EXPERIMENT 1

AIM: To prepare crystals of pure benzoic acid from an impure sample.

Theory:

Benzoic acid is a crystalline solid has high solubility in hot water.an impure sample of

benzoic acid is dissolved in hot water and then filtered to remove insoluble impurities.

Requirements:

Crude sample of benzoic acid,250 ml measuring flask,funnel,a glass rod,and a trough.

Procedure

1.take about 2-3ml of the crude sample of benzoic acidin a 250ml beaker,in another

take about 150ml of water and keep it for boiling.

2.add slowly with stirring least amount of boiling water to the beaker containing crude

sample of benzoic acid so thet it gets dissolved easily.add 0.5g of animal charcoal to the

solution and boil for a minute.

3.filter the solutionwhile hot using a fluted filter paper place d in a

stemlessfunnel.collect the clear filterate in a beaker.

4.allow the filtered solution to cool at room temp.now cool it by placing it on a beaker

filled with cold water.

5.Separate the crystals by suction using Buchner funnel.wash the crystals with water,dry

the crystals.

6.record the weight of the crystals.

Result

The crystals of benzoic acid are shining white.

Yield=………

EXPERIMENT 2

AIM: To prepare crystals of pure naphthalene from an impure sample using ethyl

alcohol as solvent

Theory

Naphthalene is a crystalline solid,insoluble in water but has high solubility in ethyl

alcohol.an impure sample of benzoic acid is dissolved in hot ethyl alcohol and then

filtered,filterate is collected in a beaker and allowed to cool slowly.shining white crystals

of naphthalene separate out.

Structure

Requirements

Crude sample of naphthalene,250mlbeakers,stemlessfunnel,a glass rod

Procedure

1.take about 2-3ml of the crude sample of naphthalene in a 250ml beaker,in another

beaker take about 50ml of ethanol and heat it on a water bath

2.add slowly with stirring least amount of hot ethyl alcohol to the beaker containing

crude sample of naphthalene so thet it gets dissolved easily.

3.filter the solutionwhile hot using a fluted filter paper place d in a

stemlessfunnel.collect the clear filterate in a beaker.

4.allow the filtered solution to cool at room temp.now cool it by placing it on a beaker

filled with cold water.

5.Separate the crystals by suction using Buchner funnel.wash the crystals with water,dry

the crystals.

6.record the weight of the crystals.

Result

The crystals of naphthalene are shining white.

Yield=…..g

EXPERIMENT 3

Aim: To determine the viscosity of a given unknown liquid with respect to water, at

laboratory temperature, by viscometer.

Requirements: Ostwald viscometer, rubber tube with screw pinch cock, stand, beaker, unknown liquid, distilled water. specific gravity bottle

Theory: The force of friction which one part of the liquid offers to another part of the

liquid is called viscosity. For measuring the viscosity coefficient, Ostwald viscometer

method is used which is based on Poiseuille’s law. According to this law, the rate of flow

of liquid through a capillary tube having viscosity coefficient, , can be expressed as

r 4tP

8vl where, v= vol. of liquid (in ml)

t= flow time (in sec.) through capillary

r= radius of the capillary (in cm)

l= length of the capillary (in cm)

P= hydrostatic pressure (in dyne/sq.cm) = viscosity coefficient (in poise). Since, the hydrostatic pressure (the driving force) of the liquid is given by P = dg h

(where h is the height of the column and d is the density of the liquid);

P t ; or, d g h t

If, 1 and 2 are the viscosity coefficients of the liquids under study, d1, d2 , are their

densities and t1 and t2 are their times of flow of equal volume of liquids through the same

capillary respectively, then 1 d1 g h t1 and 2 d2 g h t2

Hence, 1

d1t1

2 d 2t2

Here, usually the viscosity of given liquid is measured with respect to water whose

viscosity is known very accurately at different temperatures. The SI physical unit of

viscosity is the pascal-second (Pa·s), (i.e., kg·m−1

·s−1

). This means: if a fluid with a

viscosity of one Pa·s is placed between two plates, and one plate is pushed sideways with

a shear stress of one pascal, it moves a distance equal to the thickness of the layer

between the plates in one second. The cgs unit for the same is the poise (P), (named after

J. L. Marie Poiseuille). It is more commonly expressed, as centipoise (cP). [1 cP = 0.001

Pa·s]. Water at 20 °C has a viscosity of 1.0020 cP.

