Chapter 14 : The Properties of Solution

Example 14.1 Page No : 322

In [2]:
# Variables
import math 
M2 = 92. 			#gms
M1 = 78. 			#gms
pb = 118.2 			#mm
pt = 36.7 			#mm

# Calculations
n1 = M2/(M1+M2)
n2 = 1-n1
p1 = n1*pb
p2 = n2*pt
w = p1*M1/(p2*M2)

# Results
print  'partial pressure of benzene  = %.f mm'%(p1)
print  ' partial pressure of toulene  = %.1f mm'%(p2)
print  ' weight proportions  = %.2f '%(w)
partial pressure of benzene  = 64 mm
 partial pressure of toulene  = 16.8 mm
 weight proportions  = 3.22 

Example 14.2 Page No : 325

In [3]:
# Variables
vpe = 42 			#atm
p2 = 1. 			#atm

# Calculations
N2 = p2/vpe

# Results
print  'Ideal solubility of ethane  = %.3f mole fraction'%(N2)
Ideal solubility of ethane  = 0.024 mole fraction

Example 14.3 Page No : 325

In [4]:
# Variables
import math 
p1 = 25.7 			#atm
p2 = 11.84 			#atm
T1 = 173. 			#K
T2 = 153. 			#K
T3 = 25. 			#C

# Calculations
dH = math.log10(p1/p2)*4.579*T1*T2/(T1-T2)
p =  p1*10**((dH/4.576)*(273+T3-T1)/((273+T3)*T1))
s = 1./p

# Results
print  'Heat of reaction  = %d cal mole**-1'%(dH)
print  ' pressure  = %d atm'%(p)
print  ' Solubility of methane  = %.5f '%(s)

# note : rounding error is there. please check. 
Heat of reaction  = 2039 cal mole**-1
 pressure  = 309 atm
 Solubility of methane  = 0.00323 

Example 14.4 Page No : 329

In [5]:
# Variables
T1 = 20. 			#C
T2 = 80. 			#C
H1 = 4540. 			#cal mole**-1

# Calculations
n = 10**(H1*(-T2+T1)/(4.576*(273+T1)*(273+T2)))

# Results
print  'ideal solubility of napthalene  = %.3f '%(n)
ideal solubility of napthalene  = 0.266 

Example 14.5 Page No : 342

In [6]:
# Variables
R = 1.987 			#cal mole**-1 K**-1
T = 278.6 			#K
dH = 30.2 			#cal g**-1
m = 6.054 			#gms
a = 0.1263 			#degrees

# Calculations
l = R*T**2/(1000*dH)
m1 = a/l
M2 = m/m1

# Results
print  'molal depression consmath.tant  = %.2f '%(l)
print  ' molality  = %.4f '%(m1)
print  ' molecular weight of solute  = %.f gms'%(M2)
molal depression consmath.tant  = 5.11 
 molality  = 0.0247 
 molecular weight of solute  = 245 gms
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