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 \[\propto \frac{1}{Coagulating\text{ }power}\] \[Fe{{(OH)}_{3}}\] is a positively charged sol. 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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