Experiment 1
Room temperature: 26oc
Time(x103s) t t-f ln(t-f) t+dt t-t+dt ln(t-t+dt)
0.000 8.00 10.20 2.32 7.3 2.21 0.79
0.174 6.90 9.10 2.21 5.9 1.66 0.51
0.207 6.35 8.55 2.15 5.5 1.57 0.45
0.261 5.90 8.10 2.09 4.9 1.43 0.36
0.322 5.45 7.65 2.03 4.6 1.29 0.25
0.380 4.95 7.15 1.97 4.1 1.17 0.16
0.469 4.25 6.45 1.86 3.7 1.01 0.01
0.535 3.70 5.90 1.78 3.1 0.91 -0.09
0.628 3.05 5.25 1.66 2.3 0.78 -0.25
0.719 2.35 4.55 1.52 2.0 0.67 -0.40
0.785 2.10 4.30 1.46 1.6 0.60 -0.51
0.860 1.70 3.90 1.36 1.3 0.53 -0.63
0.956 1.15 3.35 1.21 0.8 0.45 -0.80
1.000 0.60 2.80 1.03 0.5 0.42 -0.87
1.186 0.30 2.50 0.92 -0.2 0.31 -1.17
1.299 -0.30 1.90 0.64 -0.5 0.25 -1.39
1.462 -0.75 1.45 0.37 -0.9 0.19 -1.66
1.632 -0.95 1.25 0.22 -1.2 0.15 -1.90
1.919 -1.40 0.80 -0.22 -1.6 0.09 -2.41
2.195 -1.75 0.45 -0.80 -1.9 0.06 -2.81
2.495 -2.05 0.15 -1.89 -2.1 0.03 -3.51
3.062 -2.15 0.05 -2.98 -2.2 0.01 -4.61
3.724 -2.20 0.03 -3.55 -2.2 0.00
4.361 -2.20 0.01 -4.62 -2.2 0.00
Experiment 2
Table :- 1. Standardization I2 with standard Na2S2O3 solution.
Types Volume Strength Initial Final Difference Strength of
of layer of soln of Burette Burette =Volm of Initial I2(aq)
Na2S2O3 Reading Reading Na2S2O3 soln
soln soln (N)
(ml) (N) (ml) (ml) (ml)
CCl4 10.0 0.05 0.1 17.4 17.3 0.085
H2O 100 0.01 0.7 11.0 10.3 0.00103
Set: 2
Chemicals: I2 (H2O) and EA
Table:-2 Standardization of extracted (H2O + I2) solution with standard
Na2S2O3 solution.
Type of Volume of Strength of Initial Final Difference Strength of
solution solution Na2S2O3 Burette Burette = Volume of Extracted-I2
soln Reading Reading Na2S2O3 soln solution
(ml) (N) (ml) (ml) (ml) (N)
After 20 0.01 11.0 15.3 4.3 0.00215
extraction
Initial 20 0.05 0.0 39.9 39.9 0.10
Calculation
Strengths of the required solutions are calculated using the relation:
V1S1 = V2S2
Set 1 :
Distribution coefficient for CCl4-H2O system,
Kd1= [I2] org / [I2] aq
= (0.085/0.00103)
=82.5
Set 2:
Distribution coefficient of H2O-EA system
Kd2= [I2] org / [I2] aq
=(0.10-0.00215)/0.00215
= 45.5
Result
Distribution coefficient of I2 on CCl4-H2O system is 82.5
Distribution coefficient of I2 on H2O-Ethyl Acetate system is 45.5
Experiment 3
Table : DATA TABLE FOR DETERMINATION OF VISCOSITY COEFFICIENT
NO. of Density of Required Average Viscosity Conc. Of Specific
obser- sample soln time time coefficient glycerin viscosity SP/C
vation =kt C SP
(gm/cc) (sec) (sec) (w/v)
1 1.0030 26.82
