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CHXII02:SOLUTIONS

319274 Osmotic pressure of a solution is 1.5 atm. The density of solution is $1.1 g / {cm}^{3}$ . Osmotic rise is ____.
$(1 a t m = 76 c m H g, d_{H g} = 13.6 g / c m^{3})$

1 1.36

2 13.6

3 1410

4 141

Explanation:

$π =$ hdg
$1.5 \times 76 \times 13.6 \times g = h \times 1.1 \times g$
$h = \frac{1.5 \times 76 \times 13.6}{1.1} cm$
$≃ 1410 cm$

CHXII02:SOLUTIONS

319275 Insulin is dissolved in suitable solvent and the osmotic pressure $(a t m) π$ , of solutions of various concentrations $(g / c m^{3})$ C is measured at $2 7^{^{\circ}} C$ . The slope of $(π)$ against C is found to be $4 .1 \times 1 0^{- 3}$ . The molecular mass of insulin is

1

6 \times 1 0^{6}

2

3 \times 10^{6}

3

6 \times 1 0^{3}

4

3 \times 1 0^{3}

Explanation:

According to Osmotic pressure,
$π = \frac{C (g / c c)}{M W} S T$
The slope of graph plot between osmotic
pressure and concentration is $\frac{S T}{M W_{2}}$
Given, Slope $= 4 .1 \times 1 0^{- 3}$
$\frac{S T}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$\frac{0 .0821 \times 300}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$M W_{2} = 6 \times 1 0^{6}$

CHXII02:SOLUTIONS

319276 Osmotic pressure of urea solution at $10^{\circ} C$ is 500 mm . Osmotic pressure of the solution becomes 105.3 mm , when it is diluted and temperature raised to $25^{\circ} C$ . The extent dilution is

1 8 times

2 5 times

3 4 times

4 7 times

Explanation:

$\frac{π_{1}}{π_{2}} = \frac{C_{1} {RT}_{1}}{C_{2} {RT}_{2}}$
$\frac{500}{105.3} = \frac{C_{1} \times 283}{C_{2} \times 298}$
$\frac{C_{1}}{C_{2}} = \frac{500}{105.3} \times \frac{298}{283} = 5$
$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{1}{5} \Rightarrow$ So, it is diluted 5 times.

CHXII02:SOLUTIONS

319277 Evaluate the following statements for their correctness.
A. The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea.
B. Azeotropic mixtures boil without change in their composition.
C. Osmosis always takes place from hypertonic to hypotonic solution.
D. The density of $32 % H_{2} S O_{4}$ solution havingmolarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.
Choose the correct answer from the options given below:

1 B and D only

2 A, B and D only

3 A and C only

4 B, D and E only

Explanation:

(A) $Δ T_{b} \propto i \times C$
(B) Azeotropic mixtures have same composition in both liquid and vapour phase.
(C) Osmosis always takes place from hypotonic to hypertonic solution.
(D) The density of $32 % H_{2} S O_{4}$ solution having molarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
(E) When KI solution is added to ${AgNO}_{3}$ solution, positively charged sol is formed due to adsorption of ${Ag}^{+}$ ions from dispersion medium. Thus, only statements B and D are correct. Hence, option (1) is correct.

JEE - 2023

CHXII02:SOLUTIONS

319278 The concentration (in $mol / L$ ) of the solution having osmotic pressure $0.0821 atm$ at $300 K$ will be

1

0 .33

2

0 .066

3

0.3 \times 10^{- 2}

4

3

Explanation:

$π = C R T$
where, $C =$ concentration,
$π = o s m o t i c p r e s s u r e,$
$R =$ gas constant, $T =$ temperature
$C = \frac{π}{R T} = \frac{0.0821}{0.0821 \times 300}$
$= 0.3 \times 10^{- 2} m o l / L$

CHXII02:SOLUTIONS

319274 Osmotic pressure of a solution is 1.5 atm. The density of solution is $1.1 g / {cm}^{3}$ . Osmotic rise is ____.
$(1 a t m = 76 c m H g, d_{H g} = 13.6 g / c m^{3})$

1 1.36

2 13.6

3 1410

4 141

Explanation:

$π =$ hdg
$1.5 \times 76 \times 13.6 \times g = h \times 1.1 \times g$
$h = \frac{1.5 \times 76 \times 13.6}{1.1} cm$
$≃ 1410 cm$

CHXII02:SOLUTIONS

319275 Insulin is dissolved in suitable solvent and the osmotic pressure $(a t m) π$ , of solutions of various concentrations $(g / c m^{3})$ C is measured at $2 7^{^{\circ}} C$ . The slope of $(π)$ against C is found to be $4 .1 \times 1 0^{- 3}$ . The molecular mass of insulin is

