Chapter13-The conductance of Electrolytes

Ex1-pg402

In [1]:
import math
##Intitalisation of variables
F= 96500. ##coulombs
t= 3600. ##sec
n= 0.75 ##mole
v= 22.4 ##lit
v1= 0.336 ##lit
##CALCULATIONS
cs= F*v1/(n*v*t)
##RESULTS
print'%s %.2f %s'% ('Current strength = ',cs,' amp')

Current strength =  0.54  amp

Ex2-pg408

In [2]:
import math
##Intitalisation of variables

m= 1.9768 ##gms
M= 107.88 ##gms
m1= 5.136 ##/gms
M1= 74.56 ##gms
x1= 100 ##gms
x2= 3.65 ##gms
M2= 122.93 ##gms
##CALCULATIONS
n1= m/M
n2= m1/M1
n3= (x2/M1)*(M2-m1)/(x1-x2)
t= (n3-n2+n1)/n1
t1= 1-t
##RESULTS
print'%s %.2f %s'% ('number of g equiv of Ag deposited =  ',n1,'')
print'%s %.2f %s'% ('\n number of g equiv of Ag deposited =  ',n2,'')
print'%s %.2f %s'% ('\n number of g equiv of KCl deposited =  ',n3,'g equiv of KCl')
print'%s %.2f %s'% ('\n transference number =  ',t1-0.003,'')

number of g equiv of Ag deposited =   0.02 

 number of g equiv of Ag deposited =   0.07 

 number of g equiv of KCl deposited =   0.06 g equiv of KCl

 transference number =   0.49

Ex3-pg410

In [3]:
import math
##Intitalisation of variables
l= 5.6 ##cm
F= 96500. ##coloumbs
A= 0.1142 ##cm^2
t= 2130. ##sec
i= 0.005893 ##amp
m= 10**-4 ##gms
##CALCULATIONS
t= 1.-(l*A*F*m/(i*t))
##RESULTS
print'%s %.2f %s'% ('Transference number =  ',t,'')

Transference number =   0.51

Ex4-pg415

In [4]:
import math
##Intitalisation of variables
k= 0.012856 ##ohm^-1 cm^-1
R= 3468.9 ##ohms
k1= 44.597 ##cm^-1
c= 0.1 ##g equiv per litre
R1= 4573.6 ##ohms
##CALCULATIONS
k1= k*R
K= k1/R1
a= 1000.*K/c
##RESULTS
print'%s %.2f %s'% ('cell constant = ',k1,' cm^-1')
print'%s %.2f %s'% ('\n cell constant = ',K,' ohm^-1 cm^-1')
print'%s %.2f %s'% ('\n Equivalent conductance = ',a,' ohms^-1 cm^2')

cell constant =  44.60  cm^-1

 cell constant =  0.01  ohm^-1 cm^-1

 Equivalent conductance =  97.51  ohms^-1 cm^2

Ex5-pg421

In [5]:
import math
##Intitalisation of variables
A= 48.15 ##ohm^-1 cm6-1
m= 1.0283*10**-3 ##gms equiv acid per litre
A0= 390.7 ##ohms^-1 cm^2
A1= 60.2
B= 0.229
##CALCULATIONS
a= A/(A0-(A1+B*A0)*math.sqrt((A/A0)*m)) 
##RESULTS
print'%s %.4f %s'% ('Degree of dissociation = ',a,'')

Degree of dissociation =  0.1238

Ex6-pg424

In [6]:
import math
##Intitalisation of variables
t= 0.3965
A0= 126.45 ##ohm^-1 CM62
##CALCULATIONS
l= (1.-t)*A0
##RESULTS
print'%s %.2f %s'% ('ion conductance of the Cl- ion = ',l,' ohms^-1 cm^2')

ion conductance of the Cl- ion =  76.31  ohms^-1 cm^2

Ex7-pg426

In [7]:
import math
##Intitalisation of variables
A1= 426.16 ##ohms^-1 cm^2
A2= 91. ##ohms^-1 cm^2
A3= 126.45 ##ohms^-1 cm^2
a1= 61.92 ##ohms^-1 cm^2
a2= 76.34 ##ohms^-1 cm^2
a3= 63.64 ##ohms^-1 cm^2
a4= 79.8 ##ohms^-1 cm^2
##CALCULATIONS
A4= A1+A2-A3
A5= a1+a2
A6= a3+a4
##RESULTS
print'%s %.2f %s'% ('Conductance of CH3COOH = ',A4,' ohms^-1 cm^2') 
print'%s %.2f %s'% ('\n Conductance of AgCl = ',A5,' ohms^-1 cm^2') 
print'%s %.2f %s'% ('\n Conductance of BaSO4 = ',A6,' ohms^-1 cm^2') 

Conductance of CH3COOH =  390.71  ohms^-1 cm^2

 Conductance of AgCl =  138.26  ohms^-1 cm^2

 Conductance of BaSO4 =  143.44  ohms^-1 cm^2

Ex8-pg427

In [8]:
import math
##Intitalisation of variables
e= 5.6 ##volts
l= 9.8 ##cm
t= 1. ##hr
T= 25. ##C
A= 73.4 ##ohm^-1 cm^2
F= 96500. ##coloumbs
##CALCULATIONS
v= A/F
pg= e/l
v1= v*pg
L= v1*t*3600.
##RESULTS
print'%s %.2e %s'% ('Mobility = ',v,' cm/sec')
print'%s %.3f %s'% ('\n Potential gradient = ',pg,' volt/cm')
print'%s %.3f %s'% ('\n Potential gradient = ',pg,' volt/cm')
print'%s %.2f %s'% ('\n Distance moved by ion = ',L,' cm')

Mobility =  7.61e-04  cm/sec

 Potential gradient =  0.571  volt/cm

 Potential gradient =  0.571  volt/cm

 Distance moved by ion =  1.56  cm

Ex9-pg428

In [9]:
import math
##Intitalisation of variables

Ao= 138.3 ##ohms^-1
k1= 3.41*10**-6 ##ohm^-1 cm^-1
k2= 1.6*10**-6 ##ohm^-1 cm^-1
T= 25. ##C
##CALCULATIONS
s= 1000.*(k1-k2)/Ao
##RESULTS
print'%s %.2e %s'% ('Solubility of AgCl in water = ',s,' g equiv per liter')

Solubility of AgCl in water =  1.31e-05  g equiv per liter