Chapter 1 : Introduction to Equilibrium

Example 1.1 Page: 9

In [3]:
import math 

m_i = 10.            #[g]
m_w = 990.           #[g]
M_i = 342.3          #[g]
M_w = 18.            #[g]


x_i = m_i/(m_i+m_w)
x_i = x_i*100.      # [in percentage]

n_i = m_i/M_i       # number of moles of sucrose
n_w = m_w/M_w       # number of moles of water
x_j = n_i/(n_i+n_w)
x_j = x_j*100       # [in percentage]

m = n_i/m_w*1000            #[molal]
x_ppm = x_i*10**(6)/100.     #[ppm]

print " sucrose concentration in terms of the mass fraction is %f%%"%(x_i)
print " sucrose concentration in terms of the mole fraction is %f%%"%(x_j)
print " sucrose concentration in terms of the molality is      %f molal"%(m)
print " sucrose concentration in terms of the ppm is           %f ppm"%(x_ppm)

 sucrose concentration in terms of the mass fraction is 1.000000%
 sucrose concentration in terms of the mole fraction is 0.053088%
 sucrose concentration in terms of the molality is      0.029509 molal
 sucrose concentration in terms of the ppm is           10000.000000 ppm

Example 1.2 Page: 9

In [4]:
import math 

T = 20.                       #[C]
d = 1.038143/1000*10.**(6)    #[kg/m**(3)]
m_i = 10.                     #[g] mass of sucrose
M_i = 342.3                   #[g/mol] molecular weight of sucrose

m = 1.00                      #[kg]
V = m/d*1000                  #[L]

m_1 = m_i/V                   #[g/L]


m_2 = (m_i/M_i)/V             #[mol/L]

print " Mass concentration of the solution is %f g/L"%(m_1)
print " Mole concentration of the solution is %f mol/L"%(m_2)

print " Molarity of the solution is           %f mol/L"%(m_2)

 Mass concentration of the solution is 10.381430 g/L
 Mole concentration of the solution is 0.030328 mol/L
 Molarity of the solution is           0.030328 mol/L
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