1

6 \times 1 0^{6}

2

3 \times 10^{6}

3

6 \times 1 0^{3}

4

3 \times 1 0^{3}

Explanation:

According to Osmotic pressure,
$π = \frac{C (g / c c)}{M W} S T$
The slope of graph plot between osmotic
pressure and concentration is $\frac{S T}{M W_{2}}$
Given, Slope $= 4 .1 \times 1 0^{- 3}$
$\frac{S T}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$\frac{0 .0821 \times 300}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$M W_{2} = 6 \times 1 0^{6}$

CHXII02:SOLUTIONS

319276 Osmotic pressure of urea solution at $10^{\circ} C$ is 500 mm . Osmotic pressure of the solution becomes 105.3 mm , when it is diluted and temperature raised to $25^{\circ} C$ . The extent dilution is

1 8 times

2 5 times

3 4 times

4 7 times

Explanation:

$\frac{π_{1}}{π_{2}} = \frac{C_{1} {RT}_{1}}{C_{2} {RT}_{2}}$
$\frac{500}{105.3} = \frac{C_{1} \times 283}{C_{2} \times 298}$
$\frac{C_{1}}{C_{2}} = \frac{500}{105.3} \times \frac{298}{283} = 5$
$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{1}{5} \Rightarrow$ So, it is diluted 5 times.

CHXII02:SOLUTIONS

319277 Evaluate the following statements for their correctness.
A. The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea.
B. Azeotropic mixtures boil without change in their composition.
C. Osmosis always takes place from hypertonic to hypotonic solution.
D. The density of $32 % H_{2} S O_{4}$ solution havingmolarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.
Choose the correct answer from the options given below:

1 B and D only

2 A, B and D only

3 A and C only

4 B, D and E only

Explanation:

(A) $Δ T_{b} \propto i \times C$
(B) Azeotropic mixtures have same composition in both liquid and vapour phase.
(C) Osmosis always takes place from hypotonic to hypertonic solution.
(D) The density of $32 % H_{2} S O_{4}$ solution having molarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
(E) When KI solution is added to ${AgNO}_{3}$ solution, positively charged sol is formed due to adsorption of ${Ag}^{+}$ ions from dispersion medium. Thus, only statements B and D are correct. Hence, option (1) is correct.

JEE - 2023

CHXII02:SOLUTIONS

319278 The concentration (in $mol / L$ ) of the solution having osmotic pressure $0.0821 atm$ at $300 K$ will be

1

0 .33

2

0 .066

3

0.3 \times 10^{- 2}

4

3

Explanation:

$π = C R T$
where, $C =$ concentration,
$π = o s m o t i c p r e s s u r e,$
$R =$ gas constant, $T =$ temperature
$C = \frac{π}{R T} = \frac{0.0821}{0.0821 \times 300}$
$= 0.3 \times 10^{- 2} m o l / L$

CHXII02:SOLUTIONS

319274 Osmotic pressure of a solution is 1.5 atm. The density of solution is $1.1 g / {cm}^{3}$ . Osmotic rise is ____.
$(1 a t m = 76 c m H g, d_{H g} = 13.6 g / c m^{3})$

1 1.36

2 13.6

3 1410

4 141

Explanation:

$π =$ hdg
$1.5 \times 76 \times 13.6 \times g = h \times 1.1 \times g$
$h = \frac{1.5 \times 76 \times 13.6}{1.1} cm$
$≃ 1410 cm$

CHXII02:SOLUTIONS

319275 Insulin is dissolved in suitable solvent and the osmotic pressure $(a t m) π$ , of solutions of various concentrations $(g / c m^{3})$ C is measured at $2 7^{^{\circ}} C$ . The slope of $(π)$ against C is found to be $4 .1 \times 1 0^{- 3}$ . The molecular mass of insulin is

1

6 \times 1 0^{6}

2

3 \times 10^{6}

3

6 \times 1 0^{3}

4

3 \times 1 0^{3}

Explanation:

According to Osmotic pressure,
$π = \frac{C (g / c c)}{M W} S T$
The slope of graph plot between osmotic
pressure and concentration is $\frac{S T}{M W_{2}}$
Given, Slope $= 4 .1 \times 1 0^{- 3}$
$\frac{S T}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$\frac{0 .0821 \times 300}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$M W_{2} = 6 \times 1 0^{6}$

CHXII02:SOLUTIONS

319276 Osmotic pressure of urea solution at $10^{\circ} C$ is 500 mm . Osmotic pressure of the solution becomes 105.3 mm , when it is diluted and temperature raised to $25^{\circ} C$ . The extent dilution is

1 8 times

2 5 times

3 4 times

4 7 times

Explanation:

$\frac{π_{1}}{π_{2}} = \frac{C_{1} {RT}_{1}}{C_{2} {RT}_{2}}$
$\frac{500}{105.3} = \frac{C_{1} \times 283}{C_{2} \times 298}$
$\frac{C_{1}}{C_{2}} = \frac{500}{105.3} \times \frac{298}{283} = 5$
$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{1}{5} \Rightarrow$ So, it is diluted 5 times.

