Among the electrolytes $N a_{2} S O_{4}, C a C l_{2}, A l_{2} (S O_{4})_{3}$ and $N H_{4} C l,$ the most effective coagulating agent for $S b_{2} S_{3}$ solution is:

N a_{2} S O_{4}

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C a C l_{2}

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A l_{2} (S O_{4})_{3}

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N H_{4} C l

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Solution

The correct option is C $A l_{2} (S O_{4})_{3}$
$S b_{2} S_{3}$ is a negative sol.
$A l_{2} (S O_{4})_{3}$ will be the most effective coagulant due to higher charge density on $A l^{3 +}$ by the Hardy-Schulze rule.
According to the Hardy-Schulze rule, the greater the valency of active ion will be its coagulating power.
Thus, the Hardy-Schulze rule states the following:
i. The ions carrying the charge opposite to that of sol particles are effective in causing coagulation of the sol.
ii. The coagulating power of an electrolyte is directly proportional to the valency of the active ions (ions causing coagulation).
The order of effectiveness of cations is as follows:

$A l^{3 +} > C a^{2 +} > N a^{+} > N H_{4}^{+}$

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Among the electrolytes Na2SO4,CaCl2,Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 solution is:

Among the electrolytes $N a_{2} S O_{4}, C a C l_{2}, A l_{2} (S O_{4})_{3}$ and $N H_{4} C l,$ the most effective coagulating agent for $S b_{2} S_{3}$ solution is: