EXPERIMENT NO: 1
STANDARDIZATION OF NaOH BY 0.05M OXALIC ACID
 Apparatus:






Burette
Pipette
Conical flask
Burette stand
Funnel
Stirrer
 Chemical Equation:
In Sodium hydroxide against oxalic acid titration, the equation is as below
 Indicator: Phenophthelene
End Point / Colour Change: Colourless to Light Pink
 Procedure:
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
Preparation of 0.05 M Standard Solution of Oxalic Acid
Clean the burette thoroughly, wash it with distilled water and finally rinse it with
sodium hydroxide solution. (Always rinse the burette with the NaOH solution).
Clamp the burette vertically in a burette stand.
Fill sodium hydroxide solution into the burette through a funnel above the zero
mark.
Remove the air gap, if any, from the nozzle of the burette by running the solution
forcefully from the burette nozzle.
Remove the funnel before noting the initial reading of the burette. Also while
noting the reading, see that no drop of the liquid is hanging at the nozzle of the
burette.
Note the initial reading by keeping the eye exactly at the same level as the
meniscus of the solution.
Pipette out 10 mL of oxalic acid solution in a washed and dried conical flask.
Always wash the pipette with water and rinse with the oxalic acid (solution to be
measured) before pipetting out the liquid.
Add 1-2 drops of phenolphthalein indicator to the conical flask.
Place the flask over the glazed tile. Titrate the acid with sodium hydroxide
solution till a very faint permanent pink color is obtained.
Add sodium hydroxide solution in small amounts initially and then dropwise.
OBSERVATION & CALCULATIONS:
According to balanced chemical equation
2NaOH + 2H2O.(COOH)2  (COONa)2 + 4H2O
BURETTE READINGS ( NaOH) :
S/NO
01
02
03
Initial volume (ml)
0
21
40
Final volume (ml)
21
40
62
<V> = 21 ml
THEORATICAL DATA:
OXALIC ACID
NaOH
M1 = 0.05M
M2 = ?
V1 = 10 ml
V2 = 21 ml
n1 = 1
n2 = 2
According to neutralization formula
M1V1 / n1
=
=> M2 = M1V1 n2/ V2 n1
=> 0.05 x 10 X 2 / 21
=> M2 = 0.04M
M2V2 / n2
Volume used (ml)
21
20
22
EXPERIMENT NO: 2
STANDARDIZATION OF H2SO4 BY USING NaOH
SOLUTION
 Apparatus:






Burette
Pipette
Conical flask
Burette stand
Funnel
Stirrer
 Chemicals:


H2SO4
NaOH
 Chemical equation:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
 Procedure
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
x.
xi.
Firstly, clean the burette with distilled water (DW), and then rinse it with sodium
hydroxide (NaOH) solution.
This ensures that the interior portion of the burette is coated with a thin layer of
NaOH solution.
Clean the pipette with DW, and rinse it with dilute solution of H2SO4.
This is done by carefully drawing about 10 ml of this solution into the pipette, and
releasing it into the sink. Do not let it enter your mouth or touch any part of your
body.
Now, fix the burette on the stand, and fill it with NaOH solution just above the
zero mL mark.
By turning on the stopcock, release some of the solution in a clean 100 ml flask to
remove any air bubbles in the burette.
Refill it up to the zero mL mark, such that the lower meniscus of the solution is
touching just above this mark.
Draw 10 ml dilute H2SO4 by a pipette, and release it into a 250 ml flask.
Add 2 – 3 drops of phenolphthalein indicator to the acid, and keep the flask just
below the burette stopcock on a white paper.
Now, hold the flask just below the burette tip, and slowly turn the stopcock so as
to release the NaOH solution in a drop-by-drop manner inside the flask.
Simultaneously, keep swirling the flask in a gentle way as sodium hydroxide starts
reacting with the acid. 10.
OBSERVATION & CALCULATIOS:
According to neutralization formula
H2SO4 + 2 NaOH
Na2SO4 + 2 H2O
BURETTE READINGS ( H2SO4) :
S/NO
01
02
03
Initial volume (ml)
0
09
19
Final volume (ml)
09
19
30
<V> = 10 ml
THEORATICAL DATA:
NaOH
H2SO4
M1 = 0.1M
M2 = ?
V1 = 10 ml
V2 = 10 ml
n1 = 1
n2 = 1
According to neutralization formula
M1V1 / n1
=
=> M2 = M1V1 n2/ V2n1
=> 0.1 x 10 / 2 x 10
=> M2 = 0.5M
M2V2 / n2
Volume used (ml)
09
10
11
EXPERIMENT NO: 3
STANDARDIZATION OF KMnO4 BY USING OXALIC ACID
 Apparatus:






Burette
Burette stand
Pipette
Conical flask
Funnel
Burner
 Chemicals:


KMnO4
Oxalic acid.
 Procedure:
i.
We took KMnO4 solution in burette upto zero mark.
ii.
We took 10 ml oxalic acid in flask by using pipette.
iii.
Start titrate until colour become purple.
iv.
Note the burette initial readings.
v.
Repeats the experiments for 3 times and notes the reading.
OBSERVATION & CALCULATIONS:
According to balanced chemical equation
KMnO4 + H2SO4 + 5 (COOH)2  K2SO4 + 2MnSO4 + 8H2O + 10 CO2
BURETTE READINGS ( KMnO4) :
S/NO
01
02
03
Initial volume (ml)
0
16.8
34.6
Final volume (ml)
16.8
34.6
53.4
<V> = 17.8 ml
THEORATICAL DATA:
OXALIC ACID
KMnO4
M1 = 0.05M
M2 = ?
V1 = 10 ml
V2 = 17.8 ml
n1 = 5
n2 = 2
According to neutralization formula
M1V1 / n1
=
M2V2 / n2
=> M2 = M1V1 n2/ V2 n1
=> 0.05 x 10 X 2 / 17.8 x 5
=> M2 = 0.01M
Volume used (ml)
16.8
17.8
18.8
EXPERIMENT NO: 4
FIND OUT AMOUNT PER LITER OF FeSO4.7H2O USING
KMnO4 SOLUTION
 Apparatus:







Burette
Burette stand
Conical flask
Pipette
Funnel
Spirit lamp
Half test tube
 Chemicals:



KMnO4
H2SO4
FeSO4
 Balanced Chemical Equation
10FeSO4·7H2O + 2KMnO4 + 8H2SO4 → 2MnSO4 + 5Fe2(SO4)3 + 78H2O + K2SO4
 Procedure:
i.
We took FeSO4 in conical flask by using pipette.
ii.
We added half test tube H2SO4 in conical flask and heated it by using spirit lamp.
iii.
We placed KMnO4 solution in burette up to zero mark.
iv.
Now we started titration and continuously shaking the flask.
v.
By appearing the color change into light purple , we stop the titration and note
the reading.
vi.
We repeated the experiment for 3 times and find the average reading.
OBSERVATION & CALCULATIONS:
According to balanced chemical equation
2 KMnO4 + 8H2SO4 + 10FeSO4.7H2O  K2SO4 + 2MnSO4 + 5 Fe2(SO4)3 + 78 H2O
BURETTE READINGS ( KMnO4) :
S/NO
01
02
03
Initial volume (ml)
0
16.6
32.2
Final volume (ml)
16.6
32.2
49.8
<V> = 16.6 ml
THEORATICAL DATA:
FeSO4.7H2O
KMnO4
M1 = ?
M2 = 0.01M
V1 = 10 ml
V2 = 16.6 ml
n1 = 10
n2 = 2
According to neutralization formula
M1V1 / n1
=
M2V2 / n2
=> M1 = M2V2n1 / V1n2
=> 0.01 x 16.6 X 10 / 10 x 2
=> M2 = 0.01M
AMOUNT / LTR:
Amount/Ltr = Molarity x Molar mass
 0.08 x 278
 Amount/Ltr = 22.24g
Volume used (ml)
16.6
15.6
17.6
EXPERIMENT NO: 5
DETERMINE THE AMOUNT OF CH3COOH IN VINEGAR
 Apparatus:




Burette
Burette stand
Conical flask
Funnel
 Chemicals:



Indicator
NaOH
Vinegar
 Procedure:
i.
We took 10ml of vinegar in conical flask.
ii.
We added some drops of phenalpthaline.
iii.
We filled the burette with NaOH up to zero mark.
iv.
We start titration until the color change inti light pink.
v.
We stop further addition and note the reading.
vi.
We repeated the experiment foe 3 time and found the averge mean of the
burette reading carefully.
PREPARATION OF SOLUTIONS:
a. 0.1M NaOH:
b.Vinegar:
n = M x Vdm3 => mg / M => M x Vml / 1000
mg = M x Vml x M / 1000
Dissolve 50g of Vinegar by weight in distil
water and dilute it in 1250ml of volumetric flask upto
the mark.
=> 0.1 x 250 x 40 / 1000 => 1g
BURETTE READINGS ( NaOH) :
S/NO
01
02
03
Initial volume (ml)
0
17.5
17.5
Final volume (ml)
17.5
17.5
35.9
<V> = 18 ml
Volume used (ml)
17.5
17.5
19
THEORATICAL DATA:
NaOH
Vinegar
M1 = 0.1M
M2 = ?
V1 = 18 ml
V2 = 10 ml
n1 = 1
n2 = 1
According to neutralization formula
M1V1 / n1
=
M2V2 / n2
=> M2 = M1V1 n2/ V2 n1
=> 0.1 x 18 / 10
=> 1.8 / 10
=> M2 = 0.18M
 MOLES OF CH3COOH / 250ml = M x Vml / 1000
=> 0.18 X 250/1000 => 0.045
 Mass of NaHCO3
n = mg/M => mg= n x M
=> 0.045 x 60 => 2.7g.
 Percentage of NaHCO3
% NaHCO3 = mg x 100 / total mass
=> 2.7 x 100 / 50 => 5.4%
EXPERIMENT NO: 6
DETERMINE THE AMOUNT OF NUMBER OF WATER OF
CRYSTALLIZATION IN OXALIC ACID.
 Apparatus:




Burette
Burette stand
Conical flask
Funnel



NaOH [0.1M]
Oxalic acid [6.3g/dm3]
Indicator
 Chemicals:
 CHEMICAL REACTIONS:
2NaOH + xH2O.(COOH)2  (COONa)2 + (x+1) H2O
 Procedure:
i.
We took 10 ml NaOH in flask .
ii.
We added 2 or 3 drops of the indicator.
iii.
We filled the burette with oxalic acid up to zero mark.
iv.
We started titration until the color change to light pink and then disappear.
v.
We note the reading carefully.
vi.
We repeated the experiment for 3 times and found the averge mean of the
burette reading carefully.
OBSERVATION & CALCULATIONS:
According to balanced chemical equation
2NaOH + xH2O.(COOH)2  (COONa)2 + (x+1) H2O
BURETTE READINGS ( OXALIC ACID) :
S/NO
01
02
03
Initial volume (ml)
0
10
18
Final volume (ml)
10
18
27
<V> = 09 ml
Volume used (ml)
10
8
9
THEORATICAL DATA:
OXALIC ACID
NaOH
M1 = ?
M2 = 0.1M
V1 = 09 ml
V2 = 10 ml
n1 = 1
n2 = 2
According to neutralization formula
M1V1 / n1
=
M2V2 / n2
=> M1 = M2V2 n1/ V1 n2
=> 0.1 x 10 X 1 / 09 x 2
=> 1/18
=> M2 = 0.05M
Amount per litre:
Amount per litre = Molarity x Molar mass
=> 6.3g = 0.05 x (90 . X18)
=> 6.3/0.05 = 90 + X18
=> 126 – 90 = X18 => X = 36/18
=> X = 2
Water of crystallization in oxalic acid is 2.
EXPERIMENT NO: 7
GRAVIMETRIC ANALYSIS OF Ba+ IN BaCl2 AS BaCrO4
 Apparatus:









Beaker
Funnel
Stand
Whatmen filter paper
Volumetric flask
Spatula
Pipette
Digital scale
Petri dish
 Chemicals:


BaCl2
K2CrO4
 Procedure:
i.
We prepare a saturated solution of BaCl2 in a beaker.
ii.
We prepare 0.1M solution of K2CrO4 solution in 100ml volumetric flask.
iii.
We isolate 20ml BaCl2 solution in another beaker and add K2CrO4 .
iv.
The yellow solution is formed BaCrO4 and precipitation.
v.
Let the precipitation of BaCrO4 in the beaker.
vi.
We filter and isolate the precipitate form BaCrO4 solution.
vii.
We weight a dry filter paper and note its weight.
viii.
After filtration we again weigh this filter along with precipitate.
ix.
We dry this filter paper and precipitate in microwave over for 2 days.
x.
After drying weight it and note its weight in grams by using the digital scale.
OBSERVATION & CALCULATIONS:
Dry paper weight = 0.6g
Wet paperweight after filteration = 11.2g
Weight of paper after drying = 3.45g
Weight of dry precipitate = 3.45 – 0.6 =
2.85g
MOLAR MASS of BaCrO4 = 253 g/mol
Atomic mass of Ba = 137 g/mol
MASS OF BA IN 1g OF BaCrO4:
Mass in 1g = 137 x 1/253
 Mass in 1g = 0.54g
MASS OF BA IN 2.58g OF BaCrO4:
Mass in 2.58g = 0.54 x 2.58
 Mass in 2.58g = 1.5g
PERCENTAGE COMPOSITION:
% of Ba in BaCrO4 = 137 x 100 / 253
 % of Ba in BaCrO4 = 54.15%
EXPERIMENT NO: 8
DETERMINE OF VISCOSITY OF GIVEN ORGANIC
LIQUIDS.
 Apparatus:







Viscometer
Beaker
Graduated cylinder
Pipette
Burette stand
Specific gravity bottle
Stop watch



