QBManager

Chemical Equilibrium

229118 The van't Hoff factor (i) for a dilute aqueous solution of sucrose is

1 zero

2 1.0

3 1.5

4 2.0

Explanation:

The dilute aqueous solution of sucrose cannot associated or dissociated.
Hence, van't Hoff factor ' $i$ ' is one
$i = 1$ (No. of association or dissociation)
$i > 1$ (for dissociation)
$i < 1$ (for association)

J and K CET-(2011)

Chemical Equilibrium

229116 The degree of ionization of $0.4 M$ acetic acid will be $(K_{a} = 1.8 \times 10^{- 5})$

1

6.71 \times 10^{- 3}

2

1.6 \times 10^{- 3}

3

0.4 \times 1.8 \times 10^{- 5}

4

1.8 \times 10^{- 5}

Explanation:

Given, Concentration (C) $= 0.4 M$
$K_{a} = 1.8 \times 10^{- 5}$
The ionisation of acetic acid is-
$\begin{aligned} ∴ {CH}_{3} COOH & ⇌ {CH}_{3} {COO}^{-} + H^{+} \\ K_{a} & = C^{2} \\ 1.8 \times 10^{- 5} & = 0.4 \times α^{2} \\ α & = \sqrt{\frac{1.8 \times 10^{- 5}}{0.4}} \\ α & = 6.71 \times 10^{- 3} \end{aligned}$

MHT CET-2012

Chemical Equilibrium

229075 For a reaction at equilibrium
$A (g) ⇌ B (g) + \frac{1}{2} C (g)$
The relation between dissociation constant $(K)$ , degree of dissociation $(α)$ and equilibrium pressure ( $p)$ is given by:

1

K = \frac{α^{\frac{1}{2}} p^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

2

K = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{(2 + α)^{\frac{1}{2}} (1 - α)}

3

K = \frac{(α p)^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

4

K = \frac{(α p)^{\frac{3}{2}}}{(1 + α) (1 - α)^{\frac{1}{2}}}

Explanation:

For a given reaction at equilibrium-
$\begin{aligned} A (g) ⇌ B (g) + \frac{1}{2} C (g) \\ Initially 1 0 0 \\ At equi. (1 - α) α \frac{α}{2} \end{aligned}$
$\begin{aligned} ∴ n_{(total)} & = 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2} \\ So, K_{p} = & \frac{(\frac{α}{1 + \frac{α}{2}} \times p) {(\frac{α / 2}{1 + \frac{α}{2}} p)}^{1 / 2}}{(\frac{1 - α}{1 + \frac{α}{2}} p)} \\ K_{p} & = \frac{α {(\frac{α}{2})}^{1 / 2} p^{1 / 2}}{(1 - α)^{1 / 2} {(1 + \frac{α}{2})}^{1 / 2}} \\ K_{p} & = \frac{α^{3 / 2} p^{1 / 2}}{(2 + α)^{1 / 2} (1 - α)} \end{aligned}$

Shift-I

Chemical Equilibrium

229076 The overall complex dissociation equilibrium constant for ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ ion is $5 \times 10^{- 12}$ the overall stability constant of the complex is

1

2 \times 10^{- 11}

2

5 \times 10^{1}

3

5 \times 10^{10}

4

2 \times 10^{11}

5

0.2 \times 10^{11}

Explanation:

Given, that-
Complex dissociation equilibrium constant $K = 5 \times 10^{- 12}$ We know,
Dissociation constant is the reciprocal of the stability constant $(β = \frac{1}{K})$
$So, overall stability constant \begin{aligned} β & = (\frac{1}{5 \times 10^{- 12}}) \\ = 0.2 \times 10^{12} \\ = 2 \times 10^{11} \end{aligned}$

Kerala CEE -03.07.2022

Chemical Equilibrium

229077 The value of Van't Hoff factor for $0.1 M$ $Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is-

1

100 %

2

92 %

3

87 %

4

74 %

Explanation:

$Ba {({NO}_{3})}_{2} ⇌ {Ba}^{2 +} + 2 {NO}_{3}^{-}$
$\begin{aligned} i = 1 + 2 α \\ 2.74 = 1 + 2 α \\ α = \frac{1.74}{2} = 0.87 = 87 % \end{aligned}$
$∴ 2.74 = 1 + 2 α$
The Van't Hoff's factor for $0.1 M Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is $87 %$ .

J and K CET-(2003)

Chemical Equilibrium

229118 The van't Hoff factor (i) for a dilute aqueous solution of sucrose is

1 zero

2 1.0

3 1.5

4 2.0

Explanation:

The dilute aqueous solution of sucrose cannot associated or dissociated.
Hence, van't Hoff factor ' $i$ ' is one
$i = 1$ (No. of association or dissociation)
$i > 1$ (for dissociation)
$i < 1$ (for association)

J and K CET-(2011)

Chemical Equilibrium

229116 The degree of ionization of $0.4 M$ acetic acid will be $(K_{a} = 1.8 \times 10^{- 5})$

1

6.71 \times 10^{- 3}

2

1.6 \times 10^{- 3}

3

0.4 \times 1.8 \times 10^{- 5}

4

1.8 \times 10^{- 5}

Explanation:

Given, Concentration (C) $= 0.4 M$
$K_{a} = 1.8 \times 10^{- 5}$
The ionisation of acetic acid is-
$\begin{aligned} ∴ {CH}_{3} COOH & ⇌ {CH}_{3} {COO}^{-} + H^{+} \\ K_{a} & = C^{2} \\ 1.8 \times 10^{- 5} & = 0.4 \times α^{2} \\ α & = \sqrt{\frac{1.8 \times 10^{- 5}}{0.4}} \\ α & = 6.71 \times 10^{- 3} \end{aligned}$

MHT CET-2012

Chemical Equilibrium

229075 For a reaction at equilibrium
$A (g) ⇌ B (g) + \frac{1}{2} C (g)$
The relation between dissociation constant $(K)$ , degree of dissociation $(α)$ and equilibrium pressure ( $p)$ is given by:

1

K = \frac{α^{\frac{1}{2}} p^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

2

K = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{(2 + α)^{\frac{1}{2}} (1 - α)}

3

K = \frac{(α p)^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

4

K = \frac{(α p)^{\frac{3}{2}}}{(1 + α) (1 - α)^{\frac{1}{2}}}

Explanation:

For a given reaction at equilibrium-
$\begin{aligned} A (g) ⇌ B (g) + \frac{1}{2} C (g) \\ Initially 1 0 0 \\ At equi. (1 - α) α \frac{α}{2} \end{aligned}$
$\begin{aligned} ∴ n_{(total)} & = 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2} \\ So, K_{p} = & \frac{(\frac{α}{1 + \frac{α}{2}} \times p) {(\frac{α / 2}{1 + \frac{α}{2}} p)}^{1 / 2}}{(\frac{1 - α}{1 + \frac{α}{2}} p)} \\ K_{p} & = \frac{α {(\frac{α}{2})}^{1 / 2} p^{1 / 2}}{(1 - α)^{1 / 2} {(1 + \frac{α}{2})}^{1 / 2}} \\ K_{p} & = \frac{α^{3 / 2} p^{1 / 2}}{(2 + α)^{1 / 2} (1 - α)} \end{aligned}$

Shift-I

Chemical Equilibrium

229076 The overall complex dissociation equilibrium constant for ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ ion is $5 \times 10^{- 12}$ the overall stability constant of the complex is

1

2 \times 10^{- 11}

2

5 \times 10^{1}

3

5 \times 10^{10}

4

2 \times 10^{11}

5

0.2 \times 10^{11}

Explanation:

Given, that-
Complex dissociation equilibrium constant $K = 5 \times 10^{- 12}$ We know,
Dissociation constant is the reciprocal of the stability constant $(β = \frac{1}{K})$
$So, overall stability constant \begin{aligned} β & = (\frac{1}{5 \times 10^{- 12}}) \\ = 0.2 \times 10^{12} \\ = 2 \times 10^{11} \end{aligned}$

Kerala CEE -03.07.2022

Chemical Equilibrium

229077 The value of Van't Hoff factor for $0.1 M$ $Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is-

1

100 %

2

92 %

3

87 %

4

74 %

Explanation:

$Ba {({NO}_{3})}_{2} ⇌ {Ba}^{2 +} + 2 {NO}_{3}^{-}$
$\begin{aligned} i = 1 + 2 α \\ 2.74 = 1 + 2 α \\ α = \frac{1.74}{2} = 0.87 = 87 % \end{aligned}$
$∴ 2.74 = 1 + 2 α$
The Van't Hoff's factor for $0.1 M Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is $87 %$ .

J and K CET-(2003)

Chemical Equilibrium

229118 The van't Hoff factor (i) for a dilute aqueous solution of sucrose is

1 zero

2 1.0

3 1.5

4 2.0

Explanation:

The dilute aqueous solution of sucrose cannot associated or dissociated.
Hence, van't Hoff factor ' $i$ ' is one
$i = 1$ (No. of association or dissociation)
$i > 1$ (for dissociation)
$i < 1$ (for association)

J and K CET-(2011)

Chemical Equilibrium

229116 The degree of ionization of $0.4 M$ acetic acid will be $(K_{a} = 1.8 \times 10^{- 5})$

1

6.71 \times 10^{- 3}

2

1.6 \times 10^{- 3}

3

0.4 \times 1.8 \times 10^{- 5}

4

1.8 \times 10^{- 5}

Explanation:

Given, Concentration (C) $= 0.4 M$
$K_{a} = 1.8 \times 10^{- 5}$
The ionisation of acetic acid is-
$\begin{aligned} ∴ {CH}_{3} COOH & ⇌ {CH}_{3} {COO}^{-} + H^{+} \\ K_{a} & = C^{2} \\ 1.8 \times 10^{- 5} & = 0.4 \times α^{2} \\ α & = \sqrt{\frac{1.8 \times 10^{- 5}}{0.4}} \\ α & = 6.71 \times 10^{- 3} \end{aligned}$

MHT CET-2012

Chemical Equilibrium

229075 For a reaction at equilibrium
$A (g) ⇌ B (g) + \frac{1}{2} C (g)$
The relation between dissociation constant $(K)$ , degree of dissociation $(α)$ and equilibrium pressure ( $p)$ is given by:

1

K = \frac{α^{\frac{1}{2}} p^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

2

K = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{(2 + α)^{\frac{1}{2}} (1 - α)}

3

K = \frac{(α p)^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

4

K = \frac{(α p)^{\frac{3}{2}}}{(1 + α) (1 - α)^{\frac{1}{2}}}

Explanation:

For a given reaction at equilibrium-
$\begin{aligned} A (g) ⇌ B (g) + \frac{1}{2} C (g) \\ Initially 1 0 0 \\ At equi. (1 - α) α \frac{α}{2} \end{aligned}$
$\begin{aligned} ∴ n_{(total)} & = 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2} \\ So, K_{p} = & \frac{(\frac{α}{1 + \frac{α}{2}} \times p) {(\frac{α / 2}{1 + \frac{α}{2}} p)}^{1 / 2}}{(\frac{1 - α}{1 + \frac{α}{2}} p)} \\ K_{p} & = \frac{α {(\frac{α}{2})}^{1 / 2} p^{1 / 2}}{(1 - α)^{1 / 2} {(1 + \frac{α}{2})}^{1 / 2}} \\ K_{p} & = \frac{α^{3 / 2} p^{1 / 2}}{(2 + α)^{1 / 2} (1 - α)} \end{aligned}$

Shift-I

Chemical Equilibrium

229076 The overall complex dissociation equilibrium constant for ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ ion is $5 \times 10^{- 12}$ the overall stability constant of the complex is

1

2 \times 10^{- 11}

2

5 \times 10^{1}

3

5 \times 10^{10}

4

2 \times 10^{11}

5

0.2 \times 10^{11}

Explanation:

Given, that-
Complex dissociation equilibrium constant $K = 5 \times 10^{- 12}$ We know,
Dissociation constant is the reciprocal of the stability constant $(β = \frac{1}{K})$
$So, overall stability constant \begin{aligned} β & = (\frac{1}{5 \times 10^{- 12}}) \\ = 0.2 \times 10^{12} \\ = 2 \times 10^{11} \end{aligned}$

Kerala CEE -03.07.2022

Chemical Equilibrium

229077 The value of Van't Hoff factor for $0.1 M$ $Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is-

1

100 %

2

92 %

3

87 %

4

74 %

Explanation:

$Ba {({NO}_{3})}_{2} ⇌ {Ba}^{2 +} + 2 {NO}_{3}^{-}$
$\begin{aligned} i = 1 + 2 α \\ 2.74 = 1 + 2 α \\ α = \frac{1.74}{2} = 0.87 = 87 % \end{aligned}$
$∴ 2.74 = 1 + 2 α$
The Van't Hoff's factor for $0.1 M Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is $87 %$ .

