates into carbon dioxide to leaven bread, and sugars in vegetables into preservative organic acids.

INTRODUCTION

Fermentation typically is the conversion of carbohydrates to alcohols and carbon dioxide or organic acids using yeasts, bacteria, or a combination thereof, under anaerobic conditions. A more restricted definition of fermentation is the chemical conversion of sugars into ethanol. The science of fermentation is known as zymology. Fermentation usually implies that the action of microorganisms is desirable, and the process is used to produce alcoholic beverages such as wine, beer, and cider. Fermentation is also employed in preservation techniques to create lactic acid in sour foods such as sauerkraut, dry sausages, and yoghurt, or vinegar for use in pickling foods.

Fermentation in food processing typically is the conversion of carbohydrates to alcohols and carbon dioxide or organic acids using yeasts, bacteria or a combination thereof, under anaerobic conditions. A more restricted definition of fermentation is the chemical conversion of sugars into ethanol. The science of fermentation is known as zymology.

Fermentation usually implies that the action of microorganisms is desirable, and the process is used to produce alcoholic beverages such as wine , beer, and cider. Fermentation is also employed in preservation techniques to create lactic acid in sour foods such as sauerkraut , dry sausages,  and yogurt, or vinegar (acetic acid) for use in pickling foods.

Uses

Food fermentation has been said to serve five main purposes:

#  Enrichment of the diet through development of a diversity of flavours, aromas, and textures in food substrates

#  Preservation of substantial amounts of food through lactic acid, alcohol, acetic acid and alkaline fermentations

#  Biological enrichment of food substrates with protein, essential amino acids, essential fatty acids, and vitamins

#  Elimination of ant nutrients.

#  A decrease in cooking times and fuel requirements

Theory

Wheat flour, gram flour, rice flour and potatoes contains starch as the major constituent. Starch present in these food materials is first brought into solution. In the presence of enzyme diastase, starch undergoes fermentation to give maltose.

Starch gives blue-violet colour with iodine whereas product of fermentation starch do not give any characteristic colour. When the fermentation is complete the reaction mixture stops giving blue-violet colour with iodine solution.

By comparing the time required for completion of fermentation of equal amounts of different substances containing starch the rates of fermentation can be compared. The enzyme diastase is obtained by germination of moist barley seeds in dark at 15 degree celsius. When the germination is complete the temperature is raised to 60 degree celsius to stop further growth. The seeds are crushed into water and filtered. The filtrate contains enzyme diastase and is called malt extract.

MATERIALS REQUIRED

#   Conical flask

#   Test tube

#   Funnel

#   Filter paper

#   Water bath

#   1 % Iodine solution

#   Yeast

#   Wheat flour

#   Gram flour

#   Rice flour

#   Potato

#   Aqueuos NaCl solution

PROCEDURE

#   Take 5 gms of wheat flour in 100 ml conical flask and add 30 ml of distilled water.

#   Boil the contents of the flask for about 5 minutes

#   Filter the above contents after cooling, the filtrate obtained is wheat flour extract.

#   To the wheat flour extract. taken in a conical flask.

Add 5 ml of 1% aq. NaCl solution.

#    Keep this flask in a water bath maintained at a temperature of 50-60 degree celsius. Add 2 ml  of malt extract.

#   After 2 minutes take 2 drops of the reaction mixture and add to diluted iodine solution.

#   Repeat step 6 after every 2 minutes. When no bluish colour is produced the fermentation is complete.

#   Record the total time taken for completion of fermentation.

#   Repeat the experiment with gram flour extract, rice flour extract, potato extract and record the observations

OBSERVATIONS

Time required for the fermentation—

# Wheat flour — 10 hours

#  Gram flour –  12.5 hours

#  Rice flour — 15 hours

#  Potato — 13 hours

BIBLIOGRAPHY

#   Chemistry manual

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S.No.		Contents	II	Page No.
I.		Introduction		4
II.		Requirements		6
III.		Procedure		7
IV.		Observation		10
V.		Conclusion		13
VI.		Bibliography		13

INTRODUCTION

An Antacid is any substance, generally a base or basic salt, which neutralizes stomach acidity. They are used to relieve acid indigestion, upset stomach, sour stomach, and heartburn.

ACTION MECHANISM

Antacids perform a neutralization reaction, i.e. they buffer gastric acid, raising the pH to reduce acidity in the stomach. When gastric hydrochloric acid reaches the nerves in the gastrointestinal mucosa, they signal pain to the central nervous system. This happens when these nerves are exposed, as in peptic ulcers. The gastric acid may also reach ulcers in the oesophagus or the duodenum.

Other mechanisms may contribute, such as the effect of aluminium ions inhibiting smooth muscle cell contraction and delaying gastric emptying.

INDICATIONS

Antacids are taken by mouth to relieve heartburn, the major symptom of gastro oesophageal reflux disease, or acid indigestion. Treatment with antacids alone is symptomatic and only justified for minor symptoms. Peptic ulcers may require H2-receptor antagonists or proton pump inhibitors.

The utility of many combinations of antacids is not clear, although the combination of magnesium and aluminium salts may prevent alteration of bowel habits.

INTERACTIONS

Altered pH or complex formation may alter the bioavailability of other drugs, such as tetracycline. Urinary excretion of certain drugs may also be affected.

PROBLEMS WITH REDUCED STOMACH ACIDITY

Reduced stomach acidity may result in an impaired ability to digest and absorb certain nutrients, such as iron and the B vitamins. Since the low pH of the stomach normally kills ingested bacteria, antacids increase the vulnerability to infection. It could also result in reduced bioavailability of some drugs. For example, the bioavailability of ketoconazole (antifungal) is reduced at high intragastric pH (low acid content).

SOME FAMOUS ANTACID BRANDS

1. Alka-Seltzer – NaHCO₃ and/or KHCO₃

2. Equate – Al(OH)₃ and Mg(OH)₂

3. Gaviscon – Al(OH)₃

4. Maalox (liquid) – Al(OH)₃ and Mg(OH)₂

5. Maalox (tablet) – CaCO₃

6. Milk of Magnesia – Mg(OH)₂

7. Pepto-Bismol – HOC₆H₄COO

8. Pepto-Bismol Children’s – CaCO₃

9. Rolaids – CaCO₃ and Mg(OH)₂

10. Tums – CaCO₃

11. Mylanta

DRUG NAMES

Some drugs used as antacids are :

1. Aluminium hydroxide

2. Magnesium hydroxide

3. Calcium carbonate

4. Sodium bicarbonate

5. Bismuth subsalicylate

6. Histamine

7. Cimetidine

8. Ranitidine

9. Omeprazole

10. Lansoprazole

REQUIREMENTS :

• Burettes
• Pipettes
• Titration flasks
• Measuring flasks
• Beakers
• Weight box
• Fractional weights
• Sodium hydroxide
• Sodium carbonate
• Hydrochloric acid
• Phenolphthalein.

PROCEDURE :

1. Prepare 1 litre of approximately  HCl solution by diluting 10 ml of the concentrated acid for one litre.

2. Similarly, make 1 litre of approximately NaOH solution by dissolving4.0g of NaOH to prepare one litre of solution.

3. Prepare  Na₂CO₃ solution by weighing exactly 1.325 g of anhydrous sodium carbonate and then dissolving it in water to prepare exactly 0.25 litres (250 ml) of solution.

4. Standardize the HCl solution by titrating it against the standard Na₂CO₃ solution using methyl orange as indicator.

5. Similarly, standardize NaOH solution by titrating it against standardized HCl solution using phenolphthalein as indicator.

6. Powder the various samples of antacid tablets and weigh 1.0 g of each.

7. Add a specific volume of standardised HCl to each of the weighed sample is taken in conical flasks. The acid should be in slight excess, so that it can neutralize all the alkaline component of the tablet.

8. Add 2 drops of phenolphthalein and warm the flask till most of powder dissolves. Filter off the insoluble material.

9. Titrate this solution against the standardised NaOH solution, till a permanent pinkish tinge is obtained. Repeat this experiment with different antacids.

OBSERVATIONS:

Standardisation of HCl solution :

Volume of Na₂CO₃ solution taken = 20.0 ml

S No. of obs.	Burette readings Initial       Final	Volume of acid used
1. 2. 3. 4. 5.	0 ml          15.0 ml 0 ml          15.1 ml 0 ml          15.0 ml 0 ml          15.0 ml 0 ml          15.0ml	15.0 ml 15.1 ml 15.0 ml 15.0 ml 15.0 ml

Concordant volume = 15.0 ml

Applying normality equation,

N₁V₁ = N₂V₂

N₁ * 15.0 =  * 20

Normality of HCl, N₁ = = 0.133 N

Standardisation of NaOH solution :

Volume of the given NaOH solution taken = 20.0 ml

S No. of obs.	Burette readings  Initial     Final	Volume of acid used
1. 2. 3. 4. 5.	0 ml          26.5 ml 0 ml          26.8 ml 0 ml          26.6 ml 0 ml          26.6 ml 0 ml          26.6ml	26.5 ml 26.8 ml 26.6 ml 26.6 ml 26.6 ml

Concordant volume = 26.6 ml

Applying normality equation,

N₁V₁ = N₂V₂

0.133 x26.6 = N₂x20

Normality of NaOH = = 0.176 N

Analysis of antacid tablet :

Weight of antacid tablet powder = 1.0 g

Volume of HCl solution added = 20.0 ml

Antacid	Vol. Of NaOH soln. Used to neutralise unused HCl	Vol. Of HCl soln. Used to neutralise 1.0 g of antacid matter
1. Gelusil 2. Digene 3. Aludrox 4. Logas 5. Ranitidine 6. Ocid 20	12.1 ml 16.0 ml 19.3 ml 24.3 ml 21.4 ml 22.7 ml	12.0 ml 16.2 ml 18.9 ml 24.4 ml 21.7 ml 21.9 ml

CONCLUSION :

The antacid which has maximum volume of HCl is used for neutralizing i.e. Ocid 20 is more effective.

BIBLIOGRAPHY

1. WEBSITES :

• www.pharmaceutical-drugmanufacturers.com

2. BOOKS :

• Comprehensive Practical Manual for class XII
• Pradeep’s New Course Chemistry
• NCERT Class XII Part II

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