QBManager

SOLUTIONS

277331 $58.5 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 mL$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
Molality of $NaCl$ solution $= \frac{n}{W} \times 1000$
$= \frac{58.5 / 58.5}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$
Molality of $C_{6} H_{12} O_{6}$ solution $= \frac{180 / 100}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$ Both the solutions have same molality but the values of (i) for $NaCl$ and glucose are 2 and 1 respectively.
$Δ T_{NaCl} = 2 \times Δ T_{C_{6} H_{12} O_{6}}$
Hence, $NaCl$ will show higher elevation of boiling.

JCECE - 2017

SOLUTIONS

277332 Which of the following aqueous solution has highest freezing point?

1 0.1 molal

{Al}_{2} {({SO}_{4})}_{3}

2 0.1 molal

{BaCl}_{2}

3 0.1 molal

{AlCl}_{3}

4 0.1 molal

{NH}_{4} Cl

Explanation:

Depression in freezing point.
$Δ T_{f} = i \times K_{f} \times m$
Where, $i = \frac{observed value}{theoretical value}$
$\begin{aligned} For {Al}_{2} {({SO}_{4})}_{3} = 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -} (i = 5) \\ {BaCl}_{2} = {Ba}^{+ 2} + 2 {Cl}^{-} (i = 3) \\ {AlCl}_{3} = {Al}^{3 +} + 3 {Cl}^{-} (i = 4) \\ {NH}_{4} Cl = {NH}_{4}^{+} + {Cl}^{-} (i = 2) \end{aligned}$
Hence, ${NH}_{4} Cl$ , has $i = 2$ which show highest freezing point.

Karnataka-CET-2017

SOLUTIONS

277333 What will be the boiling point of $30 g$ benzene containing $0.75 g$ of benzoic acid which undergoes $75 %$ dimerization? Boiling point of pure benzene $= {80.1}^{\circ} C$ and $K_{b} = 2.53 K {mol}^{- 1} kg$

1

{90.52}^{\circ} C

2

{104.35}^{\circ} C

3

{76.12}^{\circ} C

4

{80.42}^{\circ} C

Explanation:

Given data,
Weigth of benzene $= 30 g$
Weight of benzoic acid $= 0.75 g$
From formula
$\begin{aligned} Δ T_{b} & = {iK}_{b} m \\ i & = 1 - α + \frac{α}{n} \\ i & = 1 - 0.75 + \frac{0.75}{2} (\begin{array}{c} ∴ α = 0.75 \\ n = 2 \end{array}) \\ i & = 0.625 \end{aligned}$
Now putting the values in equation. (i)
$\begin{aligned} Δ T_{b} = 0.625 \times 2.53 \times \frac{0.75 / 122}{30} \times 1000 \\ Δ T_{b} = 0.324 \end{aligned}$
$Δ T_{b} = T_{b} - T_{b}^{o}$
$0.324 = T_{b} - {80.1}^{\circ} C$
$T_{b} = {80.42}^{\circ} C$

J and K CET-(2017)

SOLUTIONS

277334 If $P^{0}$ and $P$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $P$ and $P^{0}$ is

1

P^{o} = P [\frac{n_{1}}{n_{1} + n_{2}}]

2

P^{o} = P [\frac{n_{2}}{n_{1} + n_{2}}]

3

P = P^{o} [\frac{n_{2}}{n_{1} + n_{2}}]

4

P = P^{o} [\frac{n_{1}}{n_{1} + n_{2}}]

Explanation:

According to the Raoult's law -
$\frac{P^{o} - P}{P^{o}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $1 - \frac{P}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = 1 - \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
$P = P^{\circ} [\frac{n_{2}}{n_{1} + n_{2}}]$

WB-JEE-2016

SOLUTIONS

277335 $31 g$ of ethylene glycol $(C_{2} H_{6} O_{2})$ is mixed with $500 g$ of solvent $(K_{f}$ of the solvent is $2 K kg {mol}^{-}$ 1). What is the freezing point of the solution (in K)? (Freezing point of solvent $= 273 K$ )

1 272

2 271

3 270

4 274

Explanation:

From formula,
$Δ T_{f} = K_{f} \frac{W_{B} \times 1000}{M_{B} \times W_{A}} {\begin{cases} W_{B} = mass of solute = 31 g \\ W_{A} = mass of solvent = 500 g \\ M_{B} = Ethylene glycol (C_{2} H_{6} O_{2}) \end{cases}}$
$Δ T_{f} = \frac{2 \times 31 \times 1000}{62 \times 500} = 2 K$
Now, $Δ T_{f} = T_{f}^{0} - T_{f}$
$∴ T_{f} = 273 - 2$
$T_{f} = 271 K$

Kerala-CEE-2016

SOLUTIONS

277331 $58.5 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 mL$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
Molality of $NaCl$ solution $= \frac{n}{W} \times 1000$
$= \frac{58.5 / 58.5}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$
Molality of $C_{6} H_{12} O_{6}$ solution $= \frac{180 / 100}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$ Both the solutions have same molality but the values of (i) for $NaCl$ and glucose are 2 and 1 respectively.
$Δ T_{NaCl} = 2 \times Δ T_{C_{6} H_{12} O_{6}}$
Hence, $NaCl$ will show higher elevation of boiling.

JCECE - 2017

SOLUTIONS

277332 Which of the following aqueous solution has highest freezing point?

1 0.1 molal

{Al}_{2} {({SO}_{4})}_{3}

2 0.1 molal

{BaCl}_{2}

3 0.1 molal

{AlCl}_{3}

4 0.1 molal

{NH}_{4} Cl

Explanation:

Depression in freezing point.
$Δ T_{f} = i \times K_{f} \times m$
Where, $i = \frac{observed value}{theoretical value}$
$\begin{aligned} For {Al}_{2} {({SO}_{4})}_{3} = 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -} (i = 5) \\ {BaCl}_{2} = {Ba}^{+ 2} + 2 {Cl}^{-} (i = 3) \\ {AlCl}_{3} = {Al}^{3 +} + 3 {Cl}^{-} (i = 4) \\ {NH}_{4} Cl = {NH}_{4}^{+} + {Cl}^{-} (i = 2) \end{aligned}$
Hence, ${NH}_{4} Cl$ , has $i = 2$ which show highest freezing point.

Karnataka-CET-2017

SOLUTIONS

277333 What will be the boiling point of $30 g$ benzene containing $0.75 g$ of benzoic acid which undergoes $75 %$ dimerization? Boiling point of pure benzene $= {80.1}^{\circ} C$ and $K_{b} = 2.53 K {mol}^{- 1} kg$

1

{90.52}^{\circ} C

2

{104.35}^{\circ} C

3

{76.12}^{\circ} C

4

{80.42}^{\circ} C

Explanation:

Given data,
Weigth of benzene $= 30 g$
Weight of benzoic acid $= 0.75 g$
From formula
$\begin{aligned} Δ T_{b} & = {iK}_{b} m \\ i & = 1 - α + \frac{α}{n} \\ i & = 1 - 0.75 + \frac{0.75}{2} (\begin{array}{c} ∴ α = 0.75 \\ n = 2 \end{array}) \\ i & = 0.625 \end{aligned}$
Now putting the values in equation. (i)
$\begin{aligned} Δ T_{b} = 0.625 \times 2.53 \times \frac{0.75 / 122}{30} \times 1000 \\ Δ T_{b} = 0.324 \end{aligned}$
$Δ T_{b} = T_{b} - T_{b}^{o}$
$0.324 = T_{b} - {80.1}^{\circ} C$
$T_{b} = {80.42}^{\circ} C$

J and K CET-(2017)

SOLUTIONS

277334 If $P^{0}$ and $P$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $P$ and $P^{0}$ is

1

P^{o} = P [\frac{n_{1}}{n_{1} + n_{2}}]

2

P^{o} = P [\frac{n_{2}}{n_{1} + n_{2}}]

3

P = P^{o} [\frac{n_{2}}{n_{1} + n_{2}}]

4

P = P^{o} [\frac{n_{1}}{n_{1} + n_{2}}]

Explanation:

According to the Raoult's law -
$\frac{P^{o} - P}{P^{o}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $1 - \frac{P}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = 1 - \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
$P = P^{\circ} [\frac{n_{2}}{n_{1} + n_{2}}]$

WB-JEE-2016

SOLUTIONS

277335 $31 g$ of ethylene glycol $(C_{2} H_{6} O_{2})$ is mixed with $500 g$ of solvent $(K_{f}$ of the solvent is $2 K kg {mol}^{-}$ 1). What is the freezing point of the solution (in K)? (Freezing point of solvent $= 273 K$ )

1 272

2 271

3 270

4 274

Explanation:

From formula,
$Δ T_{f} = K_{f} \frac{W_{B} \times 1000}{M_{B} \times W_{A}} {\begin{cases} W_{B} = mass of solute = 31 g \\ W_{A} = mass of solvent = 500 g \\ M_{B} = Ethylene glycol (C_{2} H_{6} O_{2}) \end{cases}}$
$Δ T_{f} = \frac{2 \times 31 \times 1000}{62 \times 500} = 2 K$
Now, $Δ T_{f} = T_{f}^{0} - T_{f}$
$∴ T_{f} = 273 - 2$
$T_{f} = 271 K$

Kerala-CEE-2016

SOLUTIONS

277331 $58.5 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 mL$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
Molality of $NaCl$ solution $= \frac{n}{W} \times 1000$
$= \frac{58.5 / 58.5}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$
Molality of $C_{6} H_{12} O_{6}$ solution $= \frac{180 / 100}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$ Both the solutions have same molality but the values of (i) for $NaCl$ and glucose are 2 and 1 respectively.
$Δ T_{NaCl} = 2 \times Δ T_{C_{6} H_{12} O_{6}}$
Hence, $NaCl$ will show higher elevation of boiling.

JCECE - 2017

SOLUTIONS

277332 Which of the following aqueous solution has highest freezing point?

1 0.1 molal

{Al}_{2} {({SO}_{4})}_{3}

2 0.1 molal

{BaCl}_{2}

3 0.1 molal

{AlCl}_{3}

4 0.1 molal

{NH}_{4} Cl

Explanation:

Depression in freezing point.
$Δ T_{f} = i \times K_{f} \times m$
Where, $i = \frac{observed value}{theoretical value}$
$\begin{aligned} For {Al}_{2} {({SO}_{4})}_{3} = 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -} (i = 5) \\ {BaCl}_{2} = {Ba}^{+ 2} + 2 {Cl}^{-} (i = 3) \\ {AlCl}_{3} = {Al}^{3 +} + 3 {Cl}^{-} (i = 4) \\ {NH}_{4} Cl = {NH}_{4}^{+} + {Cl}^{-} (i = 2) \end{aligned}$
Hence, ${NH}_{4} Cl$ , has $i = 2$ which show highest freezing point.

Karnataka-CET-2017

SOLUTIONS

277333 What will be the boiling point of $30 g$ benzene containing $0.75 g$ of benzoic acid which undergoes $75 %$ dimerization? Boiling point of pure benzene $= {80.1}^{\circ} C$ and $K_{b} = 2.53 K {mol}^{- 1} kg$

1

{90.52}^{\circ} C

2

{104.35}^{\circ} C

3

{76.12}^{\circ} C

4

{80.42}^{\circ} C

Explanation:

Given data,
Weigth of benzene $= 30 g$
Weight of benzoic acid $= 0.75 g$
From formula
$\begin{aligned} Δ T_{b} & = {iK}_{b} m \\ i & = 1 - α + \frac{α}{n} \\ i & = 1 - 0.75 + \frac{0.75}{2} (\begin{array}{c} ∴ α = 0.75 \\ n = 2 \end{array}) \\ i & = 0.625 \end{aligned}$
Now putting the values in equation. (i)
$\begin{aligned} Δ T_{b} = 0.625 \times 2.53 \times \frac{0.75 / 122}{30} \times 1000 \\ Δ T_{b} = 0.324 \end{aligned}$
$Δ T_{b} = T_{b} - T_{b}^{o}$
$0.324 = T_{b} - {80.1}^{\circ} C$
$T_{b} = {80.42}^{\circ} C$

J and K CET-(2017)

SOLUTIONS

277334 If $P^{0}$ and $P$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $P$ and $P^{0}$ is

1

P^{o} = P [\frac{n_{1}}{n_{1} + n_{2}}]

2

P^{o} = P [\frac{n_{2}}{n_{1} + n_{2}}]

3

P = P^{o} [\frac{n_{2}}{n_{1} + n_{2}}]

4

P = P^{o} [\frac{n_{1}}{n_{1} + n_{2}}]

Explanation:

According to the Raoult's law -
$\frac{P^{o} - P}{P^{o}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $1 - \frac{P}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = 1 - \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
$P = P^{\circ} [\frac{n_{2}}{n_{1} + n_{2}}]$

WB-JEE-2016

SOLUTIONS

277335 $31 g$ of ethylene glycol $(C_{2} H_{6} O_{2})$ is mixed with $500 g$ of solvent $(K_{f}$ of the solvent is $2 K kg {mol}^{-}$ 1). What is the freezing point of the solution (in K)? (Freezing point of solvent $= 273 K$ )

1 272

2 271

3 270

4 274

Explanation:

From formula,
$Δ T_{f} = K_{f} \frac{W_{B} \times 1000}{M_{B} \times W_{A}} {\begin{cases} W_{B} = mass of solute = 31 g \\ W_{A} = mass of solvent = 500 g \\ M_{B} = Ethylene glycol (C_{2} H_{6} O_{2}) \end{cases}}$
$Δ T_{f} = \frac{2 \times 31 \times 1000}{62 \times 500} = 2 K$
Now, $Δ T_{f} = T_{f}^{0} - T_{f}$
$∴ T_{f} = 273 - 2$
$T_{f} = 271 K$

Kerala-CEE-2016

SOLUTIONS

277331 $58.5 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 mL$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
Molality of $NaCl$ solution $= \frac{n}{W} \times 1000$
$= \frac{58.5 / 58.5}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$
Molality of $C_{6} H_{12} O_{6}$ solution $= \frac{180 / 100}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$ Both the solutions have same molality but the values of (i) for $NaCl$ and glucose are 2 and 1 respectively.
$Δ T_{NaCl} = 2 \times Δ T_{C_{6} H_{12} O_{6}}$
Hence, $NaCl$ will show higher elevation of boiling.

JCECE - 2017

SOLUTIONS

277332 Which of the following aqueous solution has highest freezing point?

1 0.1 molal

{Al}_{2} {({SO}_{4})}_{3}

2 0.1 molal

{BaCl}_{2}

3 0.1 molal

{AlCl}_{3}

4 0.1 molal

{NH}_{4} Cl

Explanation:

Depression in freezing point.
$Δ T_{f} = i \times K_{f} \times m$
Where, $i = \frac{observed value}{theoretical value}$
$\begin{aligned} For {Al}_{2} {({SO}_{4})}_{3} = 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -} (i = 5) \\ {BaCl}_{2} = {Ba}^{+ 2} + 2 {Cl}^{-} (i = 3) \\ {AlCl}_{3} = {Al}^{3 +} + 3 {Cl}^{-} (i = 4) \\ {NH}_{4} Cl = {NH}_{4}^{+} + {Cl}^{-} (i = 2) \end{aligned}$
Hence, ${NH}_{4} Cl$ , has $i = 2$ which show highest freezing point.

