Chapter 16 : Solutions of Electrolytes

Example 16.1 Page No : 380

In [1]:
# Variables
v = 1.
m = 0.5

# Calculations
m1  = 2*m
m2  = 1*m
v1  = 2*v
v2  = 1*v
M  = (m1**2*m2)**(1/(v1+v2))

# Results
print  'mean ionic molality  = %.1f '%(m2)
print  ' mean ionic molality  = %.3f '%(M)

mean ionic molality  = 0.5 
 mean ionic molality  = 0.794

Example 16.2 Page No : 399

In [1]:
# Variables
n = 2
m = 0.01422
m1 = 0.00869
m2 = 0.025

# Calculations
M  = m2+m1
M1 = (M*m1)**(1./n)
r = m/M1

# Results
print  'mean ionic molality  = %.3f '%(r)

mean ionic molality  = 0.831

Example 16.3 Page No : 400

In [3]:
# Variables
mu = 1
mb = 2
m = 1 
m1 = 2

# Calculations
ym1 = 0.5*(mu*m**2+mu*m**2)
ym2 = 0.5*(mb*m**2+m*m1**2)
ym3 = 0.5*(mu*m1**2+mu*m1**2)

# Results
print  'ionic strength of solution  = %.f *m'%(ym1)
print  ' ionic strength of solution  = %.f *m'%(ym2)
print  ' ionic strength of solution  = %.f *m'%(ym3)

ionic strength of solution  = 1 *m
 ionic strength of solution  = 3 *m
 ionic strength of solution  = 4 *m
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