QBManager

Chemical Equilibrium

229078 Van't Hoff factor of $Ca {({NO}_{3})}_{2}$ is

1 One

2 Two

3 Three

4 Four

Explanation:

$Ca {({NO}_{3})}_{2} ⇌ {Ca}^{2 +} + 2 {NO}_{3}^{-}$
It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.

[J and K CET-(2013)]

Chemical Equilibrium

229095 $K_{2} {HgI}_{4}$ is $40 %$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is

1 1.6

2 1.8

3 2.2

4 2.0

Explanation:

The ionisation of $K_{2} {HgI}_{4}$ is given as $K_{2} {HgI}_{4} ⟶ 2 K^{+} + {HgI}_{4}^{2 -}$
$i = 1 + (n - 1) α$
$α = 0.4, n = 3$
So, $i = 1 + 2 \times 0.4 = 1.8$

[JEE Mains 2019

Chemical Equilibrium

229121 If ${BaCl}_{2}$ ionizes to an extent of $80 %$ in aqueous solution, the value of van't Hoff factor is

1 2.6

2 0.4

3 0.8

4 2.4

Explanation:

Ionisation of ${BaCl}_{2}$ is-

AP-EAMCET- (Engg.) - 2010

Chemical Equilibrium

229123 Van't Hoff factor of aq $K_{2} {SO}_{4}$ at infinite dilution has value equal to

1 1

2 2

3 3

4 between 2 and 3

Explanation:

At infinite dilution, complete dissociation will take place. So,
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}_{4}^{2 -}$
van't Hoff factor $= 2 + 1 = 3$

AMU-2009

Chemical Equilibrium

229128 In a $500 mL$ flask, the degree of dissociation of ${PCl}_{5}$ at equilibrium is $40 %$ and the initial amount is 5 moles. The value of equilibrium constant in mol $L^{- 1}$ for the decomposition of ${PCl}_{5}$ is

1 2.33

2 2.66

3 5.32

4 4.66

AP-EAMCET-2008

Chemical Equilibrium

229078 Van't Hoff factor of $Ca {({NO}_{3})}_{2}$ is

1 One

2 Two

3 Three

4 Four

Explanation:

$Ca {({NO}_{3})}_{2} ⇌ {Ca}^{2 +} + 2 {NO}_{3}^{-}$
It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.

[J and K CET-(2013)]

Chemical Equilibrium

229095 $K_{2} {HgI}_{4}$ is $40 %$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is

1 1.6

2 1.8

3 2.2

4 2.0

Explanation:

The ionisation of $K_{2} {HgI}_{4}$ is given as $K_{2} {HgI}_{4} ⟶ 2 K^{+} + {HgI}_{4}^{2 -}$
$i = 1 + (n - 1) α$
$α = 0.4, n = 3$
So, $i = 1 + 2 \times 0.4 = 1.8$

[JEE Mains 2019

Chemical Equilibrium

229121 If ${BaCl}_{2}$ ionizes to an extent of $80 %$ in aqueous solution, the value of van't Hoff factor is

1 2.6

2 0.4

3 0.8

4 2.4

Explanation:

Ionisation of ${BaCl}_{2}$ is-

AP-EAMCET- (Engg.) - 2010

Chemical Equilibrium

229123 Van't Hoff factor of aq $K_{2} {SO}_{4}$ at infinite dilution has value equal to

1 1

2 2

3 3

4 between 2 and 3

Explanation:

At infinite dilution, complete dissociation will take place. So,
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}_{4}^{2 -}$
van't Hoff factor $= 2 + 1 = 3$

AMU-2009

Chemical Equilibrium

229128 In a $500 mL$ flask, the degree of dissociation of ${PCl}_{5}$ at equilibrium is $40 %$ and the initial amount is 5 moles. The value of equilibrium constant in mol $L^{- 1}$ for the decomposition of ${PCl}_{5}$ is

1 2.33

2 2.66

3 5.32

4 4.66

AP-EAMCET-2008

Chemical Equilibrium

229078 Van't Hoff factor of $Ca {({NO}_{3})}_{2}$ is

1 One

2 Two

3 Three

4 Four

Explanation:

$Ca {({NO}_{3})}_{2} ⇌ {Ca}^{2 +} + 2 {NO}_{3}^{-}$
It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.

[J and K CET-(2013)]

Chemical Equilibrium

229095 $K_{2} {HgI}_{4}$ is $40 %$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is

1 1.6

2 1.8

3 2.2

4 2.0

Explanation:

The ionisation of $K_{2} {HgI}_{4}$ is given as $K_{2} {HgI}_{4} ⟶ 2 K^{+} + {HgI}_{4}^{2 -}$
$i = 1 + (n - 1) α$
$α = 0.4, n = 3$
So, $i = 1 + 2 \times 0.4 = 1.8$

[JEE Mains 2019

Chemical Equilibrium

229121 If ${BaCl}_{2}$ ionizes to an extent of $80 %$ in aqueous solution, the value of van't Hoff factor is

1 2.6

2 0.4

3 0.8

4 2.4

Explanation:

Ionisation of ${BaCl}_{2}$ is-

AP-EAMCET- (Engg.) - 2010

Chemical Equilibrium

229123 Van't Hoff factor of aq $K_{2} {SO}_{4}$ at infinite dilution has value equal to

1 1

2 2

3 3

4 between 2 and 3

Explanation:

At infinite dilution, complete dissociation will take place. So,
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}_{4}^{2 -}$
van't Hoff factor $= 2 + 1 = 3$

AMU-2009

Chemical Equilibrium

229128 In a $500 mL$ flask, the degree of dissociation of ${PCl}_{5}$ at equilibrium is $40 %$ and the initial amount is 5 moles. The value of equilibrium constant in mol $L^{- 1}$ for the decomposition of ${PCl}_{5}$ is

1 2.33

2 2.66

3 5.32

4 4.66

AP-EAMCET-2008

Chemical Equilibrium

229078 Van't Hoff factor of $Ca {({NO}_{3})}_{2}$ is

1 One

2 Two

3 Three

4 Four

Explanation:

$Ca {({NO}_{3})}_{2} ⇌ {Ca}^{2 +} + 2 {NO}_{3}^{-}$
It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.

[J and K CET-(2013)]

Chemical Equilibrium

229095 $K_{2} {HgI}_{4}$ is $40 %$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is

1 1.6

2 1.8

3 2.2

4 2.0

Explanation:

The ionisation of $K_{2} {HgI}_{4}$ is given as $K_{2} {HgI}_{4} ⟶ 2 K^{+} + {HgI}_{4}^{2 -}$
$i = 1 + (n - 1) α$
$α = 0.4, n = 3$
So, $i = 1 + 2 \times 0.4 = 1.8$

[JEE Mains 2019

Chemical Equilibrium

229121 If ${BaCl}_{2}$ ionizes to an extent of $80 %$ in aqueous solution, the value of van't Hoff factor is

1 2.6

2 0.4

3 0.8

4 2.4

Explanation:

Ionisation of ${BaCl}_{2}$ is-

AP-EAMCET- (Engg.) - 2010

Chemical Equilibrium

229123 Van't Hoff factor of aq $K_{2} {SO}_{4}$ at infinite dilution has value equal to

1 1

2 2

3 3

4 between 2 and 3

Explanation:

At infinite dilution, complete dissociation will take place. So,
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}_{4}^{2 -}$
van't Hoff factor $= 2 + 1 = 3$

AMU-2009

Chemical Equilibrium

229128 In a $500 mL$ flask, the degree of dissociation of ${PCl}_{5}$ at equilibrium is $40 %$ and the initial amount is 5 moles. The value of equilibrium constant in mol $L^{- 1}$ for the decomposition of ${PCl}_{5}$ is

1 2.33

2 2.66

3 5.32

4 4.66

AP-EAMCET-2008

Chemical Equilibrium

229078 Van't Hoff factor of $Ca {({NO}_{3})}_{2}$ is

1 One

2 Two

3 Three

4 Four

Explanation:

$Ca {({NO}_{3})}_{2} ⇌ {Ca}^{2 +} + 2 {NO}_{3}^{-}$
It furnishes 3 ions per formula unit. So, its Van't Hoff factor is 3.

[J and K CET-(2013)]

Chemical Equilibrium

229095 $K_{2} {HgI}_{4}$ is $40 %$ ionised in aqueous solution. The value of its Van't Hoff factor (i) is

1 1.6

2 1.8

3 2.2

4 2.0

Explanation:

The ionisation of $K_{2} {HgI}_{4}$ is given as $K_{2} {HgI}_{4} ⟶ 2 K^{+} + {HgI}_{4}^{2 -}$
$i = 1 + (n - 1) α$
$α = 0.4, n = 3$
So, $i = 1 + 2 \times 0.4 = 1.8$

[JEE Mains 2019

Chemical Equilibrium

229121 If ${BaCl}_{2}$ ionizes to an extent of $80 %$ in aqueous solution, the value of van't Hoff factor is

1 2.6

2 0.4

3 0.8

4 2.4

Explanation:

Ionisation of ${BaCl}_{2}$ is-

AP-EAMCET- (Engg.) - 2010

Chemical Equilibrium

229123 Van't Hoff factor of aq $K_{2} {SO}_{4}$ at infinite dilution has value equal to

1 1

2 2

3 3

4 between 2 and 3

Explanation:

At infinite dilution, complete dissociation will take place. So,
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}_{4}^{2 -}$
van't Hoff factor $= 2 + 1 = 3$

AMU-2009

Chemical Equilibrium

229128 In a $500 mL$ flask, the degree of dissociation of ${PCl}_{5}$ at equilibrium is $40 %$ and the initial amount is 5 moles. The value of equilibrium constant in mol $L^{- 1}$ for the decomposition of ${PCl}_{5}$ is

1 2.33

2 2.66

3 5.32

4 4.66

AP-EAMCET-2008