Amount of 0.1 M AB solution added is =10−5=5mL 0.1M AB left behind in the solution is =0.4mL 0.1M AB used up for coagulation =5−0.4=4.6mL 4.6mL of 0.1M AB contains AB=
(4.6X0.1)
1000
=0.46 milimoles As total volume of sol after mixing with electrolyte is 10mL, amount of AB required for Coagulation of 1 L of the sol=46 mill moles, hence flocculation value =46