# Chapter 4:Reactions in Aqueous Solutions¶

## Example no:4.6,Page no:148¶

In :
#Variable declaration
K2Cr2O7=294.2 #mol mass of K2Cr2O7, g
M=2.16 #Concentration of K2Cr2O7, M
V=0.250 #volume of K2Cr2O7, L

#Calculation
moles=M*V #moles of K2Cr2O7
mass=moles*K2Cr2O7

#Result
print"The mass of the K2Cr2O7 needed is :",round(mass),"g K2Cr2O7\n"

The mass of the K2Cr2O7 needed is : 159.0 g K2Cr2O7



## Example no:4.7,Page no:149¶

In :
#Variable declaration
mGlucose=3.81 #mass of Glucose, g
Glucose=180.2 #mol mass of Glucose, g
M=2.53 #Concentration of Glucose, M

#Calculation
moles=mGlucose/Glucose #moles of Glucose
V=moles/M #volume of Glucose, L

#Result
print"The volume of the Glucose needed is :",round(V*1000,2),"mL soln\n"

The volume of the Glucose needed is : 8.36 mL soln



## Example no:4.8,Page no:150¶

In :
#Variable declaration
M2=1.75 #final Concentration of H2SO4, M
V2=500 #final volume of H2SO4, mL
M1=8.61 #initial Concentration of H2SO4, M

#Calculation
V1=M2*V2/M1 #initail volume of H2SO4, mL

#Result
print"The volume of the H2SO4 needed to dilute the solution is :",round(V1),"mL"

The volume of the H2SO4 needed to dilute the solution is : 102.0 mL


## Example no:4.9,Page no:152¶

In :
#Variable declaration
mSample=0.5662 #mass of sample, g
Cl=35.5 #mol mass of Cl, g
AgCl=143.4 #mol mass of AgCl, g
mAgCl=1.0882 #mass of AgCl formed, g

#Calculation
p_Cl_AgCl=Cl/AgCl*100.0 #percent Cl in AgCl
mCl=p_Cl_AgCl*mAgCl/100.0 #mass of Cl in AgCl, g
mCl=round(mCl,3)
#the same amount of Cl is present in initial sample
p_Cl=mCl/mSample*100.0 #percent Cl in initial sample

#Result
print"The percentage of Cl in sample is :",round(p_Cl,2),"%"

The percentage of Cl in sample is : 47.51 %


## Example no:4.10,Page no:154¶

In :
#Variable declaration
mKHP=0.5468 #mass of KHP, g
KHP=204.2 #mol mass of KHP, g

#Calculation
nKHP=mKHP/KHP #moles of KHP
VNaOH=23.48 #volume of NaOH, mL
MNaOH=nKHP/VNaOH*1000 #molarity of NaOH sol, M

#Result
print"The molarity of NaOH solution is :",round(MNaOH,4),"M\n"

The molarity of NaOH solution is : 0.114 M



## Example no:4.11,Page no:155¶

In :
#Variable declaration
MNaOH=0.610 #molarity of NaOH, M
VH2SO4=20 #volume of H2SO4, mL
MH2SO4=0.245 #molarity of H2SO4, M

#Calculation
nH2SO4=MH2SO4*VH2SO4/1000 #moles of H2SO4
VNaOH=2*nH2SO4/MNaOH #Volume of NaOH, L

#Result
print"The volume of NaOH solution is :",round(VNaOH*1000,1),"mL\n"

The volume of NaOH solution is : 16.1 mL



## Example no:4.12,Page no:157¶

In :
#Variable declaration
MKMnO4=0.1327 #molarity of KMnO4, M
VKMnO4=16.42 #volume of KMnO4, mL

#Calculation
nKMnO4=MKMnO4*VKMnO4/1000
nFeSO4=5*nKMnO4
VFeSO4=25 #volume of FeSO4, mL
MFeSO4=nFeSO4/VFeSO4*1000

#Result
print"The molarity of FeSO4 solution is :",round(MFeSO4,3),"M\n"

The molarity of FeSO4 solution is : 0.436 M