Karnataka-CET-2017

SOLUTIONS

277333 What will be the boiling point of $30 g$ benzene containing $0.75 g$ of benzoic acid which undergoes $75 %$ dimerization? Boiling point of pure benzene $= {80.1}^{\circ} C$ and $K_{b} = 2.53 K {mol}^{- 1} kg$

1

{90.52}^{\circ} C

2

{104.35}^{\circ} C

3

{76.12}^{\circ} C

4

{80.42}^{\circ} C

Explanation:

Given data,
Weigth of benzene $= 30 g$
Weight of benzoic acid $= 0.75 g$
From formula
$\begin{aligned} Δ T_{b} & = {iK}_{b} m \\ i & = 1 - α + \frac{α}{n} \\ i & = 1 - 0.75 + \frac{0.75}{2} (\begin{array}{c} ∴ α = 0.75 \\ n = 2 \end{array}) \\ i & = 0.625 \end{aligned}$
Now putting the values in equation. (i)
$\begin{aligned} Δ T_{b} = 0.625 \times 2.53 \times \frac{0.75 / 122}{30} \times 1000 \\ Δ T_{b} = 0.324 \end{aligned}$
$Δ T_{b} = T_{b} - T_{b}^{o}$
$0.324 = T_{b} - {80.1}^{\circ} C$
$T_{b} = {80.42}^{\circ} C$

J and K CET-(2017)

SOLUTIONS

277334 If $P^{0}$ and $P$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $P$ and $P^{0}$ is

1

P^{o} = P [\frac{n_{1}}{n_{1} + n_{2}}]

2

P^{o} = P [\frac{n_{2}}{n_{1} + n_{2}}]

3

P = P^{o} [\frac{n_{2}}{n_{1} + n_{2}}]

4

P = P^{o} [\frac{n_{1}}{n_{1} + n_{2}}]

Explanation:

According to the Raoult's law -
$\frac{P^{o} - P}{P^{o}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $1 - \frac{P}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = 1 - \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
$P = P^{\circ} [\frac{n_{2}}{n_{1} + n_{2}}]$

WB-JEE-2016

SOLUTIONS

277335 $31 g$ of ethylene glycol $(C_{2} H_{6} O_{2})$ is mixed with $500 g$ of solvent $(K_{f}$ of the solvent is $2 K kg {mol}^{-}$ 1). What is the freezing point of the solution (in K)? (Freezing point of solvent $= 273 K$ )

1 272

2 271

3 270

4 274

Explanation:

From formula,
$Δ T_{f} = K_{f} \frac{W_{B} \times 1000}{M_{B} \times W_{A}} {\begin{cases} W_{B} = mass of solute = 31 g \\ W_{A} = mass of solvent = 500 g \\ M_{B} = Ethylene glycol (C_{2} H_{6} O_{2}) \end{cases}}$
$Δ T_{f} = \frac{2 \times 31 \times 1000}{62 \times 500} = 2 K$
Now, $Δ T_{f} = T_{f}^{0} - T_{f}$
$∴ T_{f} = 273 - 2$
$T_{f} = 271 K$

Kerala-CEE-2016

SOLUTIONS

277331 $58.5 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 mL$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Elevation in boiling point,
$Δ T_{b} = i \times K_{b} \times m$
Molality of $NaCl$ solution $= \frac{n}{W} \times 1000$
$= \frac{58.5 / 58.5}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$
Molality of $C_{6} H_{12} O_{6}$ solution $= \frac{180 / 100}{W_{H_{2} O}} \times 1000 = \frac{1000}{W_{H_{2} O}}$ Both the solutions have same molality but the values of (i) for $NaCl$ and glucose are 2 and 1 respectively.
$Δ T_{NaCl} = 2 \times Δ T_{C_{6} H_{12} O_{6}}$
Hence, $NaCl$ will show higher elevation of boiling.