CHXII02:SOLUTIONS

319277 Evaluate the following statements for their correctness.
A. The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea.
B. Azeotropic mixtures boil without change in their composition.
C. Osmosis always takes place from hypertonic to hypotonic solution.
D. The density of $32 % H_{2} S O_{4}$ solution havingmolarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.
Choose the correct answer from the options given below:

1 B and D only

2 A, B and D only

3 A and C only

4 B, D and E only

Explanation:

(A) $Δ T_{b} \propto i \times C$
(B) Azeotropic mixtures have same composition in both liquid and vapour phase.
(C) Osmosis always takes place from hypotonic to hypertonic solution.
(D) The density of $32 % H_{2} S O_{4}$ solution having molarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
(E) When KI solution is added to ${AgNO}_{3}$ solution, positively charged sol is formed due to adsorption of ${Ag}^{+}$ ions from dispersion medium. Thus, only statements B and D are correct. Hence, option (1) is correct.

JEE - 2023

CHXII02:SOLUTIONS

319278 The concentration (in $mol / L$ ) of the solution having osmotic pressure $0.0821 atm$ at $300 K$ will be

1

0 .33

2

0 .066

3

0.3 \times 10^{- 2}

4

3

Explanation:

$π = C R T$
where, $C =$ concentration,
$π = o s m o t i c p r e s s u r e,$
$R =$ gas constant, $T =$ temperature
$C = \frac{π}{R T} = \frac{0.0821}{0.0821 \times 300}$
$= 0.3 \times 10^{- 2} m o l / L$

CHXII02:SOLUTIONS

319274 Osmotic pressure of a solution is 1.5 atm. The density of solution is $1.1 g / {cm}^{3}$ . Osmotic rise is ____.
$(1 a t m = 76 c m H g, d_{H g} = 13.6 g / c m^{3})$

1 1.36

2 13.6

3 1410

4 141

Explanation:

$π =$ hdg
$1.5 \times 76 \times 13.6 \times g = h \times 1.1 \times g$
$h = \frac{1.5 \times 76 \times 13.6}{1.1} cm$
$≃ 1410 cm$

CHXII02:SOLUTIONS

319275 Insulin is dissolved in suitable solvent and the osmotic pressure $(a t m) π$ , of solutions of various concentrations $(g / c m^{3})$ C is measured at $2 7^{^{\circ}} C$ . The slope of $(π)$ against C is found to be $4 .1 \times 1 0^{- 3}$ . The molecular mass of insulin is

1

6 \times 1 0^{6}

2

3 \times 10^{6}

3

6 \times 1 0^{3}

4

3 \times 1 0^{3}

Explanation:

According to Osmotic pressure,
$π = \frac{C (g / c c)}{M W} S T$
The slope of graph plot between osmotic
pressure and concentration is $\frac{S T}{M W_{2}}$
Given, Slope $= 4 .1 \times 1 0^{- 3}$
$\frac{S T}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$\frac{0 .0821 \times 300}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$M W_{2} = 6 \times 1 0^{6}$

CHXII02:SOLUTIONS

319276 Osmotic pressure of urea solution at $10^{\circ} C$ is 500 mm . Osmotic pressure of the solution becomes 105.3 mm , when it is diluted and temperature raised to $25^{\circ} C$ . The extent dilution is

1 8 times

2 5 times

3 4 times

4 7 times

Explanation:

$\frac{π_{1}}{π_{2}} = \frac{C_{1} {RT}_{1}}{C_{2} {RT}_{2}}$
$\frac{500}{105.3} = \frac{C_{1} \times 283}{C_{2} \times 298}$
$\frac{C_{1}}{C_{2}} = \frac{500}{105.3} \times \frac{298}{283} = 5$
$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{1}{5} \Rightarrow$ So, it is diluted 5 times.

CHXII02:SOLUTIONS

319277 Evaluate the following statements for their correctness.
A. The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea.
B. Azeotropic mixtures boil without change in their composition.
C. Osmosis always takes place from hypertonic to hypotonic solution.
D. The density of $32 % H_{2} S O_{4}$ solution havingmolarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.
Choose the correct answer from the options given below:

1 B and D only

2 A, B and D only

3 A and C only

4 B, D and E only

Explanation:

(A) $Δ T_{b} \propto i \times C$
(B) Azeotropic mixtures have same composition in both liquid and vapour phase.
(C) Osmosis always takes place from hypotonic to hypertonic solution.
(D) The density of $32 % H_{2} S O_{4}$ solution having molarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
(E) When KI solution is added to ${AgNO}_{3}$ solution, positively charged sol is formed due to adsorption of ${Ag}^{+}$ ions from dispersion medium. Thus, only statements B and D are correct. Hence, option (1) is correct.

