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CHXII02:SOLUTIONS

318999 If the various terms in the given below expressions have usual meanings, the van’t Hoff factor (i) cannot be calculated by which one of the following expressions?

1

π V = \sqrt{i} n R T

2

Δ T_{f} = i K_{f} . m

3

Δ T_{f} = i K_{f} . m

4

\frac{P_{s o l v e n t}^{o} - P_{s o l u t i o n}}{P_{s o l v e n t}^{o}} = i (\frac{n}{N + n})

Explanation:

Van’t Hoff equation is
$π V = i n R T$
For depression in freezing point,
$Δ T_{f} = i \times K_{f} \times m$
For elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
For lowering of vapour pressure,
$\frac{P_{s o l v e n t}^{0} - P_{s o l u t i o n}}{P_{s o l v e n t}^{0}} = i (\frac{n}{N + n}) .$

CHXII02:SOLUTIONS

319000 Which of the following aqueous solutions has highest freezing point?

1

0 .1 m o l a l A l_{2} {(S O_{4})}_{3}

2

0 .1 m o l a l B a C l_{2}

3

0 .1 m o l a l A l C l_{3}

4

0 .1 m o l a l N H_{4} C l

Explanation:

$Δ T_{f} = i K_{f} m$
As m = 0.1 molal for all given solutions thus, lower the value of i, lower will be the depression in freezing point $(Δ T_{f})$ and higher will be the freezing point of the solution.
$F o r A l_{2} {(S O_{4})}_{3}; i = 5$
$F o r B a C l_{2}; i = 3$
$F o r A l C l_{3}; i = 4$
$F o r N H_{4} C l; i = 2$
Thus, 0.1 molal $N H_{4} C l$ will have highest freezing point.

KCET - 2017

CHXII02:SOLUTIONS

319001 The van’t Hoff factor 'I' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute

CHXII02:SOLUTIONS

319002 Which of the following aqueous solutions should have the highest boiling point ?

1

1 .0 M N a O H

2

1 .0 M N a_{2} S O_{4}

3

1 .0 M N H_{2} N O_{3}

4

1 .0 M K N O_{3}

Explanation:

$Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b} \propto i$
$i = 2 f o r N a O H; i = 3 f o r N a S O_{4}$
$i = 2 f o r N H_{4} N O_{3}; i = 2 f o r K N O_{3}$
So, elevation in boiling point will be maximum in $N a_{2} S O_{4}$ solution or $N a_{2} S O_{4}$ solution will have highest boiling point.

KCET - 2018

CHXII02:SOLUTIONS

318999 If the various terms in the given below expressions have usual meanings, the van’t Hoff factor (i) cannot be calculated by which one of the following expressions?

1

π V = \sqrt{i} n R T

2

Δ T_{f} = i K_{f} . m

3

Δ T_{f} = i K_{f} . m

4

\frac{P_{s o l v e n t}^{o} - P_{s o l u t i o n}}{P_{s o l v e n t}^{o}} = i (\frac{n}{N + n})

Explanation:

Van’t Hoff equation is
$π V = i n R T$
For depression in freezing point,
$Δ T_{f} = i \times K_{f} \times m$
For elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
For lowering of vapour pressure,
$\frac{P_{s o l v e n t}^{0} - P_{s o l u t i o n}}{P_{s o l v e n t}^{0}} = i (\frac{n}{N + n}) .$

CHXII02:SOLUTIONS

319000 Which of the following aqueous solutions has highest freezing point?

1

0 .1 m o l a l A l_{2} {(S O_{4})}_{3}

2

0 .1 m o l a l B a C l_{2}

3

0 .1 m o l a l A l C l_{3}

4

0 .1 m o l a l N H_{4} C l

Explanation:

$Δ T_{f} = i K_{f} m$
As m = 0.1 molal for all given solutions thus, lower the value of i, lower will be the depression in freezing point $(Δ T_{f})$ and higher will be the freezing point of the solution.
$F o r A l_{2} {(S O_{4})}_{3}; i = 5$
$F o r B a C l_{2}; i = 3$
$F o r A l C l_{3}; i = 4$
$F o r N H_{4} C l; i = 2$
Thus, 0.1 molal $N H_{4} C l$ will have highest freezing point.

KCET - 2017

CHXII02:SOLUTIONS

319001 The van’t Hoff factor 'I' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute

CHXII02:SOLUTIONS

319002 Which of the following aqueous solutions should have the highest boiling point ?

1

1 .0 M N a O H

2

1 .0 M N a_{2} S O_{4}

3

1 .0 M N H_{2} N O_{3}

4

1 .0 M K N O_{3}

Explanation:

$Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b} \propto i$
$i = 2 f o r N a O H; i = 3 f o r N a S O_{4}$
$i = 2 f o r N H_{4} N O_{3}; i = 2 f o r K N O_{3}$
So, elevation in boiling point will be maximum in $N a_{2} S O_{4}$ solution or $N a_{2} S O_{4}$ solution will have highest boiling point.

