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SOLUTIONS

277565 A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of ${BaCl}_{2}$ , and $NaCl$ required to coagulate the given colloid is in the order

1

NaCl > {BaCl}_{2} > {AlCl}_{3}

2

{BaCl}_{2} > {AlCl}_{3} > NaCl

3

{AlCl}_{3} = NaCl = {BaCl}_{2}

4

{AlCl}_{3} > {BaCl}_{2} > NaCl

Explanation:

According to the Hardy Schulze rule, the effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulation, greater is its power to bring about coagulation.
Hence, the electric field subjected to the coagulation power should be in the order of ${AlCl}_{3} > {BaCl}_{2} > NaCl$

Kerala-CEE-29.08.2021

SOLUTIONS

277568 Among the following the ion which will be more effective for flocculation of $Fe (OH)_{3}$ solution is

1

{PO}_{4}^{3 -}

2

{SO}_{4}^{2 -}

3

{SO}_{3}^{2 -}

4

{NO}_{3}^{-}

Explanation:

$Fe (OH)_{3}$ solution is positively charged solution. It will be coagulated or flocculation by negatively charged ions.
${PO}_{4}^{3 -}$ have maximum negative charge among the given electrolytes.
Hence, ${PO}_{4}^{3 -}$ is most effective for coagulation of $Fe (OH)_{3}$

WB-JEE-2020

SOLUTIONS

277569 The formation of association of colloidal particles by addition of electrolyte to form as insoluble precipitate is called

1 Flocculation

2 Emulsification

3 Coagulation

4 Micelle

GUJCET-2019

SOLUTIONS

277575 The coagulation values in millimoles per litre of the electrolyes used for the coagulation of ${As}_{2} S_{3}$ are given below: I. $(NaCl) = 52$ ,
II. $({BaCl}_{2}) = 0.69$ ,
III. $({MgSO}_{4}) = 0.22$
The correct order of their coagulating power is

1 I

>

II

>

III

2 II

>

I

>

III

3 III

>

II

>

I

4 III

>

I

>

II

Explanation:

Coagulation power $\propto \frac{1}{Coagulation values}$
Higher the coagulation power, lower is the coagulation values in millimoles per liter.
${MgSO}_{4} > {BaCl}_{2} > NaCl$

(NEET-II 2016)

SOLUTIONS

277577 The pair of compound which cannot exist together solution is

1

{NaHCO}_{3}

and

NaOH

2

{NaHCO}_{3}

and

H_{2} O

3

{NaHCO}_{3}

and

{Na}_{2} {CO}_{3}

4

{Na}_{2} {CO}_{3}

and

NaOH

Explanation:

${NaHCO}_{3}$ and $NaOH$ compound cannot exist together in solution as-
${NaHCO}_{3} + NaOH \to {Na}_{2} {CO}_{3} + H_{2} O$
Neutral salt

Karnataka-CET-2015

SOLUTIONS

277565 A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of ${BaCl}_{2}$ , and $NaCl$ required to coagulate the given colloid is in the order

1

NaCl > {BaCl}_{2} > {AlCl}_{3}

2

{BaCl}_{2} > {AlCl}_{3} > NaCl

3

{AlCl}_{3} = NaCl = {BaCl}_{2}

4

{AlCl}_{3} > {BaCl}_{2} > NaCl

Explanation:

According to the Hardy Schulze rule, the effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulation, greater is its power to bring about coagulation.
Hence, the electric field subjected to the coagulation power should be in the order of ${AlCl}_{3} > {BaCl}_{2} > NaCl$

Kerala-CEE-29.08.2021

SOLUTIONS

277568 Among the following the ion which will be more effective for flocculation of $Fe (OH)_{3}$ solution is

1

{PO}_{4}^{3 -}

2

{SO}_{4}^{2 -}

3

{SO}_{3}^{2 -}

4

{NO}_{3}^{-}

Explanation:

$Fe (OH)_{3}$ solution is positively charged solution. It will be coagulated or flocculation by negatively charged ions.
${PO}_{4}^{3 -}$ have maximum negative charge among the given electrolytes.
Hence, ${PO}_{4}^{3 -}$ is most effective for coagulation of $Fe (OH)_{3}$

WB-JEE-2020

SOLUTIONS

277569 The formation of association of colloidal particles by addition of electrolyte to form as insoluble precipitate is called

1 Flocculation

2 Emulsification

3 Coagulation

4 Micelle

GUJCET-2019

SOLUTIONS

277575 The coagulation values in millimoles per litre of the electrolyes used for the coagulation of ${As}_{2} S_{3}$ are given below: I. $(NaCl) = 52$ ,
II. $({BaCl}_{2}) = 0.69$ ,
III. $({MgSO}_{4}) = 0.22$
The correct order of their coagulating power is

