Chapter 2 Solutions by TEACHING CARE Online coaching and tuition classes

Chapter 2 Solutions by TEACHING CARE Online coaching and tuition classes

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A solution is a homogenous mixture of two or more substances, the composition of which may vary within limits. “A solution is a special kind of mixture in which substances are intermixed so intimately that they can not be observed as separate components”. The dispersed phase or the substance which is to be dissolved is called solute, while the dispersion medium in which the solute is dispersed to get a homogenous mixture is called the solvent. A solution is termed as binary, ternary and quartenary if it consists of two, three and four components respectively.

Solubility of as substance in the solvent is the measure of the capacity of the latter to dissolve the former at a given temperature and pressure. “Solubility of a substance may be defined as the amount of solute dissolved in 100gms of a solvent to form a saturated solution at a given temperature”. A saturated solution is a solution which contains at a given temperature as much solute as it can hold in presence of dissolveding solvent. Any solution may contain less solute than would be necessary to to saturate it. Such a solution is known as unsaturated solution. When the solution contains more solute than would be necessary to saturate it then it is termed as supersaturated solution. It is metastable.

All the three states of matter (gas , liquid or solid) may behave either as solvent or solute. Depending on the state of solute or solvent, mainly there may be following nine types of binary solutions.

SolventSoluteExample
GasGasMixture of gases, air.
GasLiquidWater vapours in air, mist.
GasSolidSublimation of a solid into a gas, smoke.
LiquidGasCO2 gas dissolved in water (aerated drinks).
LiquidLiquidMixture of miscible liquids, e.g., alcohol in water.
LiquidSolidSalt in water, sugar in water.
SolidGasAdsorption of gases over metals; hydrogen over palladium.
SolidLiquidMercury in zinc, mercury in gold, CuSO4.5H2O.
SolidSolidHomogeneous mixture of two or more metals (alloys), e.g., copper in gold, zinc in copper.

Among these solutions the most significant type of solutions are those which are in liquid phase and may be categorised as : (i) Solid in liquid solutions, (ii) Liquid in liquid solutions and (iii) Gas in liquid solutions.

  • Solid in liquid solutions : For the solid in liquid solutions the solid is referred to as The amount

 

of solute that is dissolved in

100g

of a solvent, to form a saturated solution at a particular temperature is called

 

solubility. The solubility of a solid solute in liquid depends upon

  • Nature of solute and solvent
  • Temperature : Usually the solubility of the solute increases with increase in temperature (e.g.,

KI, KNO3 , NH 4 Br ) but in some cases increase in solubility is negligible (e.g. NaCl ) and in cases of some salts

 

(e.g.,

NaOH, Na2SO4 , CeSO4 ) solubility decreases with increase in temperature.

 

 

 

Cause of miscibility of solids in liquids : The basic cause of solubility of solid solute in liquid solvent can be summed up in one line, i.e., “Similia similibus solvanter” meaning like dissolves like which implies that polar solvents dissolve polar solutes and non-polar solvents dