# Chapter 12:Physical Properties of Solutions¶

## Example no:12.2,Page no:518¶

In [39]:
#Variable declaration
msolute=0.892 #mass of solute, g
msolvent=54.6 #mass of solvent, g

#Calculation
percent=msolute/(msolute+msolvent)*100 #concentration, percent by mass

#Result
print"The concentration of KCl solution by mass is :",round(percent,2),"%"

The concentration of KCl solution by mass is : 1.61 %


## Example no:12.3,Page no:518¶

In [40]:
#Variable declaration
mass=24.4 #mass of H2SO4, g
M=98.09 #mol maass of H2SO4, g

#Calculation
n=mass/M #moles of H2SO4
massH2O=0.198 #mass of H2O, kg
m=n/massH2O #molality of H2SO4, molal

#Result
print"The molality of sulfuric acid solution is :",round(m,2),"m"

The molality of sulfuric acid solution is : 1.26 m


## Example no:12.4,Page no:520¶

In [41]:
#Variable declaration
#considering 1L solution
msolution=976 #mass of solution, g
n=2.45 #moles
CH3OH=32.04 #mol. mass of CH3OH, g

#Calculation
msolute=n*CH3OH #mass of solute, g
msolvent=(msolution-msolute)/1000 #mass of solvent, kg
m=n/msolvent #molality, molal

#Result
print"The molality of CH3OH solution is :",round(m,2),"m"

The molality of CH3OH solution is : 2.73 m


## Example no:12.5,Page no:520¶

In [32]:
#Variable declaration
#considering 100g of solution
percent=35.4 #mass percent of H3PO4
H3PO4=97.99 #mol mass of H3PO4

#Calculation
n=percent/H3PO4 #moles of H3PO4
mH2O=(100-percent)/1000 #mass of solvent
m=n/mH2O #molality of H3PO4, molal

#Result
print"the molality of H3PO4 solution is :",round(m,2),"m"

the molality of H3PO4 solution is : 5.59 m


## Example no:12.6,Page no:525¶

In [42]:
#Variable declaration
c=6.8*10**-4 #solubility of N2 in water, M
P=1 #pressure, atm

#Calculation
k=c/P #henry's constant
#for partial pressure of N2=0.78atm
P=0.78 #partial pressure of N2, atm
c=k*P #solubility of N2, M

#Result
print"The solubility of N2 gas in water is :%.1e"%c,"M"

The solubility of N2 gas in water is :5.3e-04 M


## Example no:12.7,Page no:527¶

In [43]:
#Variable declaration
H2O=18.02 #mol mass of H2O, g
V=460 #volume of water, mL
glucose=180.2 #mol. mass of glucose, g
mass=218 #mass of gllucose, g

#Calculation
n1=V/H2O #moles of water
n2=mass/glucose #moles of glucose
x1=n1/(n1+n2) #mole fraction of water
P=31.82 #vapor pressure of pure water, mmHg
P1=x1*P #vapor pressure afteraddition of glucose, mmHg
#Result
print"Vapor pressure is:",round(P1,1),"mm Hg"
print"The vapor pressure lowering is :",round(P-P1,1),"mmHg"

Vapor pressure is: 30.4 mm Hg
The vapor pressure lowering is : 1.4 mmHg


## Example no:12.8,Page no:532¶

In [44]:
#Variable declaration
mH2O=2.505 #mass of H2O, kg
mEG=651 #mass of EG, g
EG=62.07 #mol mass of EG, g

#Calculation
n=mEG/EG #moles of EG
m=n/mH2O #molality of EG
Kf=1.86 #molal freezing point depression constant, C/m
deltaTf=Kf*m #depression in freezing point, C
Kb=0.52 #molal boiling point elevation constant, C/m
deltaTb=Kb*m #elevation in boiling point, C

#Result
print"The depression in freezing point is",round(deltaTf,2),"C"
print"Elevation in boiling point is :",round(deltaTb,1),"C"

The depression in freezing point is 7.79 C
Elevation in boiling point is : 2.2 C


## Example no:12.9,Page no:536¶

In [45]:
#Variable declaration
pie=30 #osmotic pressure, atm
R=0.0821 #gas constant, L atm/K mol
T=298 #temp., K

#Calculation
M=pie/(R*T) #molar concentration, M

#Result
print"The molar concentration is :",round(M,2),"M"

The molar concentration is : 1.23 M


## Example no:12.10,Page no:537¶

In [46]:
#Variable declaration
deltaTf=1.05 #depression in freezing point, C
Kf=5.12 #molal freezing point depression constant

#Calculation
m=deltaTf/Kf #molality of solution, molal
mbenzene=301/1000.0 #mass of benzene, kg
n=m*mbenzene #moles of sapmle
msample=7.85 #mass of sample, g
molarmass=msample/n #molar mass of sample, g/mol

#Result
print"The molar mass of the sample is :",round(molarmass),"g/mol "

The molar mass of the sample is : 127.0 g/mol


## Example no:12.11,Page no:538¶

In [47]:
#Variable declaration
R=0.0821 #gas constant, L atm/K mol
T=298 #temp, K
pie=10/760.0 #osmotic pressure, atm

#Calculation
M=pie/(R*T) #molarity of the solution, M
#taking 1L of solution
mass=35 #mass of Hg, g
n=M #moles
molarmass=mass/n #molar mass of hemoglobin, g/mol

#Result
print"The molar mass of the hemoglobin is :%.2e"%molarmass,"g/mol"

The molar mass of the hemoglobin is :6.51e+04 g/mol


## Example no:12.12,Page no:540¶

In [48]:
#Variable declaration
R=0.0821 #gas constant, L atm/K mol
T=298 #temp, K
pie=0.465 #osmotic pressure, atm
M=0.01 #molarity of the solution, M

#Calculation
i=pie/(M*R*T) #vant hoff factor of KI

#Result
print"The vant hoff factor of KI at 25 C is :",round(i,2)

The vant hoff factor of KI at 25 C is : 1.9