1 I

>

II

>

III

2 II

>

I

>

III

3 III

>

II

>

I

4 III

>

I

>

II

Explanation:

Coagulation power $\propto \frac{1}{Coagulation values}$
Higher the coagulation power, lower is the coagulation values in millimoles per liter.
${MgSO}_{4} > {BaCl}_{2} > NaCl$

(NEET-II 2016)

SOLUTIONS

277577 The pair of compound which cannot exist together solution is

1

{NaHCO}_{3}

and

NaOH

2

{NaHCO}_{3}

and

H_{2} O

3

{NaHCO}_{3}

and

{Na}_{2} {CO}_{3}

4

{Na}_{2} {CO}_{3}

and

NaOH

Explanation:

${NaHCO}_{3}$ and $NaOH$ compound cannot exist together in solution as-
${NaHCO}_{3} + NaOH \to {Na}_{2} {CO}_{3} + H_{2} O$
Neutral salt

Karnataka-CET-2015

SOLUTIONS

277565 A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of ${BaCl}_{2}$ , and $NaCl$ required to coagulate the given colloid is in the order

1

NaCl > {BaCl}_{2} > {AlCl}_{3}

2

{BaCl}_{2} > {AlCl}_{3} > NaCl

3

{AlCl}_{3} = NaCl = {BaCl}_{2}

4

{AlCl}_{3} > {BaCl}_{2} > NaCl

Explanation:

According to the Hardy Schulze rule, the effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulation, greater is its power to bring about coagulation.
Hence, the electric field subjected to the coagulation power should be in the order of ${AlCl}_{3} > {BaCl}_{2} > NaCl$

Kerala-CEE-29.08.2021

SOLUTIONS

277568 Among the following the ion which will be more effective for flocculation of $Fe (OH)_{3}$ solution is

1

{PO}_{4}^{3 -}

2

{SO}_{4}^{2 -}

3

{SO}_{3}^{2 -}

4

{NO}_{3}^{-}

Explanation:

$Fe (OH)_{3}$ solution is positively charged solution. It will be coagulated or flocculation by negatively charged ions.
${PO}_{4}^{3 -}$ have maximum negative charge among the given electrolytes.
Hence, ${PO}_{4}^{3 -}$ is most effective for coagulation of $Fe (OH)_{3}$

WB-JEE-2020

SOLUTIONS

277569 The formation of association of colloidal particles by addition of electrolyte to form as insoluble precipitate is called

1 Flocculation

2 Emulsification

3 Coagulation

4 Micelle

GUJCET-2019

SOLUTIONS

277575 The coagulation values in millimoles per litre of the electrolyes used for the coagulation of ${As}_{2} S_{3}$ are given below: I. $(NaCl) = 52$ ,
II. $({BaCl}_{2}) = 0.69$ ,
III. $({MgSO}_{4}) = 0.22$
The correct order of their coagulating power is

1 I

>

II

>

III

2 II

>

I

>

III

3 III

>

II

>

I

4 III

>

I

>

II

Explanation:

Coagulation power $\propto \frac{1}{Coagulation values}$
Higher the coagulation power, lower is the coagulation values in millimoles per liter.
${MgSO}_{4} > {BaCl}_{2} > NaCl$

(NEET-II 2016)

SOLUTIONS

277577 The pair of compound which cannot exist together solution is

1

{NaHCO}_{3}

and

NaOH

2

{NaHCO}_{3}

and

H_{2} O

3

{NaHCO}_{3}

and

{Na}_{2} {CO}_{3}

4

{Na}_{2} {CO}_{3}

and

NaOH

Explanation:

${NaHCO}_{3}$ and $NaOH$ compound cannot exist together in solution as-
${NaHCO}_{3} + NaOH \to {Na}_{2} {CO}_{3} + H_{2} O$
Neutral salt

Karnataka-CET-2015

SOLUTIONS

277565 A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of ${BaCl}_{2}$ , and $NaCl$ required to coagulate the given colloid is in the order

1

NaCl > {BaCl}_{2} > {AlCl}_{3}

2

{BaCl}_{2} > {AlCl}_{3} > NaCl

3

{AlCl}_{3} = NaCl = {BaCl}_{2}

4

{AlCl}_{3} > {BaCl}_{2} > NaCl

Explanation:

According to the Hardy Schulze rule, the effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulation, greater is its power to bring about coagulation.
Hence, the electric field subjected to the coagulation power should be in the order of ${AlCl}_{3} > {BaCl}_{2} > NaCl$

Kerala-CEE-29.08.2021

SOLUTIONS

277568 Among the following the ion which will be more effective for flocculation of $Fe (OH)_{3}$ solution is

1

{PO}_{4}^{3 -}

2

{SO}_{4}^{2 -}

3

{SO}_{3}^{2 -}

4

{NO}_{3}^{-}

Explanation:

$Fe (OH)_{3}$ solution is positively charged solution. It will be coagulated or flocculation by negatively charged ions.
${PO}_{4}^{3 -}$ have maximum negative charge among the given electrolytes.
Hence, ${PO}_{4}^{3 -}$ is most effective for coagulation of $Fe (OH)_{3}$

WB-JEE-2020

SOLUTIONS

277569 The formation of association of colloidal particles by addition of electrolyte to form as insoluble precipitate is called

1 Flocculation

2 Emulsification

3 Coagulation

4 Micelle

GUJCET-2019

SOLUTIONS

277575 The coagulation values in millimoles per litre of the electrolyes used for the coagulation of ${As}_{2} S_{3}$ are given below: I. $(NaCl) = 52$ ,
II. $({BaCl}_{2}) = 0.69$ ,
III. $({MgSO}_{4}) = 0.22$
The correct order of their coagulating power is

1 I

>

II

>

III

2 II

>

I

>

III

3 III

>

II

>

I

4 III

>

I

>

II

Explanation:

Coagulation power $\propto \frac{1}{Coagulation values}$
Higher the coagulation power, lower is the coagulation values in millimoles per liter.
${MgSO}_{4} > {BaCl}_{2} > NaCl$

(NEET-II 2016)

SOLUTIONS

277577 The pair of compound which cannot exist together solution is

1

{NaHCO}_{3}

and

NaOH

2

{NaHCO}_{3}

and

H_{2} O

3

{NaHCO}_{3}

and

{Na}_{2} {CO}_{3}

4

{Na}_{2} {CO}_{3}

and

NaOH

Explanation:

${NaHCO}_{3}$ and $NaOH$ compound cannot exist together in solution as-
${NaHCO}_{3} + NaOH \to {Na}_{2} {CO}_{3} + H_{2} O$
Neutral salt

Karnataka-CET-2015

SOLUTIONS

277565 A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of ${BaCl}_{2}$ , and $NaCl$ required to coagulate the given colloid is in the order

1

NaCl > {BaCl}_{2} > {AlCl}_{3}

2

{BaCl}_{2} > {AlCl}_{3} > NaCl

3

{AlCl}_{3} = NaCl = {BaCl}_{2}

4

{AlCl}_{3} > {BaCl}_{2} > NaCl

Explanation:

According to the Hardy Schulze rule, the effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulation, greater is its power to bring about coagulation.
Hence, the electric field subjected to the coagulation power should be in the order of ${AlCl}_{3} > {BaCl}_{2} > NaCl$

Kerala-CEE-29.08.2021

SOLUTIONS

277568 Among the following the ion which will be more effective for flocculation of $Fe (OH)_{3}$ solution is

1

{PO}_{4}^{3 -}

2

{SO}_{4}^{2 -}

3

{SO}_{3}^{2 -}

4

{NO}_{3}^{-}

Explanation:

$Fe (OH)_{3}$ solution is positively charged solution. It will be coagulated or flocculation by negatively charged ions.
${PO}_{4}^{3 -}$ have maximum negative charge among the given electrolytes.
Hence, ${PO}_{4}^{3 -}$ is most effective for coagulation of $Fe (OH)_{3}$

WB-JEE-2020

SOLUTIONS

277569 The formation of association of colloidal particles by addition of electrolyte to form as insoluble precipitate is called

1 Flocculation

2 Emulsification

3 Coagulation

4 Micelle

GUJCET-2019

SOLUTIONS

277575 The coagulation values in millimoles per litre of the electrolyes used for the coagulation of ${As}_{2} S_{3}$ are given below: I. $(NaCl) = 52$ ,
II. $({BaCl}_{2}) = 0.69$ ,
III. $({MgSO}_{4}) = 0.22$
The correct order of their coagulating power is

1 I

>

II

>

III

2 II

>

I

>

III

3 III

>

II

>

I

4 III

>

I

>

II

Explanation:

Coagulation power $\propto \frac{1}{Coagulation values}$
Higher the coagulation power, lower is the coagulation values in millimoles per liter.
${MgSO}_{4} > {BaCl}_{2} > NaCl$

(NEET-II 2016)

SOLUTIONS

277577 The pair of compound which cannot exist together solution is

1

{NaHCO}_{3}

and

NaOH

2

{NaHCO}_{3}

and

H_{2} O

3

{NaHCO}_{3}

and

{Na}_{2} {CO}_{3}

4

{Na}_{2} {CO}_{3}

and

NaOH

Explanation:

${NaHCO}_{3}$ and $NaOH$ compound cannot exist together in solution as-
${NaHCO}_{3} + NaOH \to {Na}_{2} {CO}_{3} + H_{2} O$
Neutral salt

Karnataka-CET-2015