Distal water
Ethanol
Acetone
 Chemicals:
 Procedure:
i.
We take a specific gravity bottle and find its mass when it is empty.
ii.
Now we get ethanol in specific gravity bottle and find its mass.
iii.
We can find the volume of the ethanol by subtracting the mass of empty specific
gravity bottle
iv.
Then we rinse the viscometer with distal water and dry it.
v.
We filled the viscometer with water reached to the upper mark on the narrow tube
by sucking the through rubber tube.
vi.
We started the stopwatch as we release the pressure.
vii.
We noted the flow rate of water.
viii.
Now again we clear the viscometer and rinse with distal water and dry it.
ix.
We filled the viscometer with ethanol to the upper mark of narrow tube.
x.
Now we release the pressure but it the same time we started the stopwatch to note
down the flow rate of ethanol.
xi.
We repeated the same experiment for 3 times and noted the flow rate.
xii.
Now again we clear the viscometer and rinse with distal water and dry it.
xiii.
We filled the viscometer with acetone to the upper mark of narrow tube.
xiv.
Now we release the pressure but it the same time we started the stopwatch to note
down the flow rate of acetone.
DENSITY CALCULATIONS:
b. ACETONE:
a. ETAHANOL:
Mass of ampty bottle = 6.77 g
Mass of ampty bottle = 6.44 g
Mass of bottle filled with ETHANOL = 26.5 g
Mass of bottle filled with ACETONE = 24.79 g
Mass of ethanol = 19.81 g
Mass of ACETONE = 18.27 g
d = m / v => 19.81 /25
d = m / v => 18.27 /25
d = 0.792 g/cm3
d = 0.7308 g/cm3
FLOW TIME :
S/NO
WATER (H2O)
ETHANOL
ACETONE
01
S/NO
02
03
139 sec ETHANOL
382 sec
124 sec
WATER
ACETONE
142 sec
364 sec
123 sec
142 sec
373 sec
124 sec
(H2O)
<tw> = 141 sec
<tl> = 373 sec
<tl> = 123.6 sec
01
139 sec
382 sec
124 sec
VISCOSITY
CALCULATIONS:
02
142
sec
364 sec
123 sec
03
142 sec
373 sec
124 sec
<t
>
=
141
<t
>
=
373
<tl> =
w
l
A. ETHANOL:
sec
sec
123.6 sec
D = m Vscosity
/ v => 19.81
/25
of ETHANOL
= tl x dl (viscoty of water) / tw x dw
 g/cm
373 x3 0.97 x 0.8007 x 10-3 / 141 x 1
D = 0.792
 235.94 / 141 x 10-3
 Viscosity of ETHANOL = 1.67 x 10-3 kg/ms
B. ACETONE:
Vscosity of ACETONE = tl x dl (viscoty of water) / tw x dw
 123.6 x 0.73 x 0.8007 x 10-3 / 141 x 1
 123.6 x 0.73 x 0.8 / 141 x 10-3
 Viscosity of ETHANOL = 0.51 x 10-3 kg/ms
EXPERIMENT NO: 9
DETERMINE THE RHEOCHOR VALUE OF GIVEN
ORGANIC LIQUIDS
 Apparatus:







Viscometer
Beaker
Graduated cylinder
Pipette
Burette stand
Specific gravity bottle
Stop watch

Distal water

Ethanol
 Chemicals:
 Procedure:
i.
We take a specific gravity bottle and find its mass when it is empty.
ii.
Now we get ethanol in specific gravity bottle and find its mass.
iii.
The volume of the ethanol can be find out by subtracting the mass of empty
specific gravity bottle
iv.
Then we rinse the viscometer with distal water and dry it.
v.
First we find the viscosity of water which is taken as a reference.
vi.
We filled the viscometer with water reached to the upper mark on the narrow tube.
vii.
We started the stopwatch as we release the pressure from the rubber tube.
viii.
We noted the flow rate of water.
ix.
Now again we rinse the viscometer and rinse with distal water and dry it.
x.
We filled the viscometer with ethanol to the upper mark of narrow tube.
xi.
Now we release the pressure but it the same time we started the stopwatch.
xii.
We note down the flow rate of ethanol.
xiii.
We repeated the same experiment for 3 times and noted the flow rate.
DENSITY CALCULATIONS:
a.ETAHANOL:
Mass of ampty bottle = 6.77 g
Mass of bottle filled with ETHANOL = 26.5 g
Mass of ethanol = 19.81 g
d = m / v => 19.81 /25
d = 0.792 g/cm3
FLOW TIME :
S/NO
WATER (H2O)
01
02
03
S/NO
139 sec
142 sec
sec
WATER 142ETHANOL
<tw> = 141 sec
ETHANOL
382 sec
364 sec
ACETONE 373 sec
<tl> = 373 sec
(H2O)
VISCOSITY CALCULATIONS:
01
139 sec
382 sec
124 sec
02
142
sec
364
sec
123 sec
C. ETHANOL:
03
142 sec
373 sec
124 sec
<tw> =of141
<tl> = t373
<tl>of= water) / tw x dw
Vscosity
ETHANOL
x
d
(viscoty
l
l
sec
sec
123.6 sec
D = m
/ v 373
=> 19.81
x 0.97/25
x 0.8007 x 10-3 / 141 x 1
 235.94 / 141 x 10-3
D = 0.792 g/cm3
 Viscosity of ETHANOL = 1.67 x 10-3 kg/ms
 Or 1.67 centipoise
[R] = M/d viscosity1/8
=> 46 g.mol-1 x (1,39)1/8 / 0.78
=> log [R] = log 58.97 + 1/8 log (1.39)
=> log [R] = log 58.97 + 0.0178
=> log [R] = 1.7878
=> [R] = Anti log 1.7878
=> [R] = 61.34
EXPERIMENT NO: 10
DETERMINE THE SURFACE TENSION OF GIVEN
ORGANIC LIQUID
 Apparatus:






