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Detection of Adulterants in Food Stuffs

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Investigatory project on brief study of the detection of presence of adulterants in food-stuffs

Introduction


Often many of the food—stuffs bought from the market are adulterated. This causes serious diseases like cancer, asthma, ulcers, etc. Chilli powder is sometimes adulterated with brick powder, pepper is mixed with dried papaya seeds etc.


Aim


The aim of this project is to detect the presence of common adulterants in food—stuffs such as red chilli powder, pepper, turmeric, butter, sugar etc.


Requirements


1. Samples of Chilli powder, turmeric-powder, sugar etc.

2. Dil. HNO3

3. KI solution


Procedure


a) Detection of red lead in chilli powder


Add dil. HNO3 to a sample of chilli powder, shake and filter. Add KI solution to the filtrate. Yellow precipitate indicates the presence of lead salts in chilli powder. Nitric acid converts lead salt to lead nitrate which forms yellow lead iodide with potassium iodide.


Pb(NO3)2 + 2KI —> PbI2 + 2KNO3


b) Detection of yellow lead salts in turmeric powder


Add conc. HCl to a little of the given turmeric powder. Formation of a magenta colour shows the presence of yellow oxides of lead in the turmeric powder.


c) Detection of dried papaya seeds in pepper


Add a small amount of the pepper to be tested to a beaker containing water and stir. Being lighter, dried papaya floats on water while pure pepper settles to the bottom.


d) Detection of brick powder in chilli powder


Add some of the given chilli powder to water taken in a beaker. Chilli powder floats on water and brick powder settles down.


e) Detection of chalk powder and washing soda in sugar


Add diI. HCl to a little of the sugar. Brisk effervescence of CO2 shows the presence of chalk powder or washing soda in it. Carbonates react with dil. HCl to liberate carbon dioxide as brisk effervescence


Na2CO3 + 2HCl —> 2NaCI + CO2 + H2O


f) Detection of insoluble impurities in sugar


Add water to the sugar sample and stir. Pure sugar dissolves while insoluble impurities settle to the bottom.


VIVA QUESTIONS WITH ANSWERS


1. What is the common adulterant in chilli powder?


Ans: Brick powder or red lead


2. What is the adulterant in pepper?


Ans: Dry papaya seeds


3. How will you detect red lead in chilli powder?


Ans: Chilli powder is heated with dil. HNO3. To this solution potassium iodide solution is added. A yellow precipitate shows the presence of red lead.


4. What is the yellow precipitate formed?


Ans: It is lead iodide


5. How will you detect papaya seeds in pepper?


Ans: Add the sample to water taken in a beaker. If papaya seeds are present they will float on the surface of water and pepper will settle down.


6. How will you detect chalk powder in sugar?


Ans: Add dil. HCl to the sample. A brisk effervescence shows the presence of chalk powder.


7. What is the gas that evolves during the addition of dil. HCl?


Ans: Carbon dioxide

Buffer Action of Acidic and Basic Buffer

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Investigatory project on Buffer Action of Acidic and Basic Buffer

Introduction


A buffer solution is a solution, where the pH of which does not change by the addition of small amounts of an acid or a base. The process by which a buffer solution resists the change in pH by the addition of an acid or a base is called buffer action. Two types of buffers are acid buffer and basic buffer.


Acid buffer


It is a mixture of a weak acid and its salt with a strong base. For example, a mixture of acetic acid and sodium acetate. Its buffer action can be explained as follows.


CH3—COOH + OH —> CH3—COO + H2O


Here pH is maintained the same by the neutralisation of acetic acid by OH— ions added.


CH3—COO + H3O+ —> CH3—COOH + H2O


In this case the pH is maintained by the reaction between acetate ion and hydronium ion.


Basic buffer


Basic buffer is a mixture of a weak base and its salt with a strong acid. For example, a mixture of ammonium hydroxide and ammonium chloride. Buffer action of this buffer solution can be explained as follows.


NH4OH + H3O+ —> NH4+ + 2H2O


Here the pH is maintained the same due to the neutralization of NH4OH by the acid added.