J and K CET-(2003)

Chemical Equilibrium

229118 The van't Hoff factor (i) for a dilute aqueous solution of sucrose is

1 zero

2 1.0

3 1.5

4 2.0

Explanation:

The dilute aqueous solution of sucrose cannot associated or dissociated.
Hence, van't Hoff factor ' $i$ ' is one
$i = 1$ (No. of association or dissociation)
$i > 1$ (for dissociation)
$i < 1$ (for association)

J and K CET-(2011)

Chemical Equilibrium

229116 The degree of ionization of $0.4 M$ acetic acid will be $(K_{a} = 1.8 \times 10^{- 5})$

1

6.71 \times 10^{- 3}

2

1.6 \times 10^{- 3}

3

0.4 \times 1.8 \times 10^{- 5}

4

1.8 \times 10^{- 5}

Explanation:

Given, Concentration (C) $= 0.4 M$
$K_{a} = 1.8 \times 10^{- 5}$
The ionisation of acetic acid is-
$\begin{aligned} ∴ {CH}_{3} COOH & ⇌ {CH}_{3} {COO}^{-} + H^{+} \\ K_{a} & = C^{2} \\ 1.8 \times 10^{- 5} & = 0.4 \times α^{2} \\ α & = \sqrt{\frac{1.8 \times 10^{- 5}}{0.4}} \\ α & = 6.71 \times 10^{- 3} \end{aligned}$

MHT CET-2012

Chemical Equilibrium

229075 For a reaction at equilibrium
$A (g) ⇌ B (g) + \frac{1}{2} C (g)$
The relation between dissociation constant $(K)$ , degree of dissociation $(α)$ and equilibrium pressure ( $p)$ is given by:

1

K = \frac{α^{\frac{1}{2}} p^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

2

K = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{(2 + α)^{\frac{1}{2}} (1 - α)}

3

K = \frac{(α p)^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

4

K = \frac{(α p)^{\frac{3}{2}}}{(1 + α) (1 - α)^{\frac{1}{2}}}

Explanation:

For a given reaction at equilibrium-
$\begin{aligned} A (g) ⇌ B (g) + \frac{1}{2} C (g) \\ Initially 1 0 0 \\ At equi. (1 - α) α \frac{α}{2} \end{aligned}$
$\begin{aligned} ∴ n_{(total)} & = 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2} \\ So, K_{p} = & \frac{(\frac{α}{1 + \frac{α}{2}} \times p) {(\frac{α / 2}{1 + \frac{α}{2}} p)}^{1 / 2}}{(\frac{1 - α}{1 + \frac{α}{2}} p)} \\ K_{p} & = \frac{α {(\frac{α}{2})}^{1 / 2} p^{1 / 2}}{(1 - α)^{1 / 2} {(1 + \frac{α}{2})}^{1 / 2}} \\ K_{p} & = \frac{α^{3 / 2} p^{1 / 2}}{(2 + α)^{1 / 2} (1 - α)} \end{aligned}$

Shift-I

Chemical Equilibrium

229076 The overall complex dissociation equilibrium constant for ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ ion is $5 \times 10^{- 12}$ the overall stability constant of the complex is

1

2 \times 10^{- 11}

2

5 \times 10^{1}

3

5 \times 10^{10}

4

2 \times 10^{11}

5

0.2 \times 10^{11}

Explanation:

Given, that-
Complex dissociation equilibrium constant $K = 5 \times 10^{- 12}$ We know,
Dissociation constant is the reciprocal of the stability constant $(β = \frac{1}{K})$
$So, overall stability constant \begin{aligned} β & = (\frac{1}{5 \times 10^{- 12}}) \\ = 0.2 \times 10^{12} \\ = 2 \times 10^{11} \end{aligned}$

Kerala CEE -03.07.2022

Chemical Equilibrium

229077 The value of Van't Hoff factor for $0.1 M$ $Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is-

1

100 %

2

92 %

3

87 %

4

74 %

Explanation:

$Ba {({NO}_{3})}_{2} ⇌ {Ba}^{2 +} + 2 {NO}_{3}^{-}$
$\begin{aligned} i = 1 + 2 α \\ 2.74 = 1 + 2 α \\ α = \frac{1.74}{2} = 0.87 = 87 % \end{aligned}$
$∴ 2.74 = 1 + 2 α$
The Van't Hoff's factor for $0.1 M Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is $87 %$ .

