What is the normality of lead (II) nitrate if the density of its  26%(w/w) aqueous solution is 3.105 g/mL?  Take molar mass of lead (II) nitrate to be 331 g/mol. 
 

Consider 100 g of solution. It is made up of 26 g lead (II) nitrate and  74 g water. 
Volume of solution,  

V=100 g/(3.105 g/ml)=32.2061ml=0.0322 L 

V=100 g/(3.105 g/ml)=32.2061ml=0.0322 L 
Equivalent weight of lead nitrate  

=331/2=165.5 g/eq=331/2=165.5 g/eq 
Number of equivalents,  

N=26 g/(165.5 g/eq)=0.1571eqN=26 g/(165.5 g/eq)=0.1571eq 
Normality  =N/V=4.878 N=N/V=4.878 N