Question

In order to coagulate 400 ml of $Fe(OH{)}_3$ sol completely 60 ml of 0.1 M $KCl$ is required. What will be the coagulation power of $KCl$?

• A. $6$
• B. $\frac{1}{6}$
• C. $15$
• D. $\frac{1}{15}$

Answer 1

The correct option is D $\frac{1}{15}$

Coagulation value is the millimoles of an electrolyte that must be added to 1 L of a colloidal solution for complete coagulation.
Given,
For 400 mL $Fe(OH{)}_3$, millimoles of $KCl$ required $=\text{molarity}\times \text{volume(mL)}=0.1\times 60=6$
Hence,
For 1000 mL $Fe(OH{)}_3$, millimoles of KCl required
$=\frac{6}{400}\times 1000=15$
Coagulaton power $=\frac{1}{\text{coagulation value}}=\frac{1}{15}$