NH4+ + OH —> NH4OH


Here pH remains the same due to the reaction of OH ions from the base with NH4+ ions liberated from the strong electrolyte NH4Cl.


Requirements:


Acetic acid, sodium acetate, ammonium hydroxide, ammonium chloride, 0.1N hydrochloric acid, 0.1N NaOH, pH meter or pH paper, beakers etc


Aim


Aim of the investigatory project is to study the buffer actions of an acid buffer and a basic buffer.


Procedure


Preparation of Acid buffer solution


An acid buffer is prepared by dissolving 14.75 mL of N/5 acetic acid in 35.25 mL of N/5 sodium acetate solution. Note the pH of the solution


a) Buffer action of acid buffer


The acid buffer is taken in a 250mL beaker. Measure its pH using a pH meter or a pH paper. Now add 1 mL of 0.1N hydrochloric acid to the above solution. Stirr it well with a glass rod and note the pH of this solution. The experiment is repeated by using a buffer solution of acetic acid and sodium acetate as mentioned earlier and by adding 1mL of 0.1N NaOH solution. Note the pH of the solution before and after the addition of sodium hydroxide solution. Tabulate the results.


Preparation of Basic buffer solution


A basic buffer is prepared by mixing 1N solutions of NH4Cl and NH4OH. The pH of this solution is noted.


b) Buffer action of a basic buffer


The buffer solution is taken in a 250mL beaker. Measure the pH of this solution using a pH meter or a pH paper. Now add 1 mL of 0.IN HCI solution to the above buffer and note the pH again. Repeat the experiment by using the same quantity of the same buffer and then adding 1mL of 0.1N NaOH solution. Note the pH of this solution also and tabulate.


Observation


Buffer solution

pH of Buffer solution

pH of buffer solution after adding 0.1 NHCI

pH of buffer solution after adding 0.1N NaOH

Acid buffer (CH3COOH + CH3COONa)

 

 

 

Basic Buffer (NH4OH + NH4Cl)

 

 

 

 

Conclusion


The pH values of the buffer solutions remain the same after adding small amounts acid or base. This proves the buffer action of buffer solutions.


VIVA QUESTIONS WITH ANSWERS


1. What is a buffer solution?


Ans: A buffer solution is a solution the pH of which does not change by the addition of small amounts of an acid or a base.


2. How will you prepare an acid buffer?


Ans: An acid buffer is prepared by mixing a weak acid with its salt with a strong base.


3. Give the example of an acid buffer


Ans: A mixture of acetic acid and sodium acetate.


4. How does the pH of an acid buffer remain the same when acid is added to it?


Ans: The hydronium ions react with the salt of the acid to produce a weak acid.


5. What is buffer action?


Ans: The process by which a buffer solution resists the change in pH when an acid or a base is added to it is called buffer action.


6. Write equations which explain the buffer action of an acid buffer


Ans: Base reacts with weak acid as follows


CH3—COOH + OH —> CH3—COO + H20


Acid reacts with acetate ion as follows


CH3—COO + H3O+ —> CH3—COOH + H2O


7. Give the example of a basic buffer


Ans: A mixture of NH4OH and NH4CI


8. How does it act as a buffer?


Ans: When an acid is added to the buffer, it is neutralized by NH4OH


NH4OH + H+ —> NH4+ + H2O


When a base is added to the buffer, the OHions combine with NH4+ from NH4CI to form a weak base NH4OH.


NH4+ + OH —> NH4OH


9. What happens when excess acid is added to a buffer solution?


Ans: The pH value of the buffer solution decreases.


10. What happens when excess base is added to a buffer solution?


Ans: Its pH value increases.


Boiling Point Elevation Experiment Lab Report

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Lab Report on study of boiling points of solutions of various concentrations involving electrolytes and non electrolytes is as follows.


Introduction

The temperature at which the vapour pressure of the liquid is same as that the atmospheric pressure of liquid is called Boiling point. When a nonvolatile solute is added to a solvent the vapour pressure of the solution is decreased. Therefore, the boiling point of the solution is increased or elevated. Elevation in boiling point is a colligative property; ie, the boiling point depends on the number of solute particles. According to Raoult's law the elevation in boiling point is directly proportional to the molality of the solution. The relation between the elevation in boiling point and the molecular mass of the solute is given by


M = 1000 Kb WB/ ΔT WA


Where M is the molecular mass of solute, Kb the molal elevation constant of the solvent, WB the mass of the solute, WA is mass of the solvent and ΔT the elevation in boiling point.


Aim


To study the boiling points of solutions of various concentrations involving electrolytes and non—electrolytes


Requirements


Urea, NaCl, sensitive thermometer, boiling point apparatus etc.,


Procedure


Take 20gm of water in a boiling point apparatus. Fit it with a one holed cork through which a sensitive thermometer is introduced. It is clamped to an iron stand and is heated using a burner. The boiling point is noted. Now remove the thermometer and allow the liquid to cool in air and add 0.2g of urea in it and stir it with a glass rod. Introduce the thermometer in it. Heat the solution using a burner. Note the temperature at which the solution boils. Repeat the experiment by using 0.4g, 0.6g, and 0.8g of urea with the same quantity of water in each case. Note the boiling points of these solutions and tabulate. The above experiment is repeated by using 0.2g, 0.4g, 0.6g and 0.8g of sodium chloride in the same quantity of water. The boiling points of these solutions are also tabulated.


It is found that the boiling point of solution increases with increase in concentration of the solution. Compared to urea solution the boiling point of sodium chloride solution is found to be much higher.


Conclusion


Urea is a non electrolyte while sodium chloride is an electrolyte. The number of particles in a solution of an electrolyte is higher than that of a non electrolyte. Since elevation of boiling point is a colligative property, the elevation in boiling point of sodium chloride is higher than the expected value.


VIVA QUESTIONS WITH ANSWERS


1. What is meant by colligative property?


Ans: Property which depends upon the number of particles is called colligative property.


2. How is elevation in boiling point of a liquid related to the molecular mass of the solute dissolved in it?


Ans: Elevation in boiling point of a liquid is inversely proportional to the molecular mass of the solute dissolved in it.


3. Among equimolar solutions of urea and sodium chloride, which has higher boiling point?


Ans: Sodium chloride solution has higher boiling point


4. Why does sodium chloride solution have higher boiling point than glucose solution of the same concentration?


Ans: Sodium chloride being an ionic compound dissociates to give two particles per molecule. Since boiling point is a colligative property the boiling point of sodium chloride solution is higher than the expected value.


5. If you take equimolar solutions of glucose and urea which will have higher boiling point? Why?


Ans: They have the same boiling point. This is because, being covalent compounds, equimolar solutions of glucose and urea contain the same number of particles.


6. Name some examples of ionic compounds.


Ans: KCl, NH4Cl, K2SO4, Na2SO4 etc.

Foaming Capacity of Soap Viva Questions

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Foaming Capacity of Soap Viva Questions and Answers

1. What are soaps?


Soaps are sodium (Na) or potassium (K) salts of higher fatty acids such as stearic acid, palmitic acid and so on. Soap produces foam when shaken with water. This foam is responsible for the removal of dirt. A soap which produces more foam is more efficient in cleansing.


2.  Name two higher fatty acids.


Stearic acid and palmitic acid


3. Why are soaps soluble in water?


Being alkali metal salts, soaps are soluble in water.


4. How will you compare the foaming capacities of two varieties of soap?


By shaking equal masses of the two samples of soaps in equal volumes of water and then by measuring the time taken for the disappearance of the foam that is formed.


5. By knowing the foaming capacity how will you determine the quality of a soap?


The soap which has maximum foaming capacity is the best brand of soap.


6. What is the use of foam?


It is used to remove dirt.


7. What is hard water?


Hard water is the water that does not lather easily with soap.


8. How is hardness of water classified?


Temporary hardness and Permanent hardness


9. Name two sources of soft water?


Rain water, boiled water


10. Name two sources of hard water?


Ground water, sea water


11. Name the cleansing agent that can be used in hard water?


Detergents


12. Give two salts present in water?


Magnesium sulphate, Sodium chloride


13. Name two parts of a soap molecule.


Ionic (hydrophilic) and non-ionic (hydrophobic)


14. Examples of dissolved salts present in hard water?


Hydrogen carbonates, sulphates and chlorides of Ca or Mg. These, on boiling decompose and precipitates as their carbonates.