JEE - 2023

CHXII02:SOLUTIONS

319278 The concentration (in $mol / L$ ) of the solution having osmotic pressure $0.0821 atm$ at $300 K$ will be

1

0 .33

2

0 .066

3

0.3 \times 10^{- 2}

4

3

Explanation:

$π = C R T$
where, $C =$ concentration,
$π = o s m o t i c p r e s s u r e,$
$R =$ gas constant, $T =$ temperature
$C = \frac{π}{R T} = \frac{0.0821}{0.0821 \times 300}$
$= 0.3 \times 10^{- 2} m o l / L$

CHXII02:SOLUTIONS

319274 Osmotic pressure of a solution is 1.5 atm. The density of solution is $1.1 g / {cm}^{3}$ . Osmotic rise is ____.
$(1 a t m = 76 c m H g, d_{H g} = 13.6 g / c m^{3})$

1 1.36

2 13.6

3 1410

4 141

Explanation:

$π =$ hdg
$1.5 \times 76 \times 13.6 \times g = h \times 1.1 \times g$
$h = \frac{1.5 \times 76 \times 13.6}{1.1} cm$
$≃ 1410 cm$

CHXII02:SOLUTIONS

319275 Insulin is dissolved in suitable solvent and the osmotic pressure $(a t m) π$ , of solutions of various concentrations $(g / c m^{3})$ C is measured at $2 7^{^{\circ}} C$ . The slope of $(π)$ against C is found to be $4 .1 \times 1 0^{- 3}$ . The molecular mass of insulin is

1

6 \times 1 0^{6}

2

3 \times 10^{6}

3

6 \times 1 0^{3}

4

3 \times 1 0^{3}

Explanation:

According to Osmotic pressure,
$π = \frac{C (g / c c)}{M W} S T$
The slope of graph plot between osmotic
pressure and concentration is $\frac{S T}{M W_{2}}$
Given, Slope $= 4 .1 \times 1 0^{- 3}$
$\frac{S T}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$\frac{0 .0821 \times 300}{M W_{2}} = 4 .1 \times 1 0^{- 3}$
$M W_{2} = 6 \times 1 0^{6}$

CHXII02:SOLUTIONS

319276 Osmotic pressure of urea solution at $10^{\circ} C$ is 500 mm . Osmotic pressure of the solution becomes 105.3 mm , when it is diluted and temperature raised to $25^{\circ} C$ . The extent dilution is

1 8 times

2 5 times

3 4 times

4 7 times

Explanation:

$\frac{π_{1}}{π_{2}} = \frac{C_{1} {RT}_{1}}{C_{2} {RT}_{2}}$
$\frac{500}{105.3} = \frac{C_{1} \times 283}{C_{2} \times 298}$
$\frac{C_{1}}{C_{2}} = \frac{500}{105.3} \times \frac{298}{283} = 5$
$\Rightarrow \frac{C_{2}}{C_{1}} = \frac{1}{5} \Rightarrow$ So, it is diluted 5 times.

CHXII02:SOLUTIONS

319277 Evaluate the following statements for their correctness.
A. The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea.
B. Azeotropic mixtures boil without change in their composition.
C. Osmosis always takes place from hypertonic to hypotonic solution.
D. The density of $32 % H_{2} S O_{4}$ solution havingmolarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.
Choose the correct answer from the options given below:

1 B and D only

2 A, B and D only

3 A and C only

4 B, D and E only

Explanation:

(A) $Δ T_{b} \propto i \times C$
(B) Azeotropic mixtures have same composition in both liquid and vapour phase.
(C) Osmosis always takes place from hypotonic to hypertonic solution.
(D) The density of $32 % H_{2} S O_{4}$ solution having molarity 4.09 M is approximately $1.26 g {mL}^{- 1}$ .
(E) When KI solution is added to ${AgNO}_{3}$ solution, positively charged sol is formed due to adsorption of ${Ag}^{+}$ ions from dispersion medium. Thus, only statements B and D are correct. Hence, option (1) is correct.

JEE - 2023

CHXII02:SOLUTIONS

319278 The concentration (in $mol / L$ ) of the solution having osmotic pressure $0.0821 atm$ at $300 K$ will be

1

0 .33

2

0 .066

3

0.3 \times 10^{- 2}

4

3

Explanation:

$π = C R T$
where, $C =$ concentration,
$π = o s m o t i c p r e s s u r e,$
$R =$ gas constant, $T =$ temperature
$C = \frac{π}{R T} = \frac{0.0821}{0.0821 \times 300}$
$= 0.3 \times 10^{- 2} m o l / L$

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