JCECE - 2017

SOLUTIONS

277332 Which of the following aqueous solution has highest freezing point?

1 0.1 molal

{Al}_{2} {({SO}_{4})}_{3}

2 0.1 molal

{BaCl}_{2}

3 0.1 molal

{AlCl}_{3}

4 0.1 molal

{NH}_{4} Cl

Explanation:

Depression in freezing point.
$Δ T_{f} = i \times K_{f} \times m$
Where, $i = \frac{observed value}{theoretical value}$
$\begin{aligned} For {Al}_{2} {({SO}_{4})}_{3} = 2 {Al}^{3 +} + 3 {SO}_{4}^{2 -} (i = 5) \\ {BaCl}_{2} = {Ba}^{+ 2} + 2 {Cl}^{-} (i = 3) \\ {AlCl}_{3} = {Al}^{3 +} + 3 {Cl}^{-} (i = 4) \\ {NH}_{4} Cl = {NH}_{4}^{+} + {Cl}^{-} (i = 2) \end{aligned}$
Hence, ${NH}_{4} Cl$ , has $i = 2$ which show highest freezing point.

Karnataka-CET-2017

SOLUTIONS

277333 What will be the boiling point of $30 g$ benzene containing $0.75 g$ of benzoic acid which undergoes $75 %$ dimerization? Boiling point of pure benzene $= {80.1}^{\circ} C$ and $K_{b} = 2.53 K {mol}^{- 1} kg$

1

{90.52}^{\circ} C

2

{104.35}^{\circ} C

3

{76.12}^{\circ} C

4

{80.42}^{\circ} C

Explanation:

Given data,
Weigth of benzene $= 30 g$
Weight of benzoic acid $= 0.75 g$
From formula
$\begin{aligned} Δ T_{b} & = {iK}_{b} m \\ i & = 1 - α + \frac{α}{n} \\ i & = 1 - 0.75 + \frac{0.75}{2} (\begin{array}{c} ∴ α = 0.75 \\ n = 2 \end{array}) \\ i & = 0.625 \end{aligned}$
Now putting the values in equation. (i)
$\begin{aligned} Δ T_{b} = 0.625 \times 2.53 \times \frac{0.75 / 122}{30} \times 1000 \\ Δ T_{b} = 0.324 \end{aligned}$
$Δ T_{b} = T_{b} - T_{b}^{o}$
$0.324 = T_{b} - {80.1}^{\circ} C$
$T_{b} = {80.42}^{\circ} C$

J and K CET-(2017)

SOLUTIONS

277334 If $P^{0}$ and $P$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $P$ and $P^{0}$ is

1

P^{o} = P [\frac{n_{1}}{n_{1} + n_{2}}]

2

P^{o} = P [\frac{n_{2}}{n_{1} + n_{2}}]

3

P = P^{o} [\frac{n_{2}}{n_{1} + n_{2}}]

4

P = P^{o} [\frac{n_{1}}{n_{1} + n_{2}}]

Explanation:

According to the Raoult's law -
$\frac{P^{o} - P}{P^{o}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $1 - \frac{P}{P^{0}} = \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = 1 - \frac{n_{1}}{n_{1} + n_{2}}$
Or $\frac{P}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
$P = P^{\circ} [\frac{n_{2}}{n_{1} + n_{2}}]$

WB-JEE-2016

SOLUTIONS

277335 $31 g$ of ethylene glycol $(C_{2} H_{6} O_{2})$ is mixed with $500 g$ of solvent $(K_{f}$ of the solvent is $2 K kg {mol}^{-}$ 1). What is the freezing point of the solution (in K)? (Freezing point of solvent $= 273 K$ )

1 272

2 271

3 270

4 274

Explanation:

From formula,
$Δ T_{f} = K_{f} \frac{W_{B} \times 1000}{M_{B} \times W_{A}} {\begin{cases} W_{B} = mass of solute = 31 g \\ W_{A} = mass of solvent = 500 g \\ M_{B} = Ethylene glycol (C_{2} H_{6} O_{2}) \end{cases}}$
$Δ T_{f} = \frac{2 \times 31 \times 1000}{62 \times 500} = 2 K$
Now, $Δ T_{f} = T_{f}^{0} - T_{f}$
$∴ T_{f} = 273 - 2$
$T_{f} = 271 K$

Kerala-CEE-2016

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