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CHXII02:SOLUTIONS

319266 Which is the correct relation between osmotic pressure of 0.1 M NaCl solution and 0.1 M ${Na}_{2} {SO}_{4}$ solution?

1 The osmotic pressure of

{Na}_{2} {SO}_{4}

less than NaCl solution

2 The osmotic pressure of

{Na}_{2} {SO}_{4}

more than NaCl solution

3 Both have same osmotic pressure

4 None of the above

Explanation:

$π =$ iCRT
i for NaCl is 2 and for ${Na}_{2} {SO}_{4}$ is 3 .
So, $π_{{Na}_{2} {SO}_{4}} > π_{NaCl}$

CHXII02:SOLUTIONS

319267 Which of the following is not an industrial or biological importance of osmosis?

1 Movement of water from soil into plant roots and upper portion of plant

2 Salting of meat to prevent bacterial action

3 Reverse osmosis for desalination of sea water

4 Filling of ink in a fountain pen

CHXII02:SOLUTIONS

319268 A 6% solution of urea is isotonic with

1 0.05 M solution of glucose

2 6% solution of glucose

3 25% solution of glucose

4 1 M solution of glucose.

Explanation:

When a solution of urea (6%) is isotonic with a solution of glucose,
$π_{g l u c o s e} = π_{u r e a} \Rightarrow C_{g l u c o s e} = C_{u r e a}$
$∴ (\frac{w_{g} \times 100}{m_{g} \times 100}) = (\frac{w_{u} \times 1000}{m_{u} \times 100})$
$w h e r e, w_{g} = m a s s o f g l u c o s e = x g$
$m_{g} = m o l e c u l a r m a s s o f g l u c o s e = 180$
$w_{u} = m a s s o f u r e a = 6 g$
$m_{u} = m o l e c u l a r m a s s o f u r e a = 60$
$∴ \frac{x \times 100}{180 \times 100} = \frac{6 \times 1000}{60 \times 100} \Rightarrow x = 18 g$
Thus, 18 g of glucose are present in 100 mL of solution.
In other words, 1 M solution of glucose (18 g in 100 mL) is isotonic with 6% solution of urea.

KCET - 2009

CHXII02:SOLUTIONS

319269 What is the osmotic pressure of the solution obtained by mixing $300 c m^{3}$ of 2% (mass-volume) solution of urea with $300 c m^{3}$ of 3.42% solution of sucrose at $2 0^{^{\circ}} C$ ?
$(R = 0 .082 L a t m K^{- 1} m o l^{- 1})$

1 5 atm

2 5.2 atm

3 2.6 atm

4 4.5 atm

Explanation:

$100 c m^{3}$ of urea solution contains 2g of urea
$300 c m^{3}$ of urea solution contains 6g of urea
$100 c m^{3}$ of sucrose solution contains 3.42 g of sucrose
$300 c m^{3}$ of sucrose solution contains 10.26 g
$π v = [{(\frac{w}{m w})}_{u r e a} + {(\frac{w}{m w})}_{s u c r o s e}] R T$
$π \times \frac{600}{1000} = [\frac{6}{60} + \frac{10 .26}{342}] \times 0 .0821 \times 293$
$π = 5 .2 a t m$

CHXII02:SOLUTIONS

319266 Which is the correct relation between osmotic pressure of 0.1 M NaCl solution and 0.1 M ${Na}_{2} {SO}_{4}$ solution?

1 The osmotic pressure of

{Na}_{2} {SO}_{4}

less than NaCl solution

2 The osmotic pressure of

{Na}_{2} {SO}_{4}

more than NaCl solution

3 Both have same osmotic pressure

4 None of the above

Explanation:

$π =$ iCRT
i for NaCl is 2 and for ${Na}_{2} {SO}_{4}$ is 3 .
So, $π_{{Na}_{2} {SO}_{4}} > π_{NaCl}$

CHXII02:SOLUTIONS

319267 Which of the following is not an industrial or biological importance of osmosis?

1 Movement of water from soil into plant roots and upper portion of plant

2 Salting of meat to prevent bacterial action

3 Reverse osmosis for desalination of sea water

4 Filling of ink in a fountain pen

CHXII02:SOLUTIONS

319268 A 6% solution of urea is isotonic with

1 0.05 M solution of glucose

2 6% solution of glucose

3 25% solution of glucose

4 1 M solution of glucose.

Explanation:

When a solution of urea (6%) is isotonic with a solution of glucose,
$π_{g l u c o s e} = π_{u r e a} \Rightarrow C_{g l u c o s e} = C_{u r e a}$
$∴ (\frac{w_{g} \times 100}{m_{g} \times 100}) = (\frac{w_{u} \times 1000}{m_{u} \times 100})$
$w h e r e, w_{g} = m a s s o f g l u c o s e = x g$
$m_{g} = m o l e c u l a r m a s s o f g l u c o s e = 180$
$w_{u} = m a s s o f u r e a = 6 g$
$m_{u} = m o l e c u l a r m a s s o f u r e a = 60$
$∴ \frac{x \times 100}{180 \times 100} = \frac{6 \times 1000}{60 \times 100} \Rightarrow x = 18 g$
Thus, 18 g of glucose are present in 100 mL of solution.
In other words, 1 M solution of glucose (18 g in 100 mL) is isotonic with 6% solution of urea.

KCET - 2009

CHXII02:SOLUTIONS

319269 What is the osmotic pressure of the solution obtained by mixing $300 c m^{3}$ of 2% (mass-volume) solution of urea with $300 c m^{3}$ of 3.42% solution of sucrose at $2 0^{^{\circ}} C$ ?
$(R = 0 .082 L a t m K^{- 1} m o l^{- 1})$

1 5 atm

2 5.2 atm

3 2.6 atm

4 4.5 atm

Explanation:

$100 c m^{3}$ of urea solution contains 2g of urea
$300 c m^{3}$ of urea solution contains 6g of urea
$100 c m^{3}$ of sucrose solution contains 3.42 g of sucrose
$300 c m^{3}$ of sucrose solution contains 10.26 g
$π v = [{(\frac{w}{m w})}_{u r e a} + {(\frac{w}{m w})}_{s u c r o s e}] R T$
$π \times \frac{600}{1000} = [\frac{6}{60} + \frac{10 .26}{342}] \times 0 .0821 \times 293$
$π = 5 .2 a t m$

CHXII02:SOLUTIONS

319266 Which is the correct relation between osmotic pressure of 0.1 M NaCl solution and 0.1 M ${Na}_{2} {SO}_{4}$ solution?

1 The osmotic pressure of

{Na}_{2} {SO}_{4}

less than NaCl solution

2 The osmotic pressure of

{Na}_{2} {SO}_{4}

more than NaCl solution

3 Both have same osmotic pressure

4 None of the above

Explanation:

$π =$ iCRT
i for NaCl is 2 and for ${Na}_{2} {SO}_{4}$ is 3 .
So, $π_{{Na}_{2} {SO}_{4}} > π_{NaCl}$

CHXII02:SOLUTIONS

319267 Which of the following is not an industrial or biological importance of osmosis?

1 Movement of water from soil into plant roots and upper portion of plant

2 Salting of meat to prevent bacterial action

3 Reverse osmosis for desalination of sea water

4 Filling of ink in a fountain pen

CHXII02:SOLUTIONS

319268 A 6% solution of urea is isotonic with

1 0.05 M solution of glucose

2 6% solution of glucose

3 25% solution of glucose

4 1 M solution of glucose.

Explanation:

When a solution of urea (6%) is isotonic with a solution of glucose,
$π_{g l u c o s e} = π_{u r e a} \Rightarrow C_{g l u c o s e} = C_{u r e a}$
$∴ (\frac{w_{g} \times 100}{m_{g} \times 100}) = (\frac{w_{u} \times 1000}{m_{u} \times 100})$
$w h e r e, w_{g} = m a s s o f g l u c o s e = x g$
$m_{g} = m o l e c u l a r m a s s o f g l u c o s e = 180$
$w_{u} = m a s s o f u r e a = 6 g$
$m_{u} = m o l e c u l a r m a s s o f u r e a = 60$
$∴ \frac{x \times 100}{180 \times 100} = \frac{6 \times 1000}{60 \times 100} \Rightarrow x = 18 g$
Thus, 18 g of glucose are present in 100 mL of solution.
In other words, 1 M solution of glucose (18 g in 100 mL) is isotonic with 6% solution of urea.

KCET - 2009

CHXII02:SOLUTIONS

319269 What is the osmotic pressure of the solution obtained by mixing $300 c m^{3}$ of 2% (mass-volume) solution of urea with $300 c m^{3}$ of 3.42% solution of sucrose at $2 0^{^{\circ}} C$ ?
$(R = 0 .082 L a t m K^{- 1} m o l^{- 1})$

1 5 atm

2 5.2 atm

3 2.6 atm

4 4.5 atm

Explanation:

$100 c m^{3}$ of urea solution contains 2g of urea
$300 c m^{3}$ of urea solution contains 6g of urea
$100 c m^{3}$ of sucrose solution contains 3.42 g of sucrose
$300 c m^{3}$ of sucrose solution contains 10.26 g
$π v = [{(\frac{w}{m w})}_{u r e a} + {(\frac{w}{m w})}_{s u c r o s e}] R T$
$π \times \frac{600}{1000} = [\frac{6}{60} + \frac{10 .26}{342}] \times 0 .0821 \times 293$
$π = 5 .2 a t m$

CHXII02:SOLUTIONS

319266 Which is the correct relation between osmotic pressure of 0.1 M NaCl solution and 0.1 M ${Na}_{2} {SO}_{4}$ solution?

1 The osmotic pressure of

{Na}_{2} {SO}_{4}

less than NaCl solution

2 The osmotic pressure of

{Na}_{2} {SO}_{4}

more than NaCl solution

3 Both have same osmotic pressure

4 None of the above

Explanation:

$π =$ iCRT
i for NaCl is 2 and for ${Na}_{2} {SO}_{4}$ is 3 .
So, $π_{{Na}_{2} {SO}_{4}} > π_{NaCl}$

CHXII02:SOLUTIONS

319267 Which of the following is not an industrial or biological importance of osmosis?

1 Movement of water from soil into plant roots and upper portion of plant

2 Salting of meat to prevent bacterial action

3 Reverse osmosis for desalination of sea water

4 Filling of ink in a fountain pen

CHXII02:SOLUTIONS

319268 A 6% solution of urea is isotonic with

1 0.05 M solution of glucose

2 6% solution of glucose

3 25% solution of glucose

4 1 M solution of glucose.

Explanation:

When a solution of urea (6%) is isotonic with a solution of glucose,
$π_{g l u c o s e} = π_{u r e a} \Rightarrow C_{g l u c o s e} = C_{u r e a}$
$∴ (\frac{w_{g} \times 100}{m_{g} \times 100}) = (\frac{w_{u} \times 1000}{m_{u} \times 100})$
$w h e r e, w_{g} = m a s s o f g l u c o s e = x g$
$m_{g} = m o l e c u l a r m a s s o f g l u c o s e = 180$
$w_{u} = m a s s o f u r e a = 6 g$
$m_{u} = m o l e c u l a r m a s s o f u r e a = 60$
$∴ \frac{x \times 100}{180 \times 100} = \frac{6 \times 1000}{60 \times 100} \Rightarrow x = 18 g$
Thus, 18 g of glucose are present in 100 mL of solution.
In other words, 1 M solution of glucose (18 g in 100 mL) is isotonic with 6% solution of urea.

KCET - 2009

CHXII02:SOLUTIONS

319269 What is the osmotic pressure of the solution obtained by mixing $300 c m^{3}$ of 2% (mass-volume) solution of urea with $300 c m^{3}$ of 3.42% solution of sucrose at $2 0^{^{\circ}} C$ ?
$(R = 0 .082 L a t m K^{- 1} m o l^{- 1})$

1 5 atm

2 5.2 atm

3 2.6 atm

4 4.5 atm

Explanation:

$100 c m^{3}$ of urea solution contains 2g of urea
$300 c m^{3}$ of urea solution contains 6g of urea
$100 c m^{3}$ of sucrose solution contains 3.42 g of sucrose
$300 c m^{3}$ of sucrose solution contains 10.26 g
$π v = [{(\frac{w}{m w})}_{u r e a} + {(\frac{w}{m w})}_{s u c r o s e}] R T$
$π \times \frac{600}{1000} = [\frac{6}{60} + \frac{10 .26}{342}] \times 0 .0821 \times 293$
$π = 5 .2 a t m$