Stalogma meter
Beaker
Pipette
Graduated cylinder
Burette stand
Specific gravity bottle
 Chemicals:


Distal water
Ethanol
 Procedure:
i.
We take a specific gravity bottle and find its mass when it is empty.
ii.
Now we get ethanol in specific gravity bottle and find its mass.
iii.
The volume of the ethanol can be find out by subtracting the mass of empty
specific gravity bottle from total mass.
iv.
Then we find the density of ethanol.
v.
We take the stalagma meter and rinse out with distilled water and dry it.
vi.
We find the surface tension of water because water has taken as a reference.
vii.
We filled the stalagma meter with water to its upper mark.
viii.
We release the pressure and count the drops of water.
ix.
We repeated the experiment for 3 times.
x.
Rinse out the stalagma meter with ethanol.
xi.
We release the pressure and count the drops of ethanol until its reached to the
lower mark.
xii.
We repeated the same experiment for 3 times.
DENSITY CALCULATIONS:
a. ETAHANOL:
Mass of ampty bottle = 25.44 g
Mass of bottle filled with ETHANOL = 45.18 g
Mass of ethanol = 19.73 g
d = m / v => 19.73 /25
3
d = 0.78 g/cm
NOMBER
OF DROPS
:
S/NO
WATER (H2O)
01
02
03
S/NO
76
66
66
<n> = 69.3
WATER
ETHANOL
ETHANOL
150
162
161
<n> = 157.6
ACETONE
(H2O)
SURFACE TENSION CALCULATIONS:
01
139 sec
382 sec
124 sec
02
142 sec
364 sec
123 sec
a. ETHANOL:
03
142 sec
373 sec
124 sec
<tw> = 141 <tl> = 373
<tl> =
Surface tension (liquid) / surface tension
(water) = dl . nw / dw . nl
sec
sec
123.6 sec
D = m / v => 19.81 /25
surface tension (liquid) = dl . nw . surface tension (w) / dw . nl
D = 0.792 g/cm3
 0.78 x 69.3 x 71.18x10-3 / 1 x 157.6
 54.65 x 71.18 x 10-3 / 157.6
 3847.2 x 10-3 / 157.6
 Surface tension (liquid) = 24.4 x 10-3
oR
24.4 dyne.cm-3
EXPERIMENT NO: 11
DETERMINE THE PARACHOR VALUE OF THE GIVEN
ORGANIC LIQUID.
 Apparatus:
•
•
•
•
•
•
Stalogma meter
Beaker
Pipette
Graduated cylinder
Burette stand
Specific gravity bottle
 Chemicals:


Distal water
Ethanol
 Procedure:
i.
We take a specific gravity bottle and find its mass when it is empty.
ii.
Now we get ethanol in specific gravity bottle and find its mass.
iii.
The volume of the ethanol can be find out by subtracting the mass of empty
specific gravity bottle from total mass.
iv.
Then we find the density of ethanol.
v.
We take the stalagma meter and rinse out with distilled water and dry it.
vi.
We find the surface tension of water because water has taken as a reference.
vii.
We filled the stalagma meter with water until reach to its upper mark.
viii.
We release the pressure and count the drops of water until its reach to the lower
mark of stalagma meter
ix.
We repeated the experiment for 3 times.
x.
Rinse out the stalagma meter with ethanol and dry it.
xi.
We release the pressure and count the drops of ethanol until its reached to the
lower mark.
xii.
We repeated the same experiment for 3 times.
DENSITY CALCULATIONS:
b. ETAHANOL:
Mass of ampty bottle = 25.44 g
Mass of bottle filled with ETHANOL = 45.18 g
Mass of ethanol = 19.73 g
d = m / v => 19.73 /25
d = 0.78 g/cm3
NOMBER OF DROPS :
S/NO
WATER (H2O)
ETHANOL
01
76
150
S/NO02 WATER
03
(H2O)
66
ETHANOL
162
ACETONE
66
161
01
139 sec <n> = 69.3
382 sec
124<n>
sec= 157.6
02
142 sec
364 sec
123 sec
SURFACE
TENSION
CALCULATIONS:
03
142 sec
373 sec
124 sec
<tw> = 141 <tl> = 373
<tl> =
D. ETHANOL:
sec
sec
123.6 sec
D Surface
= m / v tension
=> 19.81(liquid)
/25 / surface tension (water) = d . n / d . n
l
w
w
3
D =surface
0.792 g/cm
tension (liquid) = dl . nw . surface tension (w) / dw . nl
l
PARACHOR VALUE CALCULATIONS:
[P] = M r 1/4 / d
 0.78 x 69.3 x 71.18x10-3 / 1 x 157.6
 54.65 x 71.18 x 10-3 / 157.6
 3847.2 x 10-3 / 157.6
 Surface tension (liquid) = 24.4 x 10-3 Nm-1 oR 24.4 dyne.cm-3
=> 46 (24) 1/4 / 0.78
Taking log on b.s
=> Log [P] = Log 58.97 + Log (24) 1/4
=> 1.77 + 1 (1.38) / 4
=> 1.77 + 0.345
=> 2.115
Taking anti log on B.S
=> Anti Log (2.115)
=> [P] = 130.31
EXPERIMENT NO: 12
DETERMINE THE REFRACTIVE INDEX OF GIVEN
ORGANIC LIQUID.
 Apparatus:



Refractometer
Beaker
Pipette
 Chemicals:



Ethanol
Methanol
Acetone
 Procedure:
i.
We take the refractometer to find out the refractive index of the given liquid.
ii.
We adjust the refractometer such that the border line between light and dark in the
telescope become sharp and minimum.
iii.
We find the refractive index of ethanol by putting the ethanol in the prism of the
refractometer.
iv.
Then we find the refractive index of acetone by putting it into the prism of the
refractometer.
v.
After that we wash and rinse out the refractometer.
vi.
We clear the prism and dry it.
vii.
At the last we find the refractive index of methanol by putting it into the prism of
the refractometer.
OBSERVATION & CALCULATIONS:
S/NO
01
02
03
CHEMICAL / LIQUID
WATER
ACETONE
ETHANOL
REFRACTIVE INDEX
1.335
1.344
1.358
(Fig: REFRACTOMETER)
EXPERIMENT NO: 13
DETERMINE THE MOLECULAR REFRACTIVITY OF GIVEN
ORGANIC LIQUIDS.
 Apparatus:




Refractometer
Beaker
Pipette
Specific gravity bottle
 Chemicals:



Ethanol
Methanol
Acetone
 Procedure:
i.
First of all we find the density of the liquids.
ii.
We take the empty specific gravity bottle and find its mass.
iii.
Now we putted the liquids one by one and find its masses.
iv.
After that we find the densities of the all liquids.
v.
Now we take the refractometer and adjust the refractometer such that the border
line between light and dark in the telescope become sharp and minimum.
vi.
We put some drops of ethanol in the prism of the refractometer and find its
refractive index.
vii.
Now we put some drops of acetone in the prism of the refractometer and find its
refractive index.
viii.
We put some drops of methanol in the prism of the refractometer and find its
refractive index.
ix.
After that we finded the molecular refractivity of each liquid.
OBSERVATION & CALCULATIONS:
S/NO
01
02
03
CHEMICAL / LIQUID
WATER
ACETONE
ETHANOL
REFRACTIVE INDEX
1.335
1.344
1.358
MOLECULAR REFRACTIVITY CALCULATION:
A. ETHANOL:
[R] = n2 – 1 (M) / n2 + 2 (d)
=> (1.358)2 – 1 X 46 / (1.358)2 + 2 X 0.78
=> 0.844 x 0.046 / 3.844 x 0.78
=> 0.038 / 2.99
=> [R] = 0.012
B. ACETONE
[R] = n2 – 1 (M) / n2 + 2 (d)
=> (1.344)2 – 1 X 0.058 / (1.344)2 + 2 X 0.78
=> 0.806 x 0.058 / 3.806 x 0.78
=> 0.046 / 2.98
=> [R] = 0.015
(Fig: ABBE’s REFRACTOMETER)
EXPERIMENT NO: 14
PREPARE 0.1M NaOH SOLUTION & STANDARDIZE
AGAINST KHP.
 Apparatus:







Burette,
Titration flask,
Pipette
Volumetric flask,
Wash bottle,
Digital balance.
Beaker .
 Chemicals:

Standarrd solution: 0.1M KHP as primary standard.

Analyte solution: 0.1M NaOH

Indicator:
phenalphthaline
 Chemical reactions:
CO2H
CO2Na+
+ NaOH 
CO2K+
+ H2O
CO2K+
 Procedure:
i.
We Prepared 0.1M NaOH solution by taking 1g of NaOH and dissolve in 250ml of
volumetric flask and dilute it upto the mark.
ii.
We Prepared 0.1M KHP solution by taking 2.04g of KHP in 100ml of volumetric flask
and dilute it upto the mark.
iii.
We Take 10ml KHP in titration flask (titrand) and add 2-3 drops of phenalphthaline
indicator for the detection of end point.
iv.
We Filled the burette with titrant (NaOH) upto zero mark and start adding drop by
drop into the KHP solution until the color become light pink.
v.
We Noted the reading and repeat the same procedure for three times and note the
readings from burette and find its average.
vi.
By applying neutralization formula calculate the molarity of NaOH which is 0.13M.
PREPARATION OF SOLUTIONS:
a.
0.1M NaOH:
b. 0.1M KHP:
n = M x Vdm3 => mg / M => M x Vml / 1000
n = M x Vdm3 => mg / M => M x Vml / 1000
mg = M x Vml x M / 1000
mg = M x Vml x M / 1000
=> 0.1 x 250 x 40 / 1000 => 1g
=> 0.1 x 100 x 204 / 1000 => 2.04 g
BURETTE READINGS ( NaOH) :
S/NO
01
02
03
Initial volume (ml)
0
8.5
16.5
Final volume (ml)
8.5
16.5
24.8
<V> = 8.1 ml
THEORATICAL DATA:
KHP
NaOH
M1 = 0.1M
M2 = ?
V1 = 10 ml
V2 = 8.1 ml
n1 = 1
n2 = 1
According to neutralization formula
M1V1 / n1
=
=> M2 = M1V1 n2/ V2 n1
=> 0.1 x 10 / 8.2
=> M2 = 0.12M
M2V2 / n2
Volume used (ml)
8.5
8
8
EXPERIMENT NO: 15
PREPARE 0.1M H2SO4 SOLUTION & STANDARDIZE
AGAINST Na2CO3
 Appearatus:








