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Investigatory Project on Diffusion of Solids in Liquids

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Investigatory Project and Viva Questions on Diffusion of Solids in Liquids (Class 12)

The process of intermixing of substances when they are brought into contact with each other is called diffusion. Diffusion occurs rapidly in the case of gases. In the case of liquids it is less rapid and solids do not show this property when they are brought into contact. However, diffusion of solids occurs slowly in liquids in which they are soluble. The speed of diffusion depends on (a) temperature (b) size of the particles and (c) the mass of the particles.


Aim


The aim of this investigatory project is to study the effect of temperature on rate of diffusion of a solid in a liquid by taking crystals of copper sulphate.


Materials and Requirements


250 mL beakers, Crystalline copper sulphate, Distilled water, Ice, Stop watch etc.


Procedure


Take 5g each of CuSO4. 5H20 in three 250 mL beakers. Pour 100 mL of distilled water into one of the beakers, 100 mL ice water into the second beaker and 100mL hot water into the third beaker. Cover all the three beakers with watch glasses. Note the temperature of each. Observe the development of blue colour (due to diffusion). Record the time taken for the complete dissolution and diffusion of CuSO4 in all the three cases.


No.

Temperature

Time taken in minute

1

2

3

Cold water

Distilled water

Hot water

-

-

-

 

It is found that the diffusion occurs most readily in the hot solution.


Conclusion


The rate of diffusion of a solid in a liquid is directly proportional to temperature.


VIVA QUESTIONS ON DIFFUSION OF SOLIDS IN LIQUIDS


1. What is diffusion?


Ans: Intermixing of different substances when they are brought into contact with each other is called diffusion


2. Which type of substance has maximum diffusion?


Ans: Gases have maximum diffusion.


3. What are the factors which influence diffusion?


Ans: The rate of diffusion depends on the temperature, size of particles and the mass of particles.


4. Explain effect of temperature on rate of diffusion of a solid in a liquid?


Ans: Here when the temperature increases, the rate of diffusion also increases.


5. What colour is developed when copper sulphate is dissolved in water?


Ans: Blue


6. Equal masses of crystalline copper sulphate are added to equal volumes of water taken in two beakers which are kept at two different temperatures. Which water gets the blue colour earlier?


Ans: Water which has higher temperature

Investigatory Project on Caffeine in Tea

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Determination of caffeine in tea samples Investigatory Project (Class 12)

Tea leaves contain up to 4% caffeine. They also contain tannic acid and colouring matter. The relative amounts of these substances differ in different samples of tea.


Theory:


Caffeine is more soluble in chloroform than in water. So it can be extracted from aqueous solutions by chloroform. But the tannic acid present in tea leaves must be removed before the extraction of caffeine. Tannic acid is removed as calcium tannate.


Aim:


The aim of this experiment is to determine the percentage of caffeine present in different samples of tea leaves and to find out the best tea leaves.


Requirements:


1. Different brands of tea leaves

2. Beakers

3. CaCO3

4. Chloroform


Procedure:


Take 10g each of the given sample of tea leaves and boil with about 100mL of distilled water in a beaker for about 10 minutes. Cool and filter. Add 2g CaCO3 and boil. Remove the precipitated calcium tannate by filtration. Extract the filtrate at least 3 times with chloroform. Distil off the chloroform. The residue is caffeine. Weigh it. In the ssame way, determine the amounts of caffeine in other samples of tea and tabulate.


No:

Brand name of the tea

Mass of tea (x)

Mass of caffeine (y)

% of Caffeine (y x 100)/x

1

2

3

 

 

 

 

 

Conclusion:


The tea containing the highest percentage of caffeine is ……. Hence it is the beast tea among the samples analysed.


VIVA QUESTIONS WITH ANSWERS


1. What is the percentage of caffeine in tea leaves?


Ans: About 4%


2. How will you remove tannic acid from tea leaves?


Ans: By boiling with calcium carbonate


3. What is the product formed when tannic acid is boiled with CaCO3?


Ans: Calcium tannate


4. How will you determine the quality of a sample of tea based on the caffeine content?


Ans: The sample with maximum caffeine content is the best quality tea.


