Answer
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Hint: The precipitation of colloidal solution, with the help of addition of excess of an electrolyte is known as flocculation or coagulation. The electrolyte is known as flocculating agent, and the power of an electrolyte to precipitate a colloidal arrangement is known as flocculating power.
Complete answer:
It is resolved with the help of Hardy Schulze Rule, which expresses that more prominent the valency of the ion causing coagulation, greater will be its flocculating power.
Along this, if a colloidal solution can be precipitated by an anion, at that point higher the negative charge on the anion, higher will be its flocculating power.
And, if a colloidal solution is precipitated by a cation, then greater the positive charge on the cation, greater will be its flocculating power.
As we know arsenious sulphide ${ As }_{ 2 }{ O }_{ 3 }$ sol is negatively charged owing to preferential adsorption of ${ S }^{ 2- }$ ions. Cation would be effective in causing coagulation.
Flocculating value = minimum millimole of the effective ion per litre of sol
$\dfrac { 4\times 0.005 }{ \left( 4+16 \right) \times { 10 }^{ -3 } } $
$\dfrac { 0.02 }{ 0.02 } { =1.0 }$
Therefore, the flocculating value of the effective ion is ${ 1.0 }$.
So, the correct answer is “Option C”.
Additional Information:
The presence of small amounts of appropriate electrolytes is necessary for the stability of the colloids. However, when an electrolyte is added in larger concentration, the particles of the sol take up the ions which are oppositely charged and thus, get neutralized.
Then all the neutral particles start aggregating and producing particles of larger size which are then precipitated. This process of conglomeration of colloidal particles into an insoluble precipitate by the expansion of some reasonable electrolyte is known as coagulation.
It should be noticed that the coagulation of a colloidal arrangement by an electrolyte does not happen until the additional electrolyte has certain minimum concentration in the solution.
Note: The possibility to make a mistake is that flocculating value is millimoles of electrolyte per litre volume of solution. So, you have to add both the volume of ${ BaCl }_{ 2 }$ and ${ As }_{ 2 }{ O }_{ 3 }$.
Complete answer:
It is resolved with the help of Hardy Schulze Rule, which expresses that more prominent the valency of the ion causing coagulation, greater will be its flocculating power.
Along this, if a colloidal solution can be precipitated by an anion, at that point higher the negative charge on the anion, higher will be its flocculating power.
And, if a colloidal solution is precipitated by a cation, then greater the positive charge on the cation, greater will be its flocculating power.
As we know arsenious sulphide ${ As }_{ 2 }{ O }_{ 3 }$ sol is negatively charged owing to preferential adsorption of ${ S }^{ 2- }$ ions. Cation would be effective in causing coagulation.
Flocculating value = minimum millimole of the effective ion per litre of sol
$\dfrac { 4\times 0.005 }{ \left( 4+16 \right) \times { 10 }^{ -3 } } $
$\dfrac { 0.02 }{ 0.02 } { =1.0 }$
Therefore, the flocculating value of the effective ion is ${ 1.0 }$.
So, the correct answer is “Option C”.
Additional Information:
The presence of small amounts of appropriate electrolytes is necessary for the stability of the colloids. However, when an electrolyte is added in larger concentration, the particles of the sol take up the ions which are oppositely charged and thus, get neutralized.
Then all the neutral particles start aggregating and producing particles of larger size which are then precipitated. This process of conglomeration of colloidal particles into an insoluble precipitate by the expansion of some reasonable electrolyte is known as coagulation.
It should be noticed that the coagulation of a colloidal arrangement by an electrolyte does not happen until the additional electrolyte has certain minimum concentration in the solution.
Note: The possibility to make a mistake is that flocculating value is millimoles of electrolyte per litre volume of solution. So, you have to add both the volume of ${ BaCl }_{ 2 }$ and ${ As }_{ 2 }{ O }_{ 3 }$.
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