## Example: 11.1 Page: 458¶

In [4]:
from __future__ import division
print "Example: 11.1 - Page: 458\n\n"

# Solution

#*****Data******#
a = 2.423## [g]
b = 100## [g]
Lf = 35.7## [cal/g]
Tf = 353.1## [cal/g]
delta_Tf = 0.64## [OC]
R = 2## [cal/mol K]
Mw = 32## [Molecular wt. of Sulphur, g/mol]
#*************#

M2 =  ((R*Tf**2/(1000*Lf))*(a*1000/(b)))/delta_Tf## [g/mol]
n = M2/Mw#
print "Molecular Formula of Sulphur is S%d"%(round(n))#

Example: 11.1 - Page: 458

Molecular Formula of Sulphur is S8


## Example: 11.2 Page: 459¶

In [5]:
from __future__ import division
print "Example: 11.2 - Page: 459\n\n"

# Solution

#*****Data******#
Tf = 5 + 273## [K]
Lf = 9830## [J/mol]
R = 8.314## [J/mol K]
M1 = 78## [kg/kmol]
#**************#

Kf = R*Tf**2*M1/(1000*Lf)## [kg/kmol]
print "Molal Freezing point is %.2f kg/kmol\n"%(Kf)#

Example: 11.2 - Page: 459

Molal Freezing point is 5.10 kg/kmol



## Example: 11.3 Page: 458¶

In [6]:
from __future__ import division
print "Example: 11.3 - Page: 458\n\n"

# Solution

#*****Data******#
T_melting = 40## [OC]
Tf = T_melting + 273## [K]
a = 0.172## [g]
b = 12.54## [g]
T_new = 39.25## [OC]
M2 = 135## [Molecular wt. of acetanilide, g/mol]
R = 2## [cal/mol K]
#**************#

delta_T = T_melting - T_new## [OC]
Kf = delta_T*b*M2/(1000*a)#
Lv = ((R*Tf**2/(1000)))/Kf## [cal/g]
print "Latent Heat of Fusion of phenol is %.2f cal/g\n"%(Lv)#

Example: 11.3 - Page: 458

Latent Heat of Fusion of phenol is 26.54 cal/g



## Example: 11.4 Page: 461¶

In [7]:
from __future__ import division
print "Example: 11.4 - Page: 461\n\n"

# Solution

#*****Data******#
T_boiling = 118.24## [OC]
Tb = T_boiling + 273## [K]
a = 0.4344## [g]
b = 44.16## [g]
Lv = 121## [cal/g]
T_new = 118.1## [OC]
R = 2## [cal/mol K]
#**************#

delta_Tb = T_boiling - T_new## [OC]
M2 = (R*Tb**2/(1000*Lv))*(a*1000/(b*delta_Tb))#
print "Molecular weight of anthracene is %d kg/kmol"%(round(M2))#

Example: 11.4 - Page: 461

Molecular weight of anthracene is 178 kg/kmol


## Example: 11.5 Page: 462¶

In [8]:
from __future__ import division
print "Example: 11.5 - Page: 462\n\n"

# Solution

#*****Data******#
delta_Tb = 2.3## [K]
w1 = 100## [g]
M1 = 78## [g/mol]
w2 = 13.86## [g]
M2 = 154## [g/mol]
Tb = 353.1## [K]
R = 8.314## [J/mol K]
#****************#

# Molality:
m = w2*1000/(w1*M2)## [mol/kg]
# Molal Elevation Constant:
Kb = delta_Tb/m## [K kg/mol]
# Molar Latent Heat of Vaporisation:
Lv = R*Tb**2*M1/(1000*Kb)## [J/mol]
print "Molar Latent Heat of Vaporisation is %d J/mol"%(Lv)#

Example: 11.5 - Page: 462

Molar Latent Heat of Vaporisation is 31638 J/mol


## Example: 11.6 Page: 465¶

In [9]:
from __future__ import division
print "Example: 11.6 - Page: 465\n\n"

# Solution

#*****Data******#
Temp = 50 + 273## [K]
w2 = 60## [g]
w1 = 1500## [g]
M1 = 18## [g/mol]
M2 = 180## [g/mol]
Vl = 18*10**(-6)## [Molar Volume of water, cubic m/mol]
R = 8.314## [J/mol K]
#***************#

# Mole fraction of glucose:
x2 = (w2/M2)/((w2/M2) + (w1/M1))#
# Applying Eqn. (11.45):
P = R*Temp*x2/Vl## [N/square m]
P = P/1000## [kPa]
print "Osmotic Pressure is %.2f kPa\n"%(P)#

Example: 11.6 - Page: 465

Osmotic Pressure is 594.38 kPa



## Example: 11.7 Page: 465¶

In [10]:
from __future__ import division
print "Example: 11.7 - Page: 465\n\n"

# Solution

#*****Data******#
w2 = 0.6## [g]
w3 = 1.8## [g]
Temp = 27 + 273## [K]
V1 = 100## [cubic cm]
M2 = 60## [g/mol]
M3 = 180## [g/mol]
R = 0.082## [L.atm/mol.K]
#****************#

V1 = V1/1000## [litre]
# C: Concentration per litre
C = ((w2/M2) + (w3/M3))/V1## [mol/litre]
P = C*R*Temp## [atm]
print "Osmotic Pressure of the solution is %.2f atm"%(P)#

Example: 11.7 - Page: 465

Osmotic Pressure of the solution is 4.92 atm