Question

The coagulation of 200 mL of a positive colloid took place when 0.73 g HCl was added to it without changing the volume much. The flocculation value of HCl for the colloid is:
A: 100
B: 36.5
C: 0.365
D: 150

Answer 1

Hint: Coagulation refers to a phenomenon in which small particles are combined together to form large aggregates (i.e. flocs). Coagulation is generally used to adsorb the dissolved organic matter over particulate aggregates to remove these impurities in successive liquid/solid separation processes. On the other hand, flocculation refers to a mixing process which promotes agglomeration of the particles and helps in the particle settling. The minimal concentration of an electrolyte required for the coagulation or flocculation of a solution to occur is called flocculation value

Complete step by step answer:
The value of flocculation refers to the electrolyte amount (in millimoles, (mmol)) for the complete coagulation of unit volume (1 L) of the colloidal solution.
Amount of HCl required by 200 mL of the positive colloid solution = 0.73 g HCl
Thus, number of moles of HCl $ = \dfrac{{0.73}}{{36.5}} = 0.02mol = 20mmol$
Therefore, number or millimoles of HCl required by 1000 mL or 1 L of the positive colloid solution:
$\dfrac{{20}}{{200}} \times 1000 = 100mmol$

As a result, the correct option is A i.e. The flocculation value of HCl for the colloid is 100.

Note:
In accordance with Hardy Schulze rule, more is the charge possessed by the oppositely charged ion of the added electrolyte, more effectively it can cause coagulation. We can say that the flocculation value of the electrolyte is inversely proportional to coagulating power of the effective ion, i.e. ${\text{Flocculation value }}\propto {\text{ }}\dfrac{1}{{{\text{Coagulating power}}}}$.