Flocculation value of BaCl_2 is much less tha | vedclass

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(D) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to its valence.For sol $A$,the flocculation value of

Vedclass · Accepted Answer

Sol $A$ is negatively charged and sol $B$ is positively charged.

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(D) According to the Hardy-Schulze rule,the coagulating power of an ion is directly proportional to its valence.For sol $A$,the flocculation value of $BaCl_2$ $(Ba^{2+})$ is less than $KCl$ $(K^+)$,which implies that the cation $Ba^{2+}$ is more effective at coagulation than $K^+$. Therefore,sol $A$ must be negatively charged.For sol $B$,the flocculation value of $Na_2SO_4$ $(SO_4^{2-})$ is less than $NaBr$ $(Br^-)$,which implies that the anion $SO_4^{2-}$ is more effective at coagulation than $Br^-$. Therefore,sol $B$ must be positively charged.Thus,sol $A$ is negatively charged and sol $B$ is positively charged.

Column-$I$	Column-$II$
$A$. Liquid Aerosol	$P$. Milk
$B$. Solid Sol	$Q$. Fog
$C$. Emulsion	$R$. Gel
$D$. Jelly	$S$. Alloy

Flocculation value of $BaCl_2$ is much less than that of $KCl$ for sol $A$ and flocculation value of $Na_2SO_4$ is much less than that of $NaBr$ for sol $B.$ The correct statement among the following is

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Column-$I$ Column-$II$

$A$. Liquid Aerosol $P$. Milk

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