Burette,
Titration flask,
Pipette
Volumetric flask,
Wash bottle,
Digital balance.
Beaker .
Spirit lamp.
 Chemicals:

Standard solution : 0.1M H2SO4 .

Analyte solution

Indicator
: 0.1M Na2CO3 .
: Methyl Orange
 Chemical reactions:
H2SO4 + Na2CO3
Na2SO4 + H2O + CO2
 Procedure:
i.
Prepare 0.1M H2SO4 solution by taking 1.5ml of H2SO4 in 250ml of volumetric flask
and dilute it upto the mark.
ii.
Prepare 0.1M Na2CO3 solution by taking 2.65g of Na2CO3 in 250ml of volumetric flask
and dilute it upto the mark.
iii.
Take 10ml Na2CO3 in titration flask (titrand) and add 3-5 drops of Methyl Orange
indicator and heat it for 2 minutes.
iv.
Fill the burette with titrant (H2SO4 ) upto zero mark and start adding drop by drop
into the 0.1M Na2CO3 solution until the color is change from pink to orange.
v.
Note the reading and repeat the same procedure for three times and note the
readings from burette and find its average.
vi.
By applying neutralization formula calculate the molarity of Na2CO3which is 0.07M.
PREPARATION OF SOLUTIONS:
a. 0.1M H2SO4:
b. 0.1M Na2CO3:
n = M x Vdm3 => mg / M => M x Vml / 1000
n = M x Vdm3 => mg / M => M x Vml / 1000
mg = M x Vml x M / 1000
mg = M x Vml x M / 1000
=> 0.1 x 250 x 106 / 1000 => 2.65 g
=> 0.1 x 250 x 98 / 1000 => 2.45g
H2SO4 :
solution we’d have to
Solution
As H2SO4 is in liquid form, inorder to prepare
95g
=
100g
convert mass into volume to attain accurate results.
2.4g
=
X
d=m/V
X = 100 x 2.4 / 95 = 2.75g
=> V = m / d
=> V = 2.57 / 1.84 = 1.5ml
BURETTE READINGS ( H2SO4) :
S/NO Initial volume (ml)
Final volume (ml)
Volume used (ml)
01
0
13.1
13.1
02
13.1
26.7
13.6
03
26.7
41.6
14.9
<V> = 13.9 ml
THEORATICAL DATA:
Na2CO3
H2SO4
M1 = 0.1M
M2 = ?
V1 = 10 ml
V2 = 13.9 ml
n1 = 1
n2 = 1
According to neutralization formula
M1V1 / n1
=
=> M2 = M1V1 n2/ V2 n1
=> 0.1 x 10 / 13.9
=> 1 / 13.9
=> M2 = 0.07M
M2V2 / n2
EXPERIMENT NO: 16
DETERMINE THE PERCENTAGE OF NaHCO3 IN BAKING
SODA
 Appearatus:








Burette,
Titration flask,
Pipette
Volumetric flask,
Wash bottle,
Digital balance.
Beaker .
Spirit lamp.
Solutions:


Standard solution : 0.1M H2SO4 .

