Ideal 21st Century Competitions

Chemistry

[Time:90 Mins] SOLUTION OF

Note : 1. Solutions & Colligative Properties chapter Weightage total 5 marks. 2. All questions are compulsory. 3. Draw neat and labelled diagram wherever necessary. 4. Figure to the right indicates full marks. 5. Answer to every new question must begin on a new page.

1. Write formula of van't Hoff factor I

Sol. van't Hoff factor,i

2. Write two examples of non-ideal solution showing negative deviation.Sol. (i) Chloroform and acetone,

(ii) Water and hydrochloric acid.

3. How will you represent the concentration of component 'A' in parts per million ?

Sol. ppm of A

4. Define molarity and molality. Sol. Molarity is defined as number of gram moles of solute

Molarity (M)Molality is defined as number of moles of solute present in a kilogram (1000 gram) of solvent. It is denoted by m.Molality (m)

5. Give two examples of non-ideal solutions showing positive deviation. Sol. Example of non-ideal solution showing positive deviation are :

(i)CCl4 and CHCl3, (ii) CCl4 and C6H5CH3.

6. The boiling point of solution is higher than that of boiling point of solvent. Why Sol. The vapour pressure of solution containing non

solution has to be heated to higher temperature so that its vapour pressure become equal to the atmospheric pressure. Thus, the boiling point of solution is always higher than the boiling point of solvent.

7. Determine molecular mass of non-Sol. Suppose, WBgram of non-volatile solute dissolve in W

solute is WB gram. Then, molality, m

For any non-volatile solute (from elevation in boiling point)

From eqns. (i) and (ii), we get

www.21stideal.com

emistry - Solutions & Colligative

SOLUTION OF TEST [Max.

& Colligative Properties chapter Weightage total 5 marks.

Draw neat and labelled diagram wherever necessary. Figure to the right indicates full marks. Answer to every new question must begin on a new page.

I

ideal solution showing negative deviation.

you represent the concentration of component 'A' in parts per million ?

Molarity is defined as number of gram moles of solute dissolved in a litre of solution. It is denoted by M.

Molality is defined as number of moles of solute present in a kilogram (1000 gram) of solvent. It is denoted by m.

ideal solutions showing positive deviation. ideal solution showing positive deviation are :

The boiling point of solution is higher than that of boiling point of solvent. Why ? The vapour pressure of solution containing non-volatile solute is always less than that of pure solvent. Therefore,

be heated to higher temperature so that its vapour pressure become equal to the atmospheric pressure. Thus, the boiling point of solution is always higher than the boiling point of solvent.

-volatile solute from elevation in boiling point. volatile solute dissolve in WA gram of solvent and the molecular mass of non

m will be

(i) (from elevation in boiling point)

...(ii)

1 1

Max. Marks: 35]

[1 MARK]

[1 MARK]

[1 MARK] [1/2 MARK] [1/2 MARK]

[1 MARK] [1 MARK]

[1 MARK] dissolved in a litre of solution. It is denoted by M.

[1/2 MARK]

Molality is defined as number of moles of solute present in a kilogram (1000 gram) of solvent. It is denoted by m. [1/2 MARK]

[1 MARK]

[1/2 MARK] [1/2 MARK]

[2 MARK] volatile solute is always less than that of pure solvent. Therefore,

be heated to higher temperature so that its vapour pressure become equal to the atmospheric pressure. Thus, the boiling point of solution is always higher than the boiling point of solvent. [2 MARK]

[2 MARK] gram of solvent and the molecular mass of non-volatile

[1 MARK]

[1 MARK]

Ideal 21st Century Competitions

Or

8. Define colligative properties. Gives four examples of it.Sol. The properties of solutions which depends only on the number of solute particles but are independent of their

nature are called colligative properties. Four examples of colligative properties are :(i) Relative lowering of vapour pressure,(ii) Elevation of boiling point, (iii) Depression in freezing point, (iv) Osmotic pressure.

9. What are isotonic solutions ? Sol. Such solutions which have the same osmotic pressure are known as isotonic solutions. In isotonic solutions

concentration of the solute is the same and when they are separated by semipermeable membrane osmosis does not take place.

10. Write two differences between diffusion and osmosis.Sol. (1 mark for one point)

Difference between diffusion and osmosis :

Diffusion 1. Molecules move from a region of high

concentration to low concentration.

2. Semipermeable membrane is not required3. This process takes place in gases and in

liquids. 4. Molecules of both solute and solvent

move.

5. It cannot be stopped by applying pressure from opposite direction.

11. What are ideal and non-ideal solutions ? Explain with example.Sol. Ideal solutions : Ideal solutions are those solutions in which Raoult's law can be applied completely for all

concentrations of the solutions and at all temperatures.Condition for ideal solutions are following:(i)

(ii) Example: C2H2Br + C2H5C1, C6H6 Non-ideal solutions : Solutions in which Raoult's law cannot be applied completely for all concentrations and temperatures are called non-ideal solutions. For these solutions : (i)

(ii) Example: Benzene + acetone, CHC1

12. Explain the Raoult's law for non-volatile solute. Sol. The vapour pressure of. a solution containing

solvent. i.e., Vapour pressure = Vapour pressure of solvent of solution in solution (P)

PA=KXA, (here, K is proportionality constant) For pure solvent XA = 1 and , PA = P( PoA = vapour pressure of pure solvent) Here eqn. (ii) becomes

From eqn. (ii) and (iii) PA = P oA X A

www.21stideal.com

Define colligative properties. Gives four examples of it. which depends only on the number of solute particles but are independent of their

nature are called colligative properties. Four examples of colligative properties are :(i) Relative lowering of vapour pressure,

Such solutions which have the same osmotic pressure are known as isotonic solutions. In isotonic solutions concentration of the solute is the same and when they are separated by semipermeable membrane osmosis does

Write two differences between diffusion and osmosis.

Difference between diffusion and osmosis :

Osmosis Molecules move from a region of high concentration to low concentration.

1. Molecules of solvent move from a solution of low concentration to solution of high concentration.

2. Semipermeable membrane is not required 2. Semipermeable membrane is required.3. This process takes place in gases and in 3. This process takes place only in solutions.

4. Molecules of both solute and solvent 4. Only molecules of solvent move.

5. It cannot be stopped by applying pressure 5. It can be stopped by applying pressure from opposite direction.

ideal solutions ? Explain with example. Ideal solutions are those solutions in which Raoult's law can be applied completely for all

concentrations of the solutions and at all temperatures. Condition for ideal solutions are following:

and (iii) + C6H5CH3, CC14 + SiCl4 etc.

Solutions in which Raoult's law cannot be applied completely for all concentrations and ideal solutions.

(iii) Benzene + acetone, CHC13 + HNO3 etc.

volatile solute. The vapour pressure of. a solution containing non-volatile solute is directly proportional to the mole fraction of the

Vapour pressure = Vapour pressure of solvent Mole fraction of in solution solvent

(PA) (XA) ...(i)

, (here, K is proportionality constant) ...(ii) = PA,

vapour pressure of pure solvent) PA = K ...(iii)

. ...(iv)

2 2

[2 MARK] which depends only on the number of solute particles but are independent of their

nature are called colligative properties. Four examples of colligative properties are :

[2 MARK]

[2 MARK] Such solutions which have the same osmotic pressure are known as isotonic solutions. In isotonic solutions molar concentration of the solute is the same and when they are separated by semipermeable membrane osmosis does

[2 MARK]

[2 MARK]

1. Molecules of solvent move from a solution of low concentration to solution of high

2. Semipermeable membrane is required. 3. This process takes place only in solutions.

t move.

5. It can be stopped by applying pressure from

[3 MARK] Ideal solutions are those solutions in which Raoult's law can be applied completely for all

[1 MARK] [1/2 MARK]

Solutions in which Raoult's law cannot be applied completely for all concentrations and

[1 MARK] [1/2 MARK]

[3 MARK] volatile solute is directly proportional to the mole fraction of the

[1 MARK]

Ideal 21st Century Competitions

If mole fraction of solute is xB, thenxA+xB = l ...(v) or xA = 1 xB ...(vi) Substituting value of xA from eqn. (vi) in eqn. (iv),PA= P oA XB) or PA = P oA P oA XB or P oA XB = P oA PA

or ...(vii)

Where is lowering of vapour pressure while islowering of vapour pressure. Thus, based on eqn. (vii) Raoult's law can also be defined as Relative vapour pressure of a solution containing a nonIt is clear from eqn. (vii) that relative lowering of vapour pressure depends on mole fraction and not on nature of solute. Therefore, it is a colligative property.

13. What is meant by depression in freezing point ? Solution is prepared by dissolving 1 gm NaCl in 100 gm water. If molal depression constant for water is 1.85 K kg molDepression in freezing point for NaCl solution is 0.604 K. (MP 2007 Set B

Sol. Freezing point of a substance is the temperature at which its solid and liquid phases have the same vapour pressure. If non-volatile solute is dissolved in pure liquid to constitute a solution its freezing point decreases, this decrease in freezing point is called depression of freezing point and it is denoted by Solution: Observed molecular mass of NaCl can be

Thus, observed molecular mass =30.6 and Normal molecular mass of soLet a be the extent of dissociation. (Thus, out of 1 mole NaCl mole is dissociated.

NaCl1

+

Na

Cl

In the solution No. of particles after dissociation =1

. =0.91. Thus, percentage dissociation =91.

14. What is molal elevation boiling point constant ? On dissolving phenol in benzene, two of its molecule associate to form a bigger molecules. When Z gm phenol is dissolved in 100 gm benzene, then its freezing point decreases by 0'69C Determine the extent of association of phenol. (K

Sol. Molal boiling elevation constant : It is defined as the elevation in boiling point when 1 gm of nondissolved in 1000 gm of the solvent.We know that, Tb m or Tb = KWhere, Kb is a molal elevation boiling point constant.Elevation in boiling point is directly proportional to molality of the solution.

www.21stideal.com

, then

from eqn. (vi) in eqn. (iv),

is lowering of vapour pressure while is relative lowering of vapour pressure. Thus, based on eqn. (vii) Raoult's law can also be defined as Relative vapour pressure of a solution containing a nonvolatile solute is directly proportional to mole fraction of solute.It is clear from eqn. (vii) that relative lowering of vapour pressure depends on mole fraction and not on nature of

efore, it is a colligative property.

What is meant by depression in freezing point ? Solution is prepared by dissolving 1 gm NaCl in 100 gm water. If molal depression constant for water is 1.85 K kg mol1 then determine the extent of dissociation of NaCl. Depression in freezing point for NaCl solution is 0.604 K. (MP 2007 Set B2) Freezing point of a substance is the temperature at which its solid and liquid phases have the same vapour

volatile solute is dissolved in pure liquid to constitute a solution its freezing point decreases, this decrease in freezing point is called depression of freezing point and it is denoted by

: Observed molecular mass of NaCl can be calculated by the following formula :

30.6

Thus, observed molecular mass =30.6 and Normal molecular mass of sodium chloride = 58.5.

mole is dissociated.

No. of particles after dissociation =1 + + = 1 +

Thus, percentage dissociation =91.

elevation boiling point constant ? On dissolving phenol in benzene, two of its molecule associate to form a bigger molecules. When Z gm phenol is dissolved in 100 gm benzene, then its freezing point decreases by

Determine the extent of association of phenol. (Kf= 5.12 k kg mol1). (MP 2007 Set A): It is defined as the elevation in boiling point when 1 gm of non

the solvent. = Kb m

is a molal elevation boiling point constant. Elevation in boiling point is directly proportional to molality of the solution.

14.84

3 3

[1 MARK]

lowering of vapour pressure. Thus, based on eqn. (vii) Raoult's law can also be defined as Relative lowering of volatile solute is directly proportional to mole fraction of solute.

It is clear from eqn. (vii) that relative lowering of vapour pressure depends on mole fraction and not on nature of [1 MARK]

What is meant by depression in freezing point ? Solution is prepared by dissolving 1 gm NaCl in 100 gm water. If extent of dissociation of NaCl.

[3 MARK] Freezing point of a substance is the temperature at which its solid and liquid phases have the same vapour

volatile solute is dissolved in pure liquid to constitute a solution its freezing point decreases, this decrease in freezing point is called depression of freezing point and it is denoted by Tf.

ing formula :

[1 MARK]

dium chloride = 58.5.

[1 MARK]

[1 MARK]

elevation boiling point constant ? On dissolving phenol in benzene, two of its molecule associate to form a bigger molecules. When Z gm phenol is dissolved in 100 gm benzene, then its freezing point decreases by

). (MP 2007 Set A) [3 MARK] : It is defined as the elevation in boiling point when 1 gm of non-volatile solute is

[1 MARK]

Ideal 21st Century Competitions

1 x (After association) No. of particles after associationNormal molecular mass of phenol = 6l2 + l5 + 16+l = 94

Or 5.33

Association of phenol = 5.33%.

15. What is molal freezing point depression constant on dissolving 0.4 freezing was found to be 0.124C. Calculate the molecular mass of urea. (K

Sol.: Molal freezing point depression constant of a solution is equal to thdissolving 1 gram mol of solute in 1000 gram of solvent.Molecular mass of solute by depression in freezing point :

MB Kf for water = 1.86 Mass of solute (WB) = 0.4 gm Mass of solvent (WA) = 100 gm Depression in freezing point (Tp) = 0.124MB

MB = 60 gmol1.

16. Explain in brief Berkley and Hartley's method of osmotic pressure measurement and state its uses.Sol. (1 mark for diagram)

Brkley and Hartley's method: In this method, pressure is applied over the solution to stop the flow of solvent. This pressure is equivalent to osmotic pressure. In this method, the apparatus consists of a strong veup of steel in which porous pot is fitted. In the porous pot, copper Ferro cyanide semi permeable membrane is deposited.

The porous pot is fitted with a capillary tube on one side and a water reservoir on the other spressure gauge are fitted to the steel vessel.

www.21stideal.com

Normal molecular mass of phenol = 6l2 + l5 + 16+l = 94

5.33

What is molal freezing point depression constant on dissolving 0.4 gm of urea in 100 gm water the depression in

Calculate the molecular mass of urea. (Kf for water = 1.86). (MP 2007 SetC2) Molal freezing point depression constant of a solution is equal to the depression in freezing point produced by dissolving 1 gram mol of solute in 1000 gram of solvent. Molecular mass of solute by depression in freezing point :

) = 0.124C

Explain in brief Berkley and Hartley's method of osmotic pressure measurement and state its uses.

In this method, pressure is applied over the solution to stop the flow of solvent. This pressure is equivalent to osmotic pressure. In this method, the apparatus consists of a strong veup of steel in which porous pot is fitted. In the porous pot, copper Ferro cyanide semi permeable membrane is

The porous pot is fitted with a capillary tube on one side and a water reservoir on the other spressure gauge are fitted to the steel vessel.

4 4

[1 MARK]

[1 MARK]

gm of urea in 100 gm water the depression in

[3 MARK] e depression in freezing point produced by

[1 MARK]

[1 MARK]

[1 MARK]

Explain in brief Berkley and Hartley's method of osmotic pressure measurement and state its uses. [5 MARK]

In this method, pressure is applied over the solution to stop the flow of solvent. This pressure is equivalent to osmotic pressure. In this method, the apparatus consists of a strong vessel made up of steel in which porous pot is fitted. In the porous pot, copper Ferro cyanide semi permeable membrane is

[1 MARK] The porous pot is fitted with a capillary tube on one side and a water reservoir on the other side. A piston and

[1 MARK]

Ideal 21st Century Competitions www.21stideal.com

5 5

The porous pot and steel vessel are filled with water and solution respectively. Osmosis takes place and water moves into the steel vessel from the porous pot through the semi permeable membrane this is shown by fall in water level in the capillary tube. This flow of water is stopped by applying external pressure on the solution with the help of piston. [1 MARK]

This method has the following advantages: (i) It takes comparatively lesser time to determine osmotic pressure. (ii) Concentration of solution does not change, hence better results are obtained. (iii) As high pressure is not exerted over semi permeable membrane, it does not break. (iv) High osmotic pressure can be measured. [1 MARK] OR

17. What is meant by depression in freezing point? Solution is prepared by dissolving 1 gm NaCl in 100 gm water. If molal depression constant for water is 1.85 K kg mol1 then determine the extent of dissociation of NaCl. Depression in freezing point for NaCl solution is 0.604 K. [5 MARK]

Sol. Freezing point of a substance is the temperature at which its solid and liquid phases have the same vapour pressure. If non-volatile solute is dissolved in pure liquid to constitute a solution its freezing point decreases, this decrease in freezing point is called depression of freezing point and it is denoted by Tf. [1 MARK]

Solution: Observed molecular mass of NaCl can be calculated by the following formula:

f bf

b a

K W 1000TM W =

Mb = f bf a

K W 1000 1.85 1 1000T W 0.604 100

=

= 30.6 [1 MARK] Thus, observed molecular mass = 30.6 and Normal molecular mass of sodium chloride = 58.5. Let a be the extent of dissociation. (Thus, out of 1 mole NaCl mole is dissociated.

NaCl1

+

Na

Cl

[1 MARK] In the solution

No. of particles after dissociation =1 + + = 1 +

Normal molecular massObserved molecular mass

=

Number of particles of solute after dissociationNormal number of particles of solute

=

[1 MARK]

Normal molecular mass 1Observed molecular mass 1

+ =

58.530.6

= 1 +

= 27.930.6

= 0.91. Thus, percentage dissociation = 91. [1 MARK]