A) \[NaCl\]
B) \[N{{a}_{2}}S\]
C) \[{{(N{{H}_{4}})}_{3}}P{{O}_{4}}\]
D) \[{{K}_{2}}S{{O}_{4}}\]
Correct Answer: A
Solution :
\[Flocculation\text{ }value\propto \frac{1}{Coagulating\text{ }power}\]\[Fe{{(OH)}_{3}}\]is a positively charged To coagulate\[Fe{{(OH)}_{3}},\]\[-ve\]charge electrolyte is used and greater the value of\[-ve\]charge, coagulating power will be strong. Among the given electrolytes,\[NaCl\]has lowest coagulating power. So, its flocculation value will be maximum.
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