Analyte solution

Indicator
: 1g Baking soda in 100ml
: Methyl Orange
 Chemical reactions:
H2SO4 + 2NaHCO3
Na2SO4 + 2H2O + 2CO2
 Procedure:
i.
Prepare 0.1M H2SO4 solution by taking 1.5ml of H2SO4 in 250ml of volumetric flask and
dilute it upto the mark.
ii.
Prepare 0.1M NaHCO3 solution by dissolving 1g of NaHCO3 in 100ml of volumetric flask
and dilute it upto the mark.
iii.
Take 10ml NaHCO3 in titration flask (titrand) and add 3-5 drops of Methyl Orange
indicator and heat it for 2 minutes.
iv.
Fill the burette with titrant (H2SO4 ) upto zero mark and start adding drop by drop into
the 0.1M NaHCO3 solution until the color is change from pink to light orange.
v.
Note the reading and repeat the same procedure for three times and note the readings
from burette and find its average.
vi.
By applying neutralization formula calculate the molarity of NaHCO3.
vii.
by doing further calculations determined the percentage of NaHCO3 which is 84% in
given sample.
PREPARATION OF SOLUTIONS:
a. 0.1M H2SO4:
b. NaHCO3 :
n = M x Vdm3 => mg / M => M x Vml / 1000
mg = M x Vml x M / 1000
Dissolve 1g of baking sod by weight in distil
water and dilute it in 100ml of volumetric flask upto
the mark.
=> 0.1 x 250 x 98 / 1000 => 2.45g
H2SO4 :
Solution
95g
=
100g
2.45g
=
X
As H2SO4 is in liquid form, inorder to prepare sol
convert mass into volume to attain accurate results.
d=m/V
X = 100 x 2.4 / 95 = 2.57g
=> V = m / d
=> V = 2.57 / 1.84 = 1.5ml
BURETTE READINGS ( H2SO4) :
S/NO
01
02
03
Initial volume (ml)
0.6
8
14.7
Final volume (ml)
8
14.7
21.9
<V> = 7.1 ml
Volume used (ml)
7.2
6.8
7.2
THEORATICAL DATA:
BAKING SODA
H2SO4
M1 = ?
M2 = 0.1M
V1 = 10 ml
V2 = 7.1 ml
n1 = 2
n2 = 1
 MOLES OF NaHCO3 / 100ml = M x Vml / 1000
=> 0.1 X 100/1000 => 0.01
 Mass of NaHCO3
According to neutralization formula
n = mg/M => mg= n x M
M1V1 / n1
=
M2V2 / n2
=> M2 = M1V1 n2/ V2 n1
=> 0.1 x 7.1 x 2 / 10 => 1.42 / 10=> M2 = 0.142M
=> 0.01 x 84 => 0.84g.
 Percentage of NaHCO3
% NaHCO3 = mg x 100 / total mass
=> 0.84 x 100 / 1 => 84%
EXPERIMENT NO: 17
SEPARATION OF INK MIXTURE BY CHROMATOGRAPHY
 Appearatus:

Chromatographic tank

Paper

Burette stand.
 Chemicals:

Red ink

Blue ink

Ethanol

Distill water
 Procedure:
i.
Prepare the homogenate of red ink + blue ink & water + ethanol.
ii.
Add one drop of ink mixture on paper and dip the paper in solvent.
iii.
The mixture of ethanol and water taken as solvent in chromatographic tank.
iv.
Allow the solvent to evaporate through paper which take ink mixture along with itself.
v.
After 15 minutes observe the paper on which two clear strips of red ink and blue ink
appear.
Rf VALUE CALCULATIONS:
GIVEN DATA:
Solvent front = 60 cm
Red ink distance = 50 cm
Blue ink distance = 40 cm
a. For Red ink:
Rf = red ink distance / solvent front
 50 / 60
 Rf = 0.8
b. For Blue ink:
Rf = blue ink distance / solvent front
 40 / 60
 Rf = 0.6
EXPERIMENT NO: 18
LE-CHATLIER’s PRINCIPLE, SHIFTING OF EQUILLIBRIUM
 Appearatus:

Test tubes.

Beakers.

Wash bottles.

Volumetric flask.

spitula.
 Chemicals:

Potassium di-chromate (K2Cr2O7)

Potassium chromate (KCrO4)

Sulphuric acid (H2SO4)

Sodium hydroxide (NaOH)

Distill water or Oxidane (H2O)
 Chemical equation:
KCrO4
K2Cr2O7
(YELLOWISH)
(ORANGE)
 Procedure:
i.
Take the already prepared solutions of KCrO4 and K2Cr2O7 in two test tubes among
whom KCrO4 appear yellowish and K2Cr2O7 orange in colour.
ii.
Add some amount of sulphuric acid in KCrO4 solution until appear orange in colour
which indicate in K2Cr2O7 excess.
iii.
Now add some amount of NaOH solution in potassium dichromate solution until it
appear yellowish in colour which indicates KCrO4 in excess.
iv.
The colour transition shows the equilibrium shift in chromate-dichromate which is
based upon Le-chatlier’s principle.
KCrO4
(Fig:
K2Cr2O4
EXPERIMENTAL VERIFICATION OF LECHATLIER’S
PRINCIPLE)
EXPERIMENT NO: 19
EVOPORATION IS THE PROCESS OF COOLING
 Appearatus:

Beaker

Thermometer

Hair dryer.
 Chemicals:

Spirit (solution of ethyl alchohol)
 Procedure:
i.
Take 20ml of spirit in beaker which is volatile in nature.
ii.
Now measure the initial temperature with the help of thermometer which is 170C.
iii.
Now start the hair dryer and blow the stream of air on the surface of spirit in order to
the rate of evaporation.
iv.
Continue the stream for 80 seconds and again measure the final temperature which
is now 140C.
v.
The drop in temperature shows high energy molecules escaped from the surface and
cause cooling.
AFTER APPLYING
DRYER
TEMPRATURE= 170 C
TEMPRATURE = 140 C