J and K CET-(2003)

Chemical Equilibrium

229118 The van't Hoff factor (i) for a dilute aqueous solution of sucrose is

1 zero

2 1.0

3 1.5

4 2.0

Explanation:

The dilute aqueous solution of sucrose cannot associated or dissociated.
Hence, van't Hoff factor ' $i$ ' is one
$i = 1$ (No. of association or dissociation)
$i > 1$ (for dissociation)
$i < 1$ (for association)

J and K CET-(2011)

Chemical Equilibrium

229116 The degree of ionization of $0.4 M$ acetic acid will be $(K_{a} = 1.8 \times 10^{- 5})$

1

6.71 \times 10^{- 3}

2

1.6 \times 10^{- 3}

3

0.4 \times 1.8 \times 10^{- 5}

4

1.8 \times 10^{- 5}

Explanation:

Given, Concentration (C) $= 0.4 M$
$K_{a} = 1.8 \times 10^{- 5}$
The ionisation of acetic acid is-
$\begin{aligned} ∴ {CH}_{3} COOH & ⇌ {CH}_{3} {COO}^{-} + H^{+} \\ K_{a} & = C^{2} \\ 1.8 \times 10^{- 5} & = 0.4 \times α^{2} \\ α & = \sqrt{\frac{1.8 \times 10^{- 5}}{0.4}} \\ α & = 6.71 \times 10^{- 3} \end{aligned}$

MHT CET-2012

Chemical Equilibrium

229075 For a reaction at equilibrium
$A (g) ⇌ B (g) + \frac{1}{2} C (g)$
The relation between dissociation constant $(K)$ , degree of dissociation $(α)$ and equilibrium pressure ( $p)$ is given by:

1

K = \frac{α^{\frac{1}{2}} p^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

2

K = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{(2 + α)^{\frac{1}{2}} (1 - α)}

3

K = \frac{(α p)^{\frac{3}{2}}}{{(1 + \frac{3}{2} α)}^{\frac{1}{2}} (1 - α)}

4

K = \frac{(α p)^{\frac{3}{2}}}{(1 + α) (1 - α)^{\frac{1}{2}}}

Explanation:

For a given reaction at equilibrium-
$\begin{aligned} A (g) ⇌ B (g) + \frac{1}{2} C (g) \\ Initially 1 0 0 \\ At equi. (1 - α) α \frac{α}{2} \end{aligned}$
$\begin{aligned} ∴ n_{(total)} & = 1 - α + α + \frac{α}{2} = 1 + \frac{α}{2} \\ So, K_{p} = & \frac{(\frac{α}{1 + \frac{α}{2}} \times p) {(\frac{α / 2}{1 + \frac{α}{2}} p)}^{1 / 2}}{(\frac{1 - α}{1 + \frac{α}{2}} p)} \\ K_{p} & = \frac{α {(\frac{α}{2})}^{1 / 2} p^{1 / 2}}{(1 - α)^{1 / 2} {(1 + \frac{α}{2})}^{1 / 2}} \\ K_{p} & = \frac{α^{3 / 2} p^{1 / 2}}{(2 + α)^{1 / 2} (1 - α)} \end{aligned}$

Shift-I

Chemical Equilibrium

229076 The overall complex dissociation equilibrium constant for ${[Cr {(H_{2} O)}_{6}]}^{3 +}$ ion is $5 \times 10^{- 12}$ the overall stability constant of the complex is

1

2 \times 10^{- 11}

2

5 \times 10^{1}

3

5 \times 10^{10}

4

2 \times 10^{11}

5

0.2 \times 10^{11}

Explanation:

Given, that-
Complex dissociation equilibrium constant $K = 5 \times 10^{- 12}$ We know,
Dissociation constant is the reciprocal of the stability constant $(β = \frac{1}{K})$
$So, overall stability constant \begin{aligned} β & = (\frac{1}{5 \times 10^{- 12}}) \\ = 0.2 \times 10^{12} \\ = 2 \times 10^{11} \end{aligned}$

Kerala CEE -03.07.2022

Chemical Equilibrium

229077 The value of Van't Hoff factor for $0.1 M$ $Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is-

1

100 %

2

92 %

3

87 %

4

74 %

Explanation:

$Ba {({NO}_{3})}_{2} ⇌ {Ba}^{2 +} + 2 {NO}_{3}^{-}$
$\begin{aligned} i = 1 + 2 α \\ 2.74 = 1 + 2 α \\ α = \frac{1.74}{2} = 0.87 = 87 % \end{aligned}$
$∴ 2.74 = 1 + 2 α$
The Van't Hoff's factor for $0.1 M Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is $87 %$ .

J and K CET-(2003)