Chapter 13 : Dilute solution laws

Example 13.1 Page No : 478

In [1]:
# Variables
weight = 10.     			 #weight of NaCl in grams
volume = 1. 	    		 #volume of water in litres
weight_water = 1000. 			 #  weight of water in grams (Weight = Volume*Density, density of water  = 1g/cc = 1g/ml = 1000g/l)
molwt_NaCl = 58.5 			 #molecular weight of NaCl in grams
molwt_water = 18. 			 #molecular weight of water in grams
hf = 6.002; 			 #enthalpy change of fusion in kJ/mol at 0 degree celsius
P = 101.325; 			 #pressure in kPa
T = 273.15; 			 # freezing point temperature of water at the given pressure in K
R = 8.314; 			 #universal gas constant in J/molK;

# Calculations
x2 = (weight/molwt_NaCl)/((weight/molwt_NaCl)+(weight_water/molwt_water))
delt = (R*T**2*x2)/(hf*10**3)

# Results
print ' The depression in freezing point of water when 10g of NaCl solute is added  =  %0.2f K'%(delt);
 The depression in freezing point of water when 10g of NaCl solute is added  =  0.32 K

Example 13.2 Page No : 480

In [2]:
# Variables
weight = 10.; 		    	 #weight of NaCl in grams
volume = 1.; 			     #volume of water in litres
weight_water = 1000.; 			 #  weight of water in grams (Weight = Volume*Density, density of water  = 1g/cc = 1g/ml = 1000g/l)
molwt_NaCl = 58.5; 			 #molecular weight of NaCl in grams
molwt_water = 18; 			 #molecular weight of water in grams
lat_ht = 2256.94; 			 #latent heat of vaporization in kJ/kg at 100 degree celsius (obtained from steam tables)
P = 101.325; 			 #pressure in kPa
T = 373.15; 			 #boiling point temperature of water at the given pressure in K
R = 8.314; 			 #universal gas constant in J/molK

# Calculations
x2 = 0.0031
hv = (lat_ht*molwt_water)/1000
delt = (R*T**2*x2)/(hv*10**3)

# Results
print ' The elevation in boiling point of water when 10g of NaCl solute is added  =  %0.2f K'%(delt);
 The elevation in boiling point of water when 10g of NaCl solute is added  =  0.09 K

Example 13.3 Page No : 481

In [3]:
# Variables
weight = 10.; 			 #weight of NaCl in grams
weight_water = 1000.; 			 #  weight of water in grams
molwt_NaCl = 58.5; 			 #molecular weight of NaCl in grams
molwt_water = 18.; 			 #molecular weight of water in grams
T = 300.; 			 #prevailing temperature of water in K
R = 8.314; 			 #universal gas constant in (Pa m**3)/(mol K);
v = 18*10**-6;			 #molar volume in m**3/mol

# Calculations
x2 = 0.0031
pi = ((R*T*x2)/v)*10**-3

# Results
print ' The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K  =  %0.2f kPa'%(pi);
 The osmotic pressure of a solution conatining 10g of NaCl in 1000g of water at 300K  =  429.56 kPa

Example 13.4 Page No : 483

In [5]:
import math


# Variables
temp = 20. 			 # prevailing tempearture in degree celsius
melt_temp = 80.05; 			 # melting point of naphthalene in degree celsius
hf = 18.574 			 # enthalpy of fusion in kJ/mol
R = 8.314 			 # universal gas constant in J/molK

# Calculations
t = temp+273.15
melt_t = melt_temp+273.15
x2 = math.exp(((hf*10**3)/R)*((1./melt_t)-(1./t)))

# Results
print ' The ideal solubility of naphthalene at 20 degree celsius =  %0.4f'%(x2);
 The ideal solubility of naphthalene at 20 degree celsius =  0.2737

Example 13.5 Page No : 483

In [6]:
# Variables
t = 295.43; 			 #prevailing temperature in K
sat_p = 6.05; 			 #Sasturation pressure of carbon dioxide at the prevailing temperature in MPa
p = 0.1; 			 #pressure at which solubility has to be determined in MPa

# Calculations
x2 = p/sat_p

# Results
print ' The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution\
 at 0.1MPa =  %0.4f'%(x2);
 The solubility of carbon dioxide expressed in mole fraction of carbon dioxide in solution at 0.1MPa =  0.0165