Effect of Catalyst on Rate of Reaction (Class 12)

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Investigatory Project on Effect of Catalyst on Rate of Reaction (Class 12)

Principle: A catalyst is a substance which alters the rate of a reaction without it is being consumed in the reaction. A catalyst may speed up a reaction and then it will be called a positive catalyst. It may slow down a reaction and then it will be called a negative catalyst. Their functions can be illustrated by studying the decomposition of hydrogen peroxide in the presence of various substances.


2H2O2 --> 2H2O + O2


Aim: The aim of the investigatory project is to study the influence of catalysts on the rate of reaction.


Materials Required: Hydrogen peroxide, MnO2, HCl, NaOH, Al powder, Test tubes


Procedure:


Pour 10mL portions of 20 volume hydrogen peroxide into four test-tubes. To each of the test tube add the following substances and note the rates at which bubbles evolve in each case.


1. 5 drops of 2M NaOH

2. 5 drops of 2M HCl

3. About 10 mg of MnO2

4. About 10 mg of Aluminium powder


Except in the case of adding HCl, in all other cases effervescence is seen. The briskness of the effervescence may be taken as a measure of the effectiveness of the catalyst in decomposing hydrogen peroxide. It becomes evident that while HCl acts as a negative catalyst, the other three act as positive catalysts.


Conclusion:


It is concluded from the experiment that the rate of reaction increases with a positive catalyst and decreases with a negative catalyst.


Viva Questions with Answers


1. What is a catalyst?


Ans: A substance which alters the rate of a reaction without itself undergoing any permanent chemical change is called a catalyst.


2. What is a positive catalyst?


Ans: A catalyst which increases the rate of a reaction is called a positive catalyst.


3. Give the example of a positive catalyst?


Ans: MnO2 in the dissociation of hydrogen peroxide to produce O2 and water.


4. What is a negative catalyst?


Ans: A catalyst which decreases the rate of a reaction is called a negative catalyst.


5. Give the example of a negative catalyst.


Ans: Hydrochloric acid in the dissociation of hydrogen peroxide.

Hardness of Water Experiment

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Investigatory Project and Viva Questions on Hardness of Water Experiment

Principle


Hardness of water is due to the existence of bicarbonates, chlorides and sulphates of calcium and magnesium. Temporary hardness of water is due to bicarbonates and it can be got rid off by boiling water. The bicarbonates will decompose giving insoluble carbonates which can be removed by decantation or filtration.


Ca(HCO3)2 --> CaCO3 + H2O + CO2


Hardness of water can be easily estimated by direct titration with ethylene diamine tetra acetic acid (EDTA) using the metal ion indicator eriochrome black – T


(HOOC.CH2)2.N.CH2.CH2N.(CH2.COOH)2   EDTA


EDTA is an excellent hexadentate ligand which can form very stable complexes with metallic ions by closing five rings with the metallic ion. The complex ion is formed instantaneously and so the end point of the reaction can be detected with suitable metal ion indicator. Since EDTA itself is insoluble in water, its disodium salt is usually used for titration. Its reaction with the metallic ion like Ca2+ ang Mg2+ can be represented as follows.


Na2H2Y + M2+ --> MY2-+2Na++2H+

Where, Y = (ŌOC.CH2)2.N.CH2.CH2.N.(CH2.COŌ)2


The best metal ion indicator is eriochrome black – T which gives a wine-red colour with the metallic ion. The titration is carried out at a pH = 10 and at this pH the free indicator has a deep blue colour. So the end-point of the titration is the colour change from blue to red to blue.


Reagents


1. 0.01M EDTA solution. Dissolve 3.723g of A.R. disodium ethylene diamine tetraacetate in distilled or deionised water and make it up in a 1 litre standard flask.

2. Eriochrome Black — T. Dissolve 0.2g of the substance in 15mL of triethanolamine and 5mL of absolute alcohol.

3. Buffer solution (pH = 10). Add 142mL of concentrated ammonia solution (sp.gr. 0.88 to 0.9) to 17.5g of A.R. NH4CI and dilute to 250mL.

4. Standard ZnSO4.7H20 solution. A standard 0.01M ZnSO4.7H20 solution is prepared by dissolving 2.87g of A.R. ZnSO4.7H20 in 1 litre of distilled or deionised water. This is used to standardise EDTA solution.


