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 value \[\propto \frac{\text{1}}{\text{coagulating}\,\,\text{power}}\]\[Fe{{(OH)}_{3}}\]is a positively charged To coagulate\[Fe{{(OH)}_{3}},\]\[-ve\]charged 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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