QBManager

CHXII02:SOLUTIONS

319244 $0.05 M {CuSO}_{4}$ when treated with $0.01 M K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:
supporting img
[SPM : Semi Permeable Membrane]
Due to osmosis

1 Green colour formation observed on side Y.

2 Green colour formation observed on side

X

.

3 Molarity of

{CuSO}_{4}

solution is lowered.

4 Molarity of

K_{2} {Cr}_{2} O_{7}

solution is lowered.

Explanation:

Only solvent molecules pass through the semipermeable membrane from a low concentrated solution to a more concentrated solution. The conc. of $K_{2} {Cr}_{2} O_{7}$ is less, so water molecules move from $K_{2} {Cr}_{2} O_{7}$ to ${CuSO}_{4}$ solution side, As a result conc. of ${CuSO}_{4}$ decreases. So, the correct option is (3).

JEE Main - 2024

CHXII02:SOLUTIONS

319245 Two solutions of ${KNO}_{3}$ and ${CH}_{3} COOH$ are prepared separately. Molarity of both is 0.1 M and osmotic pressures are $P_{1}$ and $P_{2}$ respectively. The correct relationship between the osmotic pressures is

1

P_{2} > P_{1}

2

P_{1} = P_{2}

3

P_{1} > P_{2}

4

\frac{P_{1}}{P_{1} + P_{2}} = \frac{P_{2}}{P_{1} + P_{2}}

Explanation:

Given that,
Osmotic pressure of $0.1 M K N O_{3} = P_{1}$
Osmotic pressure of $0.1 M C H_{3} C O O H = P_{2}$
As we know that osmotic pressure is directly proportional to the number of particles and acetic acid is a weak electrolyte so on dissociation in water ${CH}_{3} COOH$ will not give as much particles as ${KNO}_{3}$ .
Hence, $P_{1} > P_{2}$

CHXII02:SOLUTIONS

319246 Twenty grams of a substance were dissolved in 500ml of water and the osmotic pressure of the solution was found to be 600 mm of mercury at $15^{\circ} C$ . The molecular weight of the substance is

1 1120

2 1198

3 1200

4 None of these

Explanation:

$w = 20 g m; V = 500 m L = \frac{500}{1000} = 0 .5 l i t r e$
$π = 600 m m = \frac{600}{760} a t m;$
$T = 15 + 273 = 288 K; m w = ?$
According to Van’t Hoff equation,
$π V = n S T; π V = \frac{w}{m w} S T$
$∴ m = \frac{w S T}{π V} = \frac{20 \times 0 .0821 \times 288 \times 760}{600 \times 0 .5} = 1198$

CHXII02:SOLUTIONS

319247 Osmotic pressure of a sugar solution at $24^{\circ} C$ is 2.5 atmospheres. The concentration of the solution in mole per L is

1 0.0821 mol/L

2 1.082 mol/L

3 0.1025 mol/L

4 0.0827 mol/L

Explanation:

$π = 2 .5 a t m, T = 24 + 273 = 297 K,$
$S = 0 .082 l i t . a t m d e g^{- 1} m o l^{- 1}, C = ?$
$W e k n o w t h a t π = C S T o r$
$C = \frac{π}{S T} = \frac{2 .5}{0 .0821 \times 297} = 0 .1025 m o l e s / l i t r e$

CHXII02:SOLUTIONS

319248 Osmotic pressure of a solution containing $2 g$ dissolved protein per $300 {cm}^{3}$ of solution is 20 $mm$ of $Hg$ at $27^{\circ} C$ . The molecular mass of the protein is:

1

6239.6 g {mol}^{- 1}

2

12315.5 g {mol}^{- 1}

3

3692.1 g {mol}^{- 1}

4

7368.4 g {mol}^{- 1}

Explanation:

$π = \frac{n_{B} RT}{V} : n_{B} = \frac{w_{B}}{M_{B}}$
$π = \frac{w_{B}}{M_{B}} \times \frac{RT}{V}$
$∴ M_{B} = \frac{W_{B}}{V} \times \frac{RT}{π} = \frac{2 \times 0.0821 \times 300 \times 760}{0.3 \times 20}$
$= 6239.6 g {mol}^{- 1}$

CHXII02:SOLUTIONS

319244 $0.05 M {CuSO}_{4}$ when treated with $0.01 M K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:
supporting img
[SPM : Semi Permeable Membrane]
Due to osmosis

1 Green colour formation observed on side Y.

2 Green colour formation observed on side

X

.

3 Molarity of

{CuSO}_{4}

solution is lowered.

4 Molarity of

K_{2} {Cr}_{2} O_{7}

solution is lowered.

Explanation:

Only solvent molecules pass through the semipermeable membrane from a low concentrated solution to a more concentrated solution. The conc. of $K_{2} {Cr}_{2} O_{7}$ is less, so water molecules move from $K_{2} {Cr}_{2} O_{7}$ to ${CuSO}_{4}$ solution side, As a result conc. of ${CuSO}_{4}$ decreases. So, the correct option is (3).

JEE Main - 2024

CHXII02:SOLUTIONS

319245 Two solutions of ${KNO}_{3}$ and ${CH}_{3} COOH$ are prepared separately. Molarity of both is 0.1 M and osmotic pressures are $P_{1}$ and $P_{2}$ respectively. The correct relationship between the osmotic pressures is

1

P_{2} > P_{1}

2

P_{1} = P_{2}

3

P_{1} > P_{2}

4

\frac{P_{1}}{P_{1} + P_{2}} = \frac{P_{2}}{P_{1} + P_{2}}

Explanation:

Given that,
Osmotic pressure of $0.1 M K N O_{3} = P_{1}$
Osmotic pressure of $0.1 M C H_{3} C O O H = P_{2}$
As we know that osmotic pressure is directly proportional to the number of particles and acetic acid is a weak electrolyte so on dissociation in water ${CH}_{3} COOH$ will not give as much particles as ${KNO}_{3}$ .
Hence, $P_{1} > P_{2}$

CHXII02:SOLUTIONS

319246 Twenty grams of a substance were dissolved in 500ml of water and the osmotic pressure of the solution was found to be 600 mm of mercury at $15^{\circ} C$ . The molecular weight of the substance is

1 1120

2 1198

3 1200

4 None of these

Explanation:

$w = 20 g m; V = 500 m L = \frac{500}{1000} = 0 .5 l i t r e$
$π = 600 m m = \frac{600}{760} a t m;$
$T = 15 + 273 = 288 K; m w = ?$
According to Van’t Hoff equation,
$π V = n S T; π V = \frac{w}{m w} S T$
$∴ m = \frac{w S T}{π V} = \frac{20 \times 0 .0821 \times 288 \times 760}{600 \times 0 .5} = 1198$

CHXII02:SOLUTIONS

319247 Osmotic pressure of a sugar solution at $24^{\circ} C$ is 2.5 atmospheres. The concentration of the solution in mole per L is

1 0.0821 mol/L

2 1.082 mol/L

3 0.1025 mol/L

4 0.0827 mol/L

Explanation:

$π = 2 .5 a t m, T = 24 + 273 = 297 K,$
$S = 0 .082 l i t . a t m d e g^{- 1} m o l^{- 1}, C = ?$
$W e k n o w t h a t π = C S T o r$
$C = \frac{π}{S T} = \frac{2 .5}{0 .0821 \times 297} = 0 .1025 m o l e s / l i t r e$

CHXII02:SOLUTIONS

319248 Osmotic pressure of a solution containing $2 g$ dissolved protein per $300 {cm}^{3}$ of solution is 20 $mm$ of $Hg$ at $27^{\circ} C$ . The molecular mass of the protein is:

1

6239.6 g {mol}^{- 1}

2

12315.5 g {mol}^{- 1}

3

3692.1 g {mol}^{- 1}

4

7368.4 g {mol}^{- 1}

Explanation:

$π = \frac{n_{B} RT}{V} : n_{B} = \frac{w_{B}}{M_{B}}$
$π = \frac{w_{B}}{M_{B}} \times \frac{RT}{V}$
$∴ M_{B} = \frac{W_{B}}{V} \times \frac{RT}{π} = \frac{2 \times 0.0821 \times 300 \times 760}{0.3 \times 20}$
$= 6239.6 g {mol}^{- 1}$

CHXII02:SOLUTIONS

319244 $0.05 M {CuSO}_{4}$ when treated with $0.01 M K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:
supporting img
[SPM : Semi Permeable Membrane]
Due to osmosis

1 Green colour formation observed on side Y.

2 Green colour formation observed on side

X

.

3 Molarity of

{CuSO}_{4}

solution is lowered.

4 Molarity of

K_{2} {Cr}_{2} O_{7}

solution is lowered.

Explanation:

Only solvent molecules pass through the semipermeable membrane from a low concentrated solution to a more concentrated solution. The conc. of $K_{2} {Cr}_{2} O_{7}$ is less, so water molecules move from $K_{2} {Cr}_{2} O_{7}$ to ${CuSO}_{4}$ solution side, As a result conc. of ${CuSO}_{4}$ decreases. So, the correct option is (3).

JEE Main - 2024

CHXII02:SOLUTIONS

319245 Two solutions of ${KNO}_{3}$ and ${CH}_{3} COOH$ are prepared separately. Molarity of both is 0.1 M and osmotic pressures are $P_{1}$ and $P_{2}$ respectively. The correct relationship between the osmotic pressures is

1

P_{2} > P_{1}

2

P_{1} = P_{2}

3

P_{1} > P_{2}

4

\frac{P_{1}}{P_{1} + P_{2}} = \frac{P_{2}}{P_{1} + P_{2}}

Explanation:

Given that,
Osmotic pressure of $0.1 M K N O_{3} = P_{1}$
Osmotic pressure of $0.1 M C H_{3} C O O H = P_{2}$
As we know that osmotic pressure is directly proportional to the number of particles and acetic acid is a weak electrolyte so on dissociation in water ${CH}_{3} COOH$ will not give as much particles as ${KNO}_{3}$ .
Hence, $P_{1} > P_{2}$

CHXII02:SOLUTIONS

319246 Twenty grams of a substance were dissolved in 500ml of water and the osmotic pressure of the solution was found to be 600 mm of mercury at $15^{\circ} C$ . The molecular weight of the substance is

1 1120

2 1198

3 1200

4 None of these

Explanation:

$w = 20 g m; V = 500 m L = \frac{500}{1000} = 0 .5 l i t r e$
$π = 600 m m = \frac{600}{760} a t m;$
$T = 15 + 273 = 288 K; m w = ?$
According to Van’t Hoff equation,
$π V = n S T; π V = \frac{w}{m w} S T$
$∴ m = \frac{w S T}{π V} = \frac{20 \times 0 .0821 \times 288 \times 760}{600 \times 0 .5} = 1198$

CHXII02:SOLUTIONS

319247 Osmotic pressure of a sugar solution at $24^{\circ} C$ is 2.5 atmospheres. The concentration of the solution in mole per L is

1 0.0821 mol/L

2 1.082 mol/L

3 0.1025 mol/L

4 0.0827 mol/L

Explanation:

$π = 2 .5 a t m, T = 24 + 273 = 297 K,$
$S = 0 .082 l i t . a t m d e g^{- 1} m o l^{- 1}, C = ?$
$W e k n o w t h a t π = C S T o r$
$C = \frac{π}{S T} = \frac{2 .5}{0 .0821 \times 297} = 0 .1025 m o l e s / l i t r e$

CHXII02:SOLUTIONS

319248 Osmotic pressure of a solution containing $2 g$ dissolved protein per $300 {cm}^{3}$ of solution is 20 $mm$ of $Hg$ at $27^{\circ} C$ . The molecular mass of the protein is:

1

6239.6 g {mol}^{- 1}

2

12315.5 g {mol}^{- 1}

3

3692.1 g {mol}^{- 1}

4

7368.4 g {mol}^{- 1}

Explanation:

$π = \frac{n_{B} RT}{V} : n_{B} = \frac{w_{B}}{M_{B}}$
$π = \frac{w_{B}}{M_{B}} \times \frac{RT}{V}$
$∴ M_{B} = \frac{W_{B}}{V} \times \frac{RT}{π} = \frac{2 \times 0.0821 \times 300 \times 760}{0.3 \times 20}$
$= 6239.6 g {mol}^{- 1}$

CHXII02:SOLUTIONS

319244 $0.05 M {CuSO}_{4}$ when treated with $0.01 M K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:
supporting img
[SPM : Semi Permeable Membrane]
Due to osmosis

1 Green colour formation observed on side Y.

2 Green colour formation observed on side

X

.