27.15 27.03 1.082 3% 0.082 0.0273
27.12
2 1.0119 28.92
29.23 29.15 1.177 6% 0.177 0.0295
29.32
3 1.0228 30.57
31.43 31.18 1.272 9% 0.272 0.0302
31.53
4 1.0299 34.09
34.06 34.19 1.405 12% 0.405 0.0337
34.42
Calculation
Calculation of :
1= k1t = 0.040391.003027.03
=1.082
2= k2t = 0.040391.011929.15
=1.177 c.p
3= k3t = 0.040391.022831.18
=1.272 c.p
4= k4t = 0.040391.029934.19
=1.405c.p
We know,
=
1sp = (1.082-1)/1
= 0.082
1sp/c = 0.082/3
= 0.0273
2sp = (1.177-1)/1
= 0.177
2sp/c = 0.177/6
= 0.0295
3sp = (1.272-1)/1
= 0.272
3sp/c = 0.272/9
= 0.0302
4sp = (1.405-1)/1
= 0.405
4sp/c = 0.405/12
= 0.0337
Calculation of intrinsic viscosity [i]: =
where as, k=1.1210-3, = 0.6
From graph, i = 0.025
So, M = 184g/mol
Results
Intrinsic viscosity, i = 0.025
Molecular weight, M=184g/mol
Experiment 4
Molecular weight of succinic acid= 118 g
Normality of NaOH = 0.5 N
Table 1: Titration of succinic acid with standard NaOH
Temperature Volumeof Initial burette Final burette Difference(mL)
toc succinic reading(mL) reading(mL)
acid(mL)
29.5 5 0.0 13.0 13.0
34 5 13.0 26.8 13.8
39 5 26.8 42.6 15.8
44 5 14.4 31.0 16.6
Table 2:
Temperature Weight Weight Weight of
Obs. K of soln of solute solvent S lnS 1/T K-1
No. (gm) (gm) (gm)
1 302.5 5.0947 0.3835 4.7112 8.15 2.098 3.306
2 307 5.0493 0.4071 4.6472 8.77 2.172 3.257
3 312 5.0516 0.4661 4.5855 10.16 2.319 3.205
4 317 5.2271 0.4897 4.7370 10.34 2.335 3.155
Reaction
H2C4H4O4+2NaOH=Na2C4H4O4+2H2O
Calculation
Reading-1:
VA1=5ml; VB1=13.0ml; NB1=0.5N
VA1NA1=VB1NB1
NA1=(13.00.5)/(5) =1.30N
1N 1000ml59g
1.30N 5ml0.3835g
0.3835100
Solubility,S= = 8.15
4,7112
Reading-2:
VA2=5ml; VB2=13.8ml; NB1=0.5N
VA2NA2=VB2NB2
NA2=(13.80.5)/(5) =1.38N
1N 1000ml59g
1.38N 5ml0.4071g
0.4071100
Solubility,S= = 8.77
4.8593
Reading-3:
VA3=5ml; VB3=15.8ml; NB3=0.5N
VA3NA3=VB3NB3
NA3=(15.80.5)/(5) =1.58N
1N 1000ml59g
1.58N 5ml0.4661g
0.4661100
Solubility,S= = 10.16
4.8724
Reading-4:
VA4=5ml; VB4=16.6ml; NB4=0.5N
VA4NA4=VB4NB4
NA4=(16.60.5)/(5) =1.66N
1N 1000ml59g
1.66N 5ml0.4897g
0.4897100
Solubility,S= = 10.34
4.8776
The slope of the graph= -1702
We know, slope = -H/R
R = 1.98 Cal/mol.K
The value of H = -(-1702)1.98= 3370 Cal/mol= 3.37 kCal/mol
Result
HEAT OF SOLUTION , H = 3.37 kCal/mol (FROM GRAPH)
Experiment 5
Titration of two layers (CCl4 layer and H2O layer) is carried out with
standardised 0.05N Na2S2O3
solution for two different sets of solutions.
Volume of Initial Final Volume of
I2 Burette Burette Na2S2O3
Sample Layer Containing Reading Reading Solution
Solution (mL) (mL) (mL)
(mL)
I2 in H2O 50 4.6 21.9 17.3
1 I2 in CCl4 10 0.3 4.6 4.3
I2 in H2O 50 17.9 48.7 30.8
2
I2 in CCl4 20 0.3 17.9 17.6
Calculation
Set-A: Normality of Na2S2O3=0.05N; KD=82; (KI)=0.1
Concentration of I2 in CCl4
V1N1=V2N2
N2=(4.30.05)/10 = 0.0215N
M1=(0.0215/2) = 0.01075M
Concentration of I2 in aqueous layer
V1N1=V2N2
N2=(17.30.05)/50 = 0.0173N
M1=(0.0173/2) = 8.6510-3M =y
[I2]=x=(0.01075)/82=1.31110-4molL-1
[KI3]= z=y-x=8.5210-3molL-1
z
Ke= = 710
x{(KI)z}
Set-B: Normality of Na2S2O3=0.05N; KD=82; (KI)=0.1
Concentration of I2 in CCl4
V1N1=V2N2
N2=(17.6 0.05)/20 = 0.044N
M1=(0.0329/2) =0.022M
Concentration of I2 in aqueous layer
V1N1=V2N2
N2=(30.80.05)/50 = 0.0308N
M1=(0.0308/2) =0.0154M =y
[I2]=x=(0.022)/82=2.68310-4molL-1
[KI3]= z=y-x=0.0151molL-1
z
Ke= = 663
x{(KI)z}
Results
For System 1, Equilibrium constant, Ke =710
For System 2, Equilibrium constant, Ke =663
Experiment 6
Data for temperature with time:
Time (s) Temperature (oC)
0 34.1
2.38 34.2
3.72 34.6
5.39 34.7
6.58 34.8
9.2 35.1
10.99 35.2
16.55 35.3
20.18 35.3
29.18 35.3
39.22 35.3
46.28 35.3
49.6 35.3
87.53 35.2
108.29 35.2
128.57 35.2
147.29 35.2
176 35.2
203 35.2
223 35.2
258 35.2
290 35.2
327 35.2
361 35.2
517 35.2
686 35.2
852 35.2
988 35.2
1078 35.1
1211 35.1
1383 35.1
1548 35.1
1683 35.1
1548 35.1
1683 35.1
1861 35.1
Calculation
Mass of the beaker = 519.3 g
Mass of beaker with solution = 313.5 g
MS = (490-270) g
=205.8g
QS = 1 calg-1
T =1.2 (from graph)
K = 23 cal/
Q = MsQsT + KT
= (205.8g 1calg-1/ 1.2) + (23 cal/1.23)
= 274.56 cal
Here, N1=0.10N
V1=200ml
No. of moles, n= (2000.10)/1000 = 0.020 mol
H = Q/ n
= 274.56 cal/0.020 mol
=13.73 kcal/mol
Result
Heat of neutralization, H= 13.73 kcal/mol
Experiment 7
Conductance of pure H2O = 0.02810-3 ohm-1cm-1
Conductance of BaSO4 solution= 0.24710-3 ohm-1cm-1
Table: Specific and equivalent conductance of KCl solution of various
concentration
Chemical Concentration C Specific Equilibrium
species C(M) conductance L conductance
(mL) (ohmcm2eqv-1)
0.001 0.032 0.178 178
0.005 0.071 0.706 141.2
0.01 0.1 1.218 121.8
KCL 0.05 0.224 5.53 110.8
0.1 0.316 10.28 102.8
0.25 0.5 23.2 92.8
0.5 0.707 35.8 71.6
Calculation
Cell conductance is given ; 1mS
So, specific conductance = observed conductance
Solubility of BaSO4:
Ls(H2O) = 0.006 mS = 0.00610-3 -1
Ls(BaSO4) = 0.276 mS = 0.27610-3 -1
(BaSO4)= 127.2 + 160
= 287.2 -1cm2/eq
So, C(BaSO4) = (Ls 1000)/
= (0.270 10-3 1000)/287.2
= 9.4 10-4 M
Solubility product:
ksp = C2 = 8.8410-7 mol2L2
Equivalent conductance at infinite dilution,
From the graph;o = 136.8 -1cm2/eq
Results
Equivalent conductance of KCl solution at infinite dilution,
o (KCl) = 136.8 -1cm2/eq
The solubility product of BaSO4, Ksp =8.8410-7 mol2L2
