Match the items in List-I with the items in L | vedclass

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(B) . Negatively charged sol: When $FeCl_3$ is added to excess $NaOH$ solution, it forms a negatively charged $Fe(OH)_3$ sol due to the adsorption of

Vedclass · Accepted Answer

$A-III, B-I, C-IV, D-II$

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(B) . Negatively charged sol: When $FeCl_3$ is added to excess $NaOH$ solution, it forms a negatively charged $Fe(OH)_3$ sol due to the adsorption of $OH^-$ ions. Thus, $A-III$.$B$. Milk: Milk is a well-known example of an emulsion (liquid in liquid). Thus, $B-I$.$C$. Gold number: It is a measure of the protective power of a lyophilic colloid. Thus, $C-IV$.$D$. Colloidal antimony: It is used in the treatment of the disease kala-azar. Thus, $D-II$.Therefore, the correct matching is $A-III, B-I, C-IV, D-II$.

List-$I$	List-$II$
$A$. Negatively charged sol	$I$. Emulsion
$B$. Milk	$II$. Kala-azar
$C$. Gold number	$III$. $FeCl_3$ solution is added to excess $NaOH$ solution
$D$. Colloidal antimony	$IV$. Protection of colloids

Match the items in List-$I$ with the items in List-$II$.
List-$I$ List-$II$
$A$. Negatively charged sol $I$. Emulsion
$B$. Milk $II$. Kala-azar
$C$. Gold number $III$. $FeCl_3$ solution is added to excess $NaOH$ solution
$D$. Colloidal antimony $IV$. Protection of colloids

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Match the items in List-$I$ with the items in List-$II$.List-$I$List-$II$$A$. Negatively charged sol$I$. Emulsion$B$. Milk$II$. Kala-azar$C$. Gold number$III$. $FeCl_3$ solution is added to excess $NaOH$ solution$D$. Colloidal antimony$IV$. Protection of colloids