3 Molarity of

{CuSO}_{4}

solution is lowered.

4 Molarity of

K_{2} {Cr}_{2} O_{7}

solution is lowered.

Explanation:

Only solvent molecules pass through the semipermeable membrane from a low concentrated solution to a more concentrated solution. The conc. of $K_{2} {Cr}_{2} O_{7}$ is less, so water molecules move from $K_{2} {Cr}_{2} O_{7}$ to ${CuSO}_{4}$ solution side, As a result conc. of ${CuSO}_{4}$ decreases. So, the correct option is (3).

JEE Main - 2024

CHXII02:SOLUTIONS

319245 Two solutions of ${KNO}_{3}$ and ${CH}_{3} COOH$ are prepared separately. Molarity of both is 0.1 M and osmotic pressures are $P_{1}$ and $P_{2}$ respectively. The correct relationship between the osmotic pressures is

1

P_{2} > P_{1}

2

P_{1} = P_{2}

3

P_{1} > P_{2}

4

\frac{P_{1}}{P_{1} + P_{2}} = \frac{P_{2}}{P_{1} + P_{2}}

Explanation:

Given that,
Osmotic pressure of $0.1 M K N O_{3} = P_{1}$
Osmotic pressure of $0.1 M C H_{3} C O O H = P_{2}$
As we know that osmotic pressure is directly proportional to the number of particles and acetic acid is a weak electrolyte so on dissociation in water ${CH}_{3} COOH$ will not give as much particles as ${KNO}_{3}$ .
Hence, $P_{1} > P_{2}$

CHXII02:SOLUTIONS

319246 Twenty grams of a substance were dissolved in 500ml of water and the osmotic pressure of the solution was found to be 600 mm of mercury at $15^{\circ} C$ . The molecular weight of the substance is

1 1120

2 1198

3 1200

4 None of these

Explanation:

$w = 20 g m; V = 500 m L = \frac{500}{1000} = 0 .5 l i t r e$
$π = 600 m m = \frac{600}{760} a t m;$
$T = 15 + 273 = 288 K; m w = ?$
According to Van’t Hoff equation,
$π V = n S T; π V = \frac{w}{m w} S T$
$∴ m = \frac{w S T}{π V} = \frac{20 \times 0 .0821 \times 288 \times 760}{600 \times 0 .5} = 1198$

CHXII02:SOLUTIONS

319247 Osmotic pressure of a sugar solution at $24^{\circ} C$ is 2.5 atmospheres. The concentration of the solution in mole per L is

1 0.0821 mol/L

2 1.082 mol/L

3 0.1025 mol/L

4 0.0827 mol/L

Explanation:

$π = 2 .5 a t m, T = 24 + 273 = 297 K,$
$S = 0 .082 l i t . a t m d e g^{- 1} m o l^{- 1}, C = ?$
$W e k n o w t h a t π = C S T o r$
$C = \frac{π}{S T} = \frac{2 .5}{0 .0821 \times 297} = 0 .1025 m o l e s / l i t r e$

CHXII02:SOLUTIONS

319248 Osmotic pressure of a solution containing $2 g$ dissolved protein per $300 {cm}^{3}$ of solution is 20 $mm$ of $Hg$ at $27^{\circ} C$ . The molecular mass of the protein is:

1

6239.6 g {mol}^{- 1}

2

12315.5 g {mol}^{- 1}

3

3692.1 g {mol}^{- 1}

4

7368.4 g {mol}^{- 1}

Explanation:

$π = \frac{n_{B} RT}{V} : n_{B} = \frac{w_{B}}{M_{B}}$
$π = \frac{w_{B}}{M_{B}} \times \frac{RT}{V}$
$∴ M_{B} = \frac{W_{B}}{V} \times \frac{RT}{π} = \frac{2 \times 0.0821 \times 300 \times 760}{0.3 \times 20}$
$= 6239.6 g {mol}^{- 1}$

CHXII02:SOLUTIONS

319244 $0.05 M {CuSO}_{4}$ when treated with $0.01 M K_{2} {Cr}_{2} O_{7}$ gives green colour solution of ${Cu}_{2} {Cr}_{2} O_{7}$ . The two solutions are separated as shown below:
supporting img
[SPM : Semi Permeable Membrane]
Due to osmosis

1 Green colour formation observed on side Y.

2 Green colour formation observed on side

X

.

3 Molarity of

{CuSO}_{4}

solution is lowered.

4 Molarity of

K_{2} {Cr}_{2} O_{7}

solution is lowered.

Explanation:

Only solvent molecules pass through the semipermeable membrane from a low concentrated solution to a more concentrated solution. The conc. of $K_{2} {Cr}_{2} O_{7}$ is less, so water molecules move from $K_{2} {Cr}_{2} O_{7}$ to ${CuSO}_{4}$ solution side, As a result conc. of ${CuSO}_{4}$ decreases. So, the correct option is (3).

JEE Main - 2024

CHXII02:SOLUTIONS

319245 Two solutions of ${KNO}_{3}$ and ${CH}_{3} COOH$ are prepared separately. Molarity of both is 0.1 M and osmotic pressures are $P_{1}$ and $P_{2}$ respectively. The correct relationship between the osmotic pressures is

1

P_{2} > P_{1}

2

P_{1} = P_{2}

3

P_{1} > P_{2}

4

\frac{P_{1}}{P_{1} + P_{2}} = \frac{P_{2}}{P_{1} + P_{2}}

Explanation:

Given that,
Osmotic pressure of $0.1 M K N O_{3} = P_{1}$
Osmotic pressure of $0.1 M C H_{3} C O O H = P_{2}$
As we know that osmotic pressure is directly proportional to the number of particles and acetic acid is a weak electrolyte so on dissociation in water ${CH}_{3} COOH$ will not give as much particles as ${KNO}_{3}$ .
Hence, $P_{1} > P_{2}$

CHXII02:SOLUTIONS

319246 Twenty grams of a substance were dissolved in 500ml of water and the osmotic pressure of the solution was found to be 600 mm of mercury at $15^{\circ} C$ . The molecular weight of the substance is

1 1120

2 1198

3 1200

4 None of these

Explanation:

$w = 20 g m; V = 500 m L = \frac{500}{1000} = 0 .5 l i t r e$
$π = 600 m m = \frac{600}{760} a t m;$
$T = 15 + 273 = 288 K; m w = ?$
According to Van’t Hoff equation,
$π V = n S T; π V = \frac{w}{m w} S T$
$∴ m = \frac{w S T}{π V} = \frac{20 \times 0 .0821 \times 288 \times 760}{600 \times 0 .5} = 1198$

CHXII02:SOLUTIONS

319247 Osmotic pressure of a sugar solution at $24^{\circ} C$ is 2.5 atmospheres. The concentration of the solution in mole per L is

1 0.0821 mol/L

2 1.082 mol/L

3 0.1025 mol/L

4 0.0827 mol/L

Explanation:

$π = 2 .5 a t m, T = 24 + 273 = 297 K,$
$S = 0 .082 l i t . a t m d e g^{- 1} m o l^{- 1}, C = ?$
$W e k n o w t h a t π = C S T o r$
$C = \frac{π}{S T} = \frac{2 .5}{0 .0821 \times 297} = 0 .1025 m o l e s / l i t r e$

CHXII02:SOLUTIONS

319248 Osmotic pressure of a solution containing $2 g$ dissolved protein per $300 {cm}^{3}$ of solution is 20 $mm$ of $Hg$ at $27^{\circ} C$ . The molecular mass of the protein is:

1

6239.6 g {mol}^{- 1}

2

12315.5 g {mol}^{- 1}

3

3692.1 g {mol}^{- 1}

4

7368.4 g {mol}^{- 1}

Explanation:

$π = \frac{n_{B} RT}{V} : n_{B} = \frac{w_{B}}{M_{B}}$
$π = \frac{w_{B}}{M_{B}} \times \frac{RT}{V}$
$∴ M_{B} = \frac{W_{B}}{V} \times \frac{RT}{π} = \frac{2 \times 0.0821 \times 300 \times 760}{0.3 \times 20}$
$= 6239.6 g {mol}^{- 1}$