The flocculation values (mmol/L) for a sol are: NaCl 55, Na2SO4 54, CaCl2 0.70 and MgSO4 0.72. Which conclusion is correct?

The sol is negative, and Ca^2+ is about 79 times as powerful as Na+ in this sol

The sol is negative, and Ca^2+ is about 2 times as powerful as Na+ in this sol

The sol is positive, and Ca^2+ is about 79 times as powerful as Na+ in this sol

The sol is positive, and Cl- is about 79 times as powerful as SO4^2- in this sol

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JEE Advanced levelSingle correct optionDifficulty 3 of 3sc206

The flocculation values (mmol/L) for a sol are: $NaCl$ 55, $Na X_{2} SO X_{4}$ 54, $CaCl X_{2}$ 0.70 and $MgSO X_{4}$ 0.72. Which conclusion is correct?

AThe sol is negative, and $Ca X^{2 +}$ is about 79 times as powerful as $Na X^{+}$ in this sol
BThe sol is negative, and $Ca X^{2 +}$ is about 2 times as powerful as $Na X^{+}$ in this sol
CThe sol is positive, and $Ca X^{2 +}$ is about 79 times as powerful as $Na X^{+}$ in this sol
DThe sol is positive, and $Cl X^{-}$ is about 79 times as powerful as $SO X_{4} X^{2 -}$ in this sol

Answer: A

Changing the anion (

Cl X^{-}

SO X_{4} X^{2 -}

) barely changes the flocculation value (55 vs 54, 0.70 vs 0.72), while changing the cation charge from 1 to 2 changes it hugely. So the cation is the coagulating ion and the sol is negative. Relative power of

Ca X^{2 +}

Na X^{+}

= 55/0.70 = 78.6

, about 79.

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The flocculation values (mmol/L) for a sol are: NaCl 55, NaX2​SOX4​ 54, CaClX2​ 0.70 and MgSOX4​ 0.72. Which conclusion is correct?

Answer: A

More questions on Coagulation, Hardy-Schulze rule, protective colloids

The flocculation values (mmol/L) for a sol are: $NaCl$ 55, $Na X_{2} SO X_{4}$ 54, $CaCl X_{2}$ 0.70 and $MgSO X_{4}$ 0.72. Which conclusion is correct?