Objective of the Experiment: 


The objective of the Experiment is to determine the hardness of water from different sources


Procedure


Collect water from different sources like tap—water, wells in different localities, from different rivers, lakes etc. 50mL portion of water from each sample is pipetted out into a clean beaker. Add 1 mL of the buffer solution and 3-4 drops of Eriochrome Black T indicator. Titrate with 0.01 M EDTA solution taken in a burette until the colour of the solution in the beaker changes form wine-red to pure blue with no reddish—tinge. Repeat the experiment with other samples of water. Total hardness can be calculated from these values.


Temporary and Permanent Hardness


Pipette out 250mL of a sample of water into a clean beaker. Boil it for about 30 minutes, cool it and filter it into a 250mL flask. After filtration, make it upto 250mL. Take 50mL from it and titrate with EDTA solution as before. From this permanent hardness can be calculated.


Calculation


Usually hardness is expressed in gm of CaCO3 in million parts of water. ie, CaCO3 in ppm. 1mL of 0.01 M EDTA = 0.001g of CaCO3. Suppose a 50mL sample of water requires Vcm3 of 0.01 M EDTA. Then the hardness of water is given by


(V x .001 x 106/50) ppm of CaCO3 or (V x 103/50) ppm of CaCO3


The values obtained from different sources of water are tabulated as follows to compare the permanent and temporary hardness in these samples.

Sample of water

Volume of water taken (mL)

Volume of 0.01M EDTA required (VmL)

Hardness of water in CaCO3 ppm

Vx103/50

Volume of EDTA required after boiling (V1mL)

Permanent hardness in CaCO3 ppm V1x103/50

Temporary hardness in CaCO3 ppm (V-V1) x 103 / 50

1

2

3

4

5

6

50

50

50

50

50

50

 

 

 

 

 

 

HARDNESS OF WATER EXPERIMENT VIVA QUESTIONS WITH ANSWERS


1. What is hard water?


Ans: Water that does not lather easily with soap is called hard water.


2. Which are the substances that make water hard?


Ans: Bicarbonates, chlorides and sulphates of calcuim and magnesium make water hard.


3. How is hardness of water classified?


Ans: Hardness of water is classified into temporary hardness and permanent hardness.


4. What is temporary hardness due to?


Ans: It is due to the presence of bicarbonates of Ca and Mg. These, on boiling decompose and precipitate as their carbonates.


5. How is hardness of water estimated?


Ans: Hardness of water is estimated by E.D.T.A. titration.


6. What is E.D.TA?


Ans: The chemical name of E.D.T.A. is ethylene diamine tetraacetic acid


7. What is co-ordination number of E.D.T.A.?


Ans: Co-ordination number of E.D.T.A. is six.


8. Give the name of the metal indicator used in the E.D.T.A titration of water?


Ans: The metal indicator used is eriochrome black —T


9. What is the colour change expected at the equivalence point in the E.D.TA. titration using eriochrome black — T?


Ans: The colour change is from red to blue.


10. What is the red colour due to?


Ans: It is due to the complex formed between the metal and eriochrome black—T


11. What is the blue colour due to?


Ans: It is due to free E.D.T.A. at pH =10


12. What is the buffer used to get pH =10


Ans: NH4OH/NH4Cl mixture is used to get pH= 10


13. How is hardness of water usually expressed?


Ans: It is expressed in parts/weight of CaCO3 in million parts of water ie, in ppm.

Paper Chromatography

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Investigatory Project and Viva Questions on Paper Chromatography


Introduction


Chromatography is a very efficient method for the separation and purification of different components present in a mixture. The method is based on the distribution of different components in a mixture between a fixed or stationary phase and a mobile or moving phase. There are a number of chromatographic techniques based on the nature of the stationary phase. The stationary phase may be an adsorbent column, a paper or a glass plate coated with a thin layer of an adsorbent, through which the mobile phase moves. The mobile phase may be a liquid or a gas. The separation takes place due to the differential migration of the components which in turn depends on the relative affinity of the components towards the stationary and mobile phases. The component which is less strongly held by a stationary phase moves faster in a mobile phase.


Aim


Separation of pigments present in leaves and flowers by paper chromatography and the determination of Rf values of the separated components.


Paper chromatography:


This method is used for the separation and identification of the components present in a mixture especially when they are present in small amounts. In paper chromatography the stationary and mobile phases are liquids. The stationary phase is water present in filter paper (cellulose) and the mobile phase is a pure solvent or a mixture of solvents. Therefore, the basic principle of separation in paper chromatography is partition of the components of a mixture in two liquids.


Some flowers or leaves are taken and they are ground well in a mortar using a pestle. About 3 mL of acetone is added to it and it is again ground. The coloured clean solution is decanted. A Whatman filter paper No. 1 is taken and is cut to get a size of approximately 5cm x 30cm (depending on the size of the jar used). A line is drawn using a pencil at a distance of about 2 cm away from one end of the paper. This line is known as the reference line. About 2 to 3 drops of the extract of the leaves or flowers are placed at the centre of the reference line. It is then dried using a drier.

About 40-50 mL of the mobile liquid (acetone or alcohol) is taken in a gas jar. The filter paper is hanged into the gas jar in such a way that filter paper tip touches the liquid and the reference line is above the surface of the mobile liquid. The gas jar is now closed with a lid which has a slit on it. The upper end of the filter paper strip passes through the slit. The filter paper is allowed to stand for about 2 hours.


The solvent moves up on the filter paper. Along with the solvent the different constituents also move with varying speeds according to their partition coefficients. The coloured components are finally separated from each other. When the solvent front traverses about 75% of the filter paper, the paper is removed and the solvent front is marked on the paper. The chromatogram is dried and the distances travelled by the solvent front and the different components from the reference line are measured. From these the Rf values of different components are calculated. Rf value (retention factor) is defined as the ratio of distance travelled by a particular component from the origin line to the distance travelled by the solvent from the origin line.


Rf = Distance travelled by a component from the origin line/Distance travelled by the solvent from the origin line


The Rf value depends on the nature of the substance, nature of the mobile and stationary phases and the temperature. Rf values can be used to identify the components in a mixture.


Calculation


Distance travelled by solvent = a cm

Distance travelled by component X = b cm

Distance travelled by component Y = c cm

Rf of component X = b/a

Rf of component Y = c/a


PAPER CHROMATOGRAPHY VIVA QUESTIONS WITH ANSWERS.


1. What is chromatography?


Ans: It is a technique for separating components of a mixture by their relative movements in a mobile phase against a stationary phase.


2. Why do different substances travel at different speeds in the mobile phase?


Ans: Different substances have different degree of attachments towards the stationery phase. Substances which are attached strongly will move slowly in the mobile phase. Those which are attached weakly will move faster in the mobile phase.


3. What is adsorption chromatography?


Ans: If a substance is held to the surface of another substance by surface forces it is called adsorption. Chromatography based on adsorption is called adsorption chromatography.


4. What is partition chromatography?


Ans: Here one substance penetrates into another substance. There is absorption rather that adsorption. For example a substance between two liquids. Chromatography based on absorption is called partition chromatography.


5. Why is paper chromatography considered as partition chromatography?


Ans: Here the substance concerned is partitioned between the mobile phase and the water present in the paper.


6. Mention two substances that are commonly used as stationary phase in adsorption chromatography. 


Ans: (1) Aluminium Oxide (2) Chalk


7. What is Rf factor?


Ans: It is the ratio of the distance travelled by a solute substance to the total distance travelled by the solvent front.


8. In paper chromatography Rf factors help to identify the substance. How?


Ans: If the Rf  factors of different substances concerned are known, then from the relative positions of various substances they can be identified.

Resonance Column Experiment (Class 11) Readings

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Resonance Column Experiment (Class 11) Readings

Experiment – 1


Aim


i. To find the velocity of sound in air at room temperature and hence at 0°C using a resonance column apparatus.

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


Apparatus


Resonance column apparatus, Tuning forks, Rubber hammer, Meter Scale etc.


