Question

The coagulation values in millimoles per litre of the electrolytes used for the coagulation of $As_2S_3$ are given below :
I. $(NaCl) = 52$,
II. $(BaCl_2) = 0.69$,
III. $(MgSO_4) = 0.22$
The correct order of the their coagulating power is

• A. I > II > III
• B. II > I > III
• C. III > II > I
• D. III > I > II

Answer 1

The Correct Option is C

The question involves determining the coagulating power of different electrolytes used for coagulating As_2S_3. The coagulation power of an electrolyte is inversely related to its coagulation value; lower coagulation value corresponds to higher coagulating power.

Let's analyze the coagulation values given:

1. For NaCl, the coagulation value is 52 millimoles per litre.
2. For BaCl_2, the coagulation value is 0.69 millimoles per litre.
3. For MgSO_4, the coagulation value is 0.22 millimoles per litre.

According to Hardy-Schulze rule, the coagulating power of an electrolyte increases with an increase in the valency of its oppositely charged ion. Hence, the smaller the coagulation value, the higher the coagulating power.

Arranging the given electrolytes in order of their coagulation values from highest to lowest:

• MgSO_4 = 0.22
• BaCl_2 = 0.69
• NaCl = 52

This clearly implies the order of their coagulating power is:

• MgSO_4 (highest coagulating power)
• BaCl_2
• NaCl (lowest coagulating power)

Thus, the correct order of the coagulating power is:

III > II > I

Therefore, the correct answer is III > II > I.