KCET - 2018

CHXII02:SOLUTIONS

318999 If the various terms in the given below expressions have usual meanings, the van’t Hoff factor (i) cannot be calculated by which one of the following expressions?

1

π V = \sqrt{i} n R T

2

Δ T_{f} = i K_{f} . m

3

Δ T_{f} = i K_{f} . m

4

\frac{P_{s o l v e n t}^{o} - P_{s o l u t i o n}}{P_{s o l v e n t}^{o}} = i (\frac{n}{N + n})

Explanation:

Van’t Hoff equation is
$π V = i n R T$
For depression in freezing point,
$Δ T_{f} = i \times K_{f} \times m$
For elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
For lowering of vapour pressure,
$\frac{P_{s o l v e n t}^{0} - P_{s o l u t i o n}}{P_{s o l v e n t}^{0}} = i (\frac{n}{N + n}) .$

CHXII02:SOLUTIONS

319000 Which of the following aqueous solutions has highest freezing point?

1

0 .1 m o l a l A l_{2} {(S O_{4})}_{3}

2

0 .1 m o l a l B a C l_{2}

3

0 .1 m o l a l A l C l_{3}

4

0 .1 m o l a l N H_{4} C l

Explanation:

$Δ T_{f} = i K_{f} m$
As m = 0.1 molal for all given solutions thus, lower the value of i, lower will be the depression in freezing point $(Δ T_{f})$ and higher will be the freezing point of the solution.
$F o r A l_{2} {(S O_{4})}_{3}; i = 5$
$F o r B a C l_{2}; i = 3$
$F o r A l C l_{3}; i = 4$
$F o r N H_{4} C l; i = 2$
Thus, 0.1 molal $N H_{4} C l$ will have highest freezing point.

KCET - 2017

CHXII02:SOLUTIONS

319001 The van’t Hoff factor 'I' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute

CHXII02:SOLUTIONS

319002 Which of the following aqueous solutions should have the highest boiling point ?

1

1 .0 M N a O H

2

1 .0 M N a_{2} S O_{4}

3

1 .0 M N H_{2} N O_{3}

4

1 .0 M K N O_{3}

Explanation:

$Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b} \propto i$
$i = 2 f o r N a O H; i = 3 f o r N a S O_{4}$
$i = 2 f o r N H_{4} N O_{3}; i = 2 f o r K N O_{3}$
So, elevation in boiling point will be maximum in $N a_{2} S O_{4}$ solution or $N a_{2} S O_{4}$ solution will have highest boiling point.

KCET - 2018

CHXII02:SOLUTIONS

318999 If the various terms in the given below expressions have usual meanings, the van’t Hoff factor (i) cannot be calculated by which one of the following expressions?

1

π V = \sqrt{i} n R T

2

Δ T_{f} = i K_{f} . m

3

Δ T_{f} = i K_{f} . m

4

\frac{P_{s o l v e n t}^{o} - P_{s o l u t i o n}}{P_{s o l v e n t}^{o}} = i (\frac{n}{N + n})

Explanation:

Van’t Hoff equation is
$π V = i n R T$
For depression in freezing point,
$Δ T_{f} = i \times K_{f} \times m$
For elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
For lowering of vapour pressure,
$\frac{P_{s o l v e n t}^{0} - P_{s o l u t i o n}}{P_{s o l v e n t}^{0}} = i (\frac{n}{N + n}) .$

CHXII02:SOLUTIONS

319000 Which of the following aqueous solutions has highest freezing point?

1

0 .1 m o l a l A l_{2} {(S O_{4})}_{3}

2

0 .1 m o l a l B a C l_{2}

3

0 .1 m o l a l A l C l_{3}

4

0 .1 m o l a l N H_{4} C l

Explanation:

$Δ T_{f} = i K_{f} m$
As m = 0.1 molal for all given solutions thus, lower the value of i, lower will be the depression in freezing point $(Δ T_{f})$ and higher will be the freezing point of the solution.
$F o r A l_{2} {(S O_{4})}_{3}; i = 5$
$F o r B a C l_{2}; i = 3$
$F o r A l C l_{3}; i = 4$
$F o r N H_{4} C l; i = 2$
Thus, 0.1 molal $N H_{4} C l$ will have highest freezing point.

KCET - 2017

CHXII02:SOLUTIONS

319001 The van’t Hoff factor 'I' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute

CHXII02:SOLUTIONS

319002 Which of the following aqueous solutions should have the highest boiling point ?

1

1 .0 M N a O H

2

1 .0 M N a_{2} S O_{4}

3

1 .0 M N H_{2} N O_{3}

4

1 .0 M K N O_{3}

Explanation:

$Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b} \propto i$
$i = 2 f o r N a O H; i = 3 f o r N a S O_{4}$
$i = 2 f o r N H_{4} N O_{3}; i = 2 f o r K N O_{3}$
So, elevation in boiling point will be maximum in $N a_{2} S O_{4}$ solution or $N a_{2} S O_{4}$ solution will have highest boiling point.

KCET - 2018

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