Procedure:

1. Note the laboratory temperature. 2. Wash the specific gravity bottle with distilled water and dry. 3. Take the weight of the empty & filled (with distilled water) specific gravity bottle

(with stopper). Then, weigh the filled with specific gravity bottle h unknown given

liquid. Use the data for measuring the densities. 4. Clean and rinse the viscometer properly with distilled water. Fix the viscometer

vertically on the stand and filled with specific amount (say 20ml) of mixture (every time

take the same volume). 5. Time of flows were recorded for each solutions (water and the given liquid). 6. Take 3 to 4 readings.

Observations:

1. Laboratory temperature=…. C

2. Density measurement:

Weight of empty R.D.bottle (w1) =…g.

Weight of R.D.bottle with water (w2) =…g.

Weight of R.D. bottle with liquid (w3) =…g.

So, weight of water (ww) = (w2-w1) =…g.

Sl no. Flow times (sec)

t1 t2 t3 mean

1

2

3

4

Calculations:

1. Determination of the density of the liquid (dl): Density of liquid (dl) Density of water (dw)

Density of liquid (dl ) w

l d w

ww

= Weight of liquid (wl)

Weight of water (ww)

(Take density of water =1.0g/ml at 25 C)

2. Determination of the viscosity of the liquid (l )

Viscosity of the liquid, tl dl

l w tw

d w

Result: The viscosity of the given liquid with respect to water at laboratory

temperature was found to be ……..cP.

______

Experiment - 4

Aim:

Determine the surface tension of a given liquid at room temp using

stalgmometer by drop number method

Requirements:

Stalgmometer, specific gravity bottle, a small rubber tube with a screw pinch

cork, distilled water, experimental liquid.

Theory:

In the drop number method, the number of drops formed by equal volumes of

two liquid is counted. If m1 and m2 is the mass of one drop of each of the liquid

having densities d1 and d2 respectively. If n1 and n2 is the number of drops

formed by volume v of the two liquids, then their surface tensions are related as

¥1/¥2= (d1/d2)*(n2/n1)

One of the liquid is water its surface tension and density are known.Then te

surface tension of the given liquid can be calculated.

Procedure:

1. Clean the stalgmometer with chromic acid mix, wash with water and dry it

2. Attach a small piece of rubber tube having a screw pinch cock at the upper

end of the stalgmometer.

3. Immerse the lower end of the stalgmometer in distilled water and suck the

water 1-2cm above mark A. adjust the pinch cork so that 10-15 drops fall per

minute

4.Clamp the stalgmometer allow the water drops to fall and start counting the

number of drops when the meniscus crosses the upper mark A and stop

counting when the meniscus passes mark B

5. Repeat the exercise to take three to four readings

6. Rinse the stalgmometer with alcohol and dry it

7. Suck the given liquid in the stalgmometer and count the drops as in case of

water

8. Take a clean dry weighing bottle weighs it with water as well as with liquid.

9. Note the temp of water taken in a beaker.

Observations:

Room temp=t0C

Density of water=dw

Surface tension of water=¥ dynes/cm

No of drops From a Fixed Volume Mean

Liquid 1…. 2…… 3….. nl=

Water 1…. 2……. 3…… nw=

Weight of empty specific gravity bottle=w1gram

Weight of specific gravity bottle+water=w2 gram

Weight of empty sp.gravity bottle+liquid=w3gram

Weight of water= (w2-w1)gram

Weight of liquid= (w3-w1)gram

Calculations:

Density of the liquid

Dl= (w3-w1)/(w2 –w1) *dw

Surface tension of liquid=

¥l/¥w = (dl/dw)*(nw/nl)* ¥w

Result

The surface tension of liquid is ………dynes/cm.

Experiment -5

Aim- Determine the surface tension of a given liquid by means of stalgmometer

using drop weight method.

Requirements- Stalgmometer , wide mouthed weighing bottle , a small rubber

tube with screw pinch cork, distilled water, experimental liquid.

Theory-

The size of the drop of liquid forming on a capillary end depends upon the

surface tension of the liquid. The drop falls when the total surface 2πr¥ is equal

to its weight (w)

i.e. w= mg= 2πr¥

if two liquids having surface tension ¥1 and ¥2 are allowed to fall through the

same capillary then

w1/w2 = m1/m2 = ¥1/¥2

Procedure-

1. Clean the stalagmometer with chromic acid mixture , wash with water

then with alcohol and dry it.

2. Attach a small piece of rubber tube having a screw pinch cork at the

upper end of the stalgmometer.

3. Immerse the lower end of the stalgmometer in distilled water and suck the

water 1-2cm above the mark A. Adjust the pinch cork so that 10-15 drops

fall per minute.

4. Clamp the stalgmometer and allow 10 drops of water to fall into the pre-

weighed wide mouthed weighing bottle.

5. Weigh the weighing bottle along with 10 drops of water. Repeat the

experiment three to four times.

6. Rinse the stalgmometer with the given liquid, dry it. Fill it with the

experimental liquid and the repeat the above steps.

Observations-

Room temperature = toC\

Density of water = dw

Surface tension of water= ¥w

Weight of empty weighing bottle=w1g

Weight of weighing bottle+ 10 drops of water=w2 gram

Weight of bottle+ 10 drops of liquid=w3gram

Weight of water= (w2-w1)gram

Weight of liquid= (w3-w1)gram

Calculations-

¥l/¥w= (w3-w1)/(w2- w1)

Result-

The surface tension of the given liquid is …………………. Dynes/cm

Experiment -6

Aim: Determine the heat of neutralization of strong acid (HCl) with a strong

base (NaOH) calorimetrically.

Theory:

Heat of neutralization may be defined as the amount of heat evolved when one

gram equivalent of the acid is completely neutralized with one gram equivalent

of the base or vice versa in dilute solutions. Acc.to Arrhenius theory strong acid

and base are completely ionized in aqueous solutions; hence the neutralization

of a strong acid (HCl) and strong base may be defined as

(H++Cl

-)+ (Na

++OH

-) Na

+ +Cl

- + H2O +57.3KJ

Apparatus:

Thermos flask with tight fitted cork, two 250ml beakers, two 100ml measuring

flask,1/10degree thermometer.

Chemicals:

0.2MHCl, 0.2 NaOH, distilled water.

Procedure:

To determine the water equivalent of calorimeter

1. Clean the calorimeter, take 100ml of water in it and note its temperature

2. In a separate beaker take 100ml of water and heat it to 15-200C higher

than the room temp.

3. Add this hot water quickly to the cold water in the calorimeter. Replace

the stopper, stir with the stirrer and note the temp of the mixture.

To determine the heat of neutralization of HCl with NaOH

4. Take 100ml of 0.2m HCl in a clean calorimeter and note its temp

5. Take 100ml of 0.2m NaOH. The temperature must be equal to that of

HCl, otherwise wait for sometime till both the solution attain same temp.

6. Pour 100ml of 0.2m NaOH solution quickly into the HCl sol in the

calorimeter taking care to avoid splashing. Stir well and note the highest

temp attained.

Observations:

For water equivalent of calorimeter

Volume of cold water taken=100ml

Temp of cold water =t0c

Volume of hot water=100ml

Temp of hot water=t10C

Temp of the mixture=t20C

Density of water=1 g/ml

For heat of neutralization of HCl with NaOH

Volume of 0.2m HCl taken=100ml

Volume of 0.2 m NaOH taken=100ml

Initial temp of HCl and NaOH=t30C

Final temp after neutralization=t40C

Calculations

To calculate the water equivalent

Rise in the temp of cold water=(t2-t0)0C

Fall in temp of hot water=(t1-t2)0C

Let W be the water equivalent of the calorimeter and its contents

Acc to principle of calorimetry

Heat lost by hot water=heat gained by calorimeter and cold water

W= [100(t1-t2)/(t2-t0) -100] g

To calculate the enthalpy of neutralization

Initial temp of HCl and NaOH before mixing=

Final temp after mixing=t40C

Rise in temp=(t4-t3)0C

Total volume of mixture in the calorimeter=200ml

Heat of neutralization of 1000ml of 1M HCl=

(200+W)*(t4-t3)*4.184/0.2*100 kJ =Q KJ

Enthalpy of neutralization of HCl with NaOH=q KJ/mol

Result

Heat of neutralization of HCl and NaOH=…….kJ

Actual value= -57.3kJ

%error =………

Precautions

1. Use very accurate 1/10th

degree thermometer.

2. The bulb of the thermometer should not touch the bottom of the

calorimeter.

3. The solution should be thoroughly stirred while noting the temp.

4. The solution should not splash out when poured from beaker in to

flask.

Experiment - 7

Aim: Determine the heat of neutralization of weak acid (CH3COOH) with a

strong base (NaOH) calorimetrically.

Theory:

Heat of neutralization may be defined as the amount of heat evolved when one

gram equivalent of the acid is completely neutralized with one gram equivalent

of the base or vice versa in dilute solutions. However if the acid involved in

neutralization is weak ,then the heat evolved is less than 57.3 KJ because some

amount of heat is used in dissociation of weak acid.

(CH3COO- +H

+)+ (Na

++OH

-) Na

+ +

CH3COO

- + H2O

Apparatus:

Thermos flask with tight fitted cork, two 250ml beakers, two 100ml measuring

flask,1/10degree thermometer.

Chemicals:

0.2M CH3COOH, 0.2 NaOH, distilled water.

Procedure:

To determine the water equivalent of calorimeter

1. Clean the calorimeter, take 100ml of water in it and note its temperature

2. In a separate beaker take 100ml of water and heat it to 15-200C higher

than the room temp.

3. Add this hot water quickly to the cold water in the calorimeter. Replace

the stopper, stir with the stirrer and note the temp of the mixture.

To determine the heat of neutralization of CH3COOH with NaOH

4. Take 100ml of 0.2m CH3COOH in a clean calorimeter and note its

temperature.

5. Take 100ml of 0.2m NaOH. The temperature must be equal to that of

HCl, otherwise wait for sometime till both the solution attain same temp.

6. Pour 100ml of 0.2m NaOH solution quickly into the sol CH3COOH in

the calorimeter taking care to avoid splashing. Stir well and note the

highest temp attained.

Observations:

For water equivalent of calorimeter

Volume of cold water taken=100ml

Temp of cold water =t0c

Volume of hot water=100ml

Temp of hot water=t10C

Temp of the mixture=t20C

Density of water=1 g/ml

For heat of neutralization of CH3COOH with NaOH

Volume of 0.2m CH3COOH taken=100ml

Volume of 0.2 m NaOH taken=100ml

Initial temp of CH3COOH and NaOH=t30C

Final temp after neutralization=t40C

Calculations

To calculate the water equivalent

Rise in the temp of cold water=(t2-t0)0C

Fall in temp of hot water=(t1-t2)0C

Let W be the water equivalent of the calorimeter and its contents

Acc to principle of calorimetry

Heat lost by hot water=heat gained by calorimeter and cold water

W= [100(t1-t2)/(t2-t0) -100] g

To calculate the enthalpy of neutralization

Initial temp of CH3COOH and NaOH before mixing=

Final temp after mixing=t40C

Rise in temp=(t4-t3)0C

Total volume of mixture in the calorimeter=200ml

Heat of neutralization of 1000ml of 1M CH3COOH=

(200+W)*(t4-t3)*4.184/0.2*100 kJ =Q KJ

Enthalpy of neutralization of CH3COOH with NaOH=q KJ/mol

Result

Heat of neutralization of CH3COOH and NaOH=…….kJ

Precautions

1. Use very accurate 1/10th

degree thermometer.

2. The bulb of the thermometer should not touch the bottom of the

calorimeter.

3. The solution should be thoroughly stirred while noting the temp.

4. The solution should not splash out when poured from beaker in to

flask.