15. What is a micelle?

When soap molecules are dissolved in water they form a shape as shown in figure. The shape formed is called micelle. The ionic ends will points outside the water and the non ionic end will points towards the oil.


16. Aim of Foaming capacity of different soaps experiment.


The main objective of Foaming capacity of different soaps experiment is to compare the foaming capacities of different brands of soaps available in the market.


17. Apparatus required for Foaming capacity of different soaps experiment.


250 mL conical flasks, Numbered test tubes, Different types of soaps, Stop watch


18. Working Principle of Foaming capacity of different soaps experiment.


Solutions of different soaps are prepared by dissolving equal masses of the soaps in equal volumes of water. They are shaken vigorously to produce foam and then they are allowed to stand. The time taken for the disappearance of the foam is measured for the different samples. The soap, which takes maximum time for the disappearance of the foam has the maximum foaming capacity and it is the best soap.


19. Procedure of Foaming capacity of different soaps experiment.


Dissolve 0.5g each of the different samples of the soap in 50 mL of distilled water taken in different conical flasks labeled 1, 2, 3, 4,….. etc. Now take test tubes labeled 1, 2, 3, 4,…. Etc and add 10 mL distilled water to each of them. Add 1 mL each of the soap solution from conical flasks 1, 2, 3, 4,….. etc to test tubes 1, 2, 3, 4,….. etc respectively. Cork test tube 1 and shake for one minute. Place the test tube in the test tube stand and note the time in which the foam just disappears. Similarly, note the time for the disappearance of the foam in the other test tubes and record the data.

Determination of Melting Point and Boiling Point Experiment

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a. Determination of Melting Point Experiment

 

The melting point of a substance is defined as the temperature at which a substance changes from the solid state to the liquid state. A pure substance has a sharp melting point. Therefore, melting point is a characteristic physical constant of a substance.

 

Procedure

The most convenient apparatus for the determination of melting point is Thiele's tube. It is a hard glass test tube which is fused with the two ends of a V—shaped glass tube at the middle and at the bottom as shown in the figure. The tube can be closed with a one holed cork which carries a sensitive mercury thermometer. The cork is provided with a side slit. About three fourth of Thiele's tube is filled with a high boiling liquid like conc. sulphuric acid or liquid paraffin.

 

The substance, the melting point of which is to be determined, is powdered well. A capillary tube of about 5cm length, one end of which is fused, is taken and the powdered substance is introduced into it so as to get a column of about 0.5 cm.

 

The bulb of the thermometer is wetted with tube liquid paraffin or conc. sulphuric acid. Now the capillary tube is allowed to cling to the thermometer in such a way that the bulb of the thermometer and the specimen are as close as possible. The apparatus is now closed with the cork in such a way that the thermometer is introduced into the liquid. The bulb of the thermometer should be at a level of the upper side tube. The open end of the capillary tube should be above the liquid level. The apparatus is clamped to an iron stand and it is heated gently by waving the flame at the side tube. Due to the movement of the heated liquid molecules, the liquid will be uniformly heated up. Development of high pressure during heating is prevented by the escape of vapours through the slit on the cork. During heating the mercury thread slowly rises and at a particular time the solid substance in the capillary tube suddenly shrinks and melts. The temperature at this time is noted as the melting point of the substance.

 

Result : Melting point of the substance = ____°C

 

b. Determination of Boiling Point Experiment

Boiling point of a liquid is defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure exerted on the surface of the liquid. Since the atmospheric pressure varies at different places, the boiling point of a liquid also varies. Boiling point is a characteristic physical property of a liquid. The boiling point of a liquid varies with the presence of impurities in it. The boiling point of a liquid can be determined in a boiling point apparatus. The apparatus consists of a pyrex glass tube which is provided with a side tube. The side tube is connected to an adapter which in turn is introduced into a receiver.

 

The liquid, the boiling point of which is to be determined, is taken in the apparatus. A few porcelain pieces are introduced into the liquid to promote uniform heating and to prevent bumping of the liquid. The mouth of the apparatus is closed with a one holed cork carrying a thermometer.

 

The boiling point apparatus is slowly heated. The liquid boils and the vapours escape through the exit. The steady reading on the thermometer at which the liquid distills constantly gives the boiling point of the liquid. The first few drops of the condensed liquid are discarded and the rest of the liquid is collected. The apparatus is then cooled and the experiment is repeated with the distilled liquid to get concordant results.

 

Results: Boiling point of the liquid = ____°C


Sonometer Experiment with Readings (Class 11)

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Sonometer Experiment with Readings (Class 11)

Sonometer Experiment - I


Aim:


i. To show that n x l is a constant

ii. To find the unknown frequency of the given tuning fork.


Apparatus:


Sonometer, Tuning forks. Slotted weights, Rubber hammer etc.


The sonometer consists of a hollow rectangular wooden box with a peg at one end and a smooth vertical pulley at the other end. A wire is attached to the peg and other end carries a weight hanger which passes over the two bridges A and B and the pulley.


Theory:


The frequency (n) of transverse vibration of a string is inversely proportional to the length of the vibrating segment of the wire when tension (T) and linear density (m) are constants.

i.e., n ∝ 1/l

n x l = K, a constant

Unknown frequency n’ = K/l’

Where, l’ = resonating length for the tuning fork whose frequency is to be determined.


Procedure


1. Verification of the law:


The sonometer wire is stretched with a suitable mass M (say 2 kg) by placing on the weight hanger. The bridges A and B are kept close to each other. A thin paper rider is placed on the string between the bridges. The tuning fork of frequency ‘n’ is excited and its stem is pressed on the sonometer box. The bridges are adjusted until the paper rider vibrates with maximum amplitude and is thrown off. The length of the wire between the bridges (l) is measured. The experiment is repeated and the mean length (l) is found out. Then n x l is calculated. The experiment Is repeated with other tuning forks of different frequencies without changing the tension. Each case n x l is found to be a constant (K).

A graph is drawn with frequency ‘n’ along X-axis and ‘1/l’ along the Y-axis The graph will be a straight line.


(ii) To determine unknown frequency


Using the tuning fork whose frequency is to be determined, resonating length ‘l’ is measured. If n’ is its frequency, then n’ x l’ = K. Thus n’ = K/l’ can be calculated.


Observation Table and Readings


To study the relation between n and l


Frequency of tuning fork

Length of wire vibrating in unison

1/l (cm-1)

nl

1

2

Mean (cm)

288

23

23

23

0.0434

6624

320

17.5

17.7

17.6

0.05681

7497.6

480

15.4

15.6

15.5

0.0645

7440

Unknown Frequency

13.1

13.5

13.3

0.0751

 

 

Unknown frequency, x = nl/l’ = 6812.4/13.3 = 512.18 Hz


Results


i. n x l is found to be a constant.

ii. Frequency of the given tuning fork = 512.18 Hz

 

Sonometer Experiment - II


Aim:


To find the relation between length and tension, for a constant frequency and diameter of a stretched string.


Apparatus:


Sonometer, slotted weights, a tuning fork, paper rider, steel wire, etc.


Theory:


For fixed frequency of a string, the square root of the tension T of the string is directly proportional to the length l of the string. i.e., √T l. If the tension is in kgwt, M, then, M l, i.e., M l2 or M/l2 is a constant.


Procedure:


To study the relation between length and tension for a constant frequency.


For a fixed frequency, T/l is a constant where T is the tension of the string and l is the length of the vibrating segment of the string. If tension is in kg. wt (M),


M/l = a constant


Therefore, M/l2 = K, a constant i.e., M l2


The sonometer wire is stretched by placing a mass or 1 kg (M) on the weight hanger. The bridges A and B are placed close to each other. A light paper rider is placed on the string between the bridges. The tuning fork is excited and its stem is pressed on the sonometer box between A and B. The length of the vibrating segment is adjusted so that the paper rider vibrates vigorously. This happen when the frequency of the vibrating segment is same as that of the tuning fork. The length of the vibrating segment is measured. The experiment is repeated and the mean l is found out.


The experiment is repeated by keeping weights of 1.5 kg, 2kg … etc., on the weight hanger and the resonating length is found out in each case. A graph is drawn with M along the X-axis and l2 along the Y-axis. The graph is found to be a straight line. (M/l2) is calculated.


Observation Table and Readings


To study the relation between l and T


Frequency of the tuning fork, n = 288 Hz


Tension (M kgwt)

Length of wire vibrating in unison

l2 (cm2)

M/l2

1

2

Mean l (cm)

1

14.54

14.56

14.55

211.7

0.0047

1.5

15.5

15.7

15.6

243.66

0.0616

2

17.5

17.8

17.65

311.52

0.0064

Unknown Mass

21.4

21.6

21.5

462.25

 

 

Unknown mass, M/l2 = 0.006

M = l2 x 0.006

= 462.25 x 0.006 = 2.8 kg

 

Result:


i. l2 — M graph is drawn. The graph is found to be a straight line. Also, (M/l2) is found to be a constant. So, it is concluded that T l2.

ii. Unknown mass = 2.8 kg

Viscosity Experiment (Class 11)

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Viscosity Experiment for Class 11


Aim:


To determine the coefficient of viscosity of a viscous liquid by measuring the terminal velocity of a given spherical body.


Apparatus:


A tall glass jar, viscous liquid (ex: castor oil or diesel), 6 or 7 identical lead shots, screw gauge, stop watch etc.


Principle:


Terminal velocity, v= [2(ρ - σ)g/9η] x r2

Therefore, coefficient of viscosity of a viscous liquid, η = 2r2(ρ σ)g/9v

where, ρ – density of lead shot

σ – density of viscous liquid

r – radius of lead shot

v – terminal velocity of lead shot

Also, v = distance travelled by lead shot/time taken = AB/t

 

Procedure:


Two marks, A and B, at a measured distance S (say, 10 cm apart) are marked one vertically below the other on the outside surface of the jar by tying cotton threads around it. The mark B should be sufficiently well down the jar to ensure that the spherical body acquires terminal velocity when it fall through the viscous liquid in the jar before passing A. The jar is filled with the given liquid.


The radius r of a lead shot is measured with the screw gauge. It is dropped gently into the liquid. The time of descend t of the lead shot between A and B is noted using the stop-watch. The terminal velocity v of the lead shot is calculated (v = S/t). The experiment is repeated with lead shots of different radii.


Plot a graph between the terminal velocity v along the y-axis and the square of the radius r2 of lead shots along the x-axis. The graph is a straight line.


The terminal velocity is given by the equation v= [2(ρ - σ)g/9η] x r2


The slope m of the v — r2 graph gives 2(ρ - σ)g/9η. From this η, the coefficient of viscosity of the liquid is calculated.


η = 2(ρ σ)g/9 m


where ρ is the density lead shots and σ is the density of the liquid. (They are noted from the physical table).


Observation Table and Readings

(a) Radius of the lead shots


Pitch of the screw = _____ mm

Least count = ____ mm

Zero correction = ____ div

 

Lead Shot

Trial

PSR (mm)

Observed HSR

Corrected HSR

Diameter (mm)

Mean Diameter

Mean radius

1

1

2

3

 

 

 

 

 

 

2

1

2

3

 

 

 

 

 

 

3

1

2

3

 

 

 

 

 

 

4

1

2

3

 

 

 

 

 

 

 

Distance between A and B = S = ___


Lead shot

r2(m2)

Time t(s)

Terminal velocity

1

 

 

 

2

 

 

 

3

 

 

 

4

 

 

 

 

Density of the material of lead shot, ρ = ___ kg/m3

Density of the liquid, σ = ___ kg/m3

Slope of v – r2 graph, m = BC/AB = ___

g = 9.8 m/s2

η = 2(ρ σ)g/9m = ___ Nsm-2

 

Result:


Coefficient of viscosity of the liquid = ___ Nsm-2