Principle:


i. The velocity of sound at room temperature, Vt = 2n(l2-l1)

where, n = Frequency of the tuning fork

l1 = First resonating length

l2 = Second resonating length

ii. The velocity of sound at 0°C is given by V0 = Vt(273/(273+t))

or V0 = Vt – 0.6t

where, t = Room temperature°C

iii. Unknown frequency, n’ = Vt/(2(l2’-l1’))


Procedure


Length of the air column is kept very small. A tuning fork of known frequency (n) is excited and held horizontally over the mouth of the inner tube. The length of air column in the inner tube is slowly increased by raising the tube till a booming sound is heard. The length of the air column is measured (l1). Keeping the tuning fork excited at the mouth of the tube the length of the air column is measured (l1). Keeping the tuning fork excited at the mouth of the tube the length of the air column is increased further. The length (l2) of the air column is measured when the booming sound is heard (l2>3l1). The experiment is repeated and the mean values of l1 and l2 are found out. The velocity of sound at room temperature is calculated. The experiment is repeated for different tuning forks and the mean value of Vt is found. From this the velocity of sound at 0°C is calculated.


Using the tuning forks of known frequencies the mean value of Vt is calculated. Then using the tuning fork of unknown frequency, the first and second resonating lengths (l1’ and l2’) are measured. The unknown frequency (n’) can be calculated.


Observations and Readings


Frequency of tuning fork (n)

First resonance length (l1)

Second resonance length (l2)

Vt

1(cm)

2(cm)

Mean (l1’)

1(cm)

2(cm)

Mean (l2’)

512

16.5

16.5

16.5

50.5

50.5

50.5

34816

480

17.5

17.5

17.5

53.5

53.5

53.5

34560

Unknown

22.5

22.5

22.5

68.5

68.5

68.5

 

 

Mean Vt = 34688 cm/s = 346.88 m/s

Room temperature, t = 30°C

Velocity of sound at 0°C, Vo = Vt – 0.6t = 346.88 – 0.6 x 30 = 328.88 m/s

Unknown Frequency, n’ = Vt/(2(l2’-l1’)) = 403.3 Hz


Results:


1. Velocity of sound at room temperature = 346.88 m/s

2. Velocity of sound at 0°C = 328.88 m/s

3. Unknown frequency = 403.3 Hz


Experiment - 2


Aim: To compare the frequencies of two tuning forks and also to determine the end correction.


Principle:


Let l1 and l2 are the first and second resonance length with a tuning fork of frequency n1, l1’ and l2’ respectively are the first and second resonance length with another tuning fork of frequency n2. Then,


i. Ratio of frequencies, n1/n2 = (l2’-l1’)/(l2-l1)

ii. The end correction is given by, e = (l2-3l1)/2


Procedure:


The length of the air column is kept very small. The first tuning fork of frequency n1 is excited and is held horizontally close to the mouth of the inner tube. The inner tube is slowly raised until maximum sound is heard. The length of air column is measured as l1. Then the inner tube is further raised, keeping the vibrating fork at the mouth of the tube, till the maximum sound is heard. The length of air column noted as l2. Repeating this procedure for another tuning fork of frequency n2 and the resonance lengths are measured as l1’ and l2’. From this n1:n2 is found out. The end correction is also calculated.


Observations and Readings


1. To compare the frequencies


Frequency of tuning fork (n)

First resonance length

Second resonance length

(l2’-l1’) /

(l2-l1)

1(cm)

2(cm)

Mean

1(cm)

2(cm)

Mean

n1 = 512

16.5

16.5

l1 = 16.5

50.5

50.5

l2 = 50.5

1.06

n2 = 480

17.5

17.5

l1’ = 17.5

53.5

53.5

l2’ = 53.5

 

Ratio of frequencies, n1/n2 = 1.06

(l2’-l1’)/(l2-l1) = 1.06


2. To find the end correction


Tuning fork

First resonance length (l1)

Second resonance length (l2)

e

1(cm)

2(cm)

Mean

1(cm)

2(cm)

Mean

First

16.5

16.5

16.5

50.5

50.5

50.5

0.50

Second

17.5

17.5

17.5

53.5

53.5

53.5

0.50

 

Mean e = 0.5 cm = 0.5 x 10-2 m


Results:


i. Frequencies of two tuning forks are compared.

ii. End correction = 0.5 x 10-2 m