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 :
\[Floclation\,value\propto \frac{1}{coagulating\,power}\] \[\text{Fe(OH}{{\text{)}}_{\text{3}}}\] is a positively charged To coagulate \[\text{Fe(OH}{{\text{)}}_{\text{3}}},\]\[~-\text{ve}\] charged electrolyte is used and greater the value of \[-\text{ve}\] charge, coagulating power will be strong. Among the given electrolytes, \[\text{NaCl}\] has lowest coagulating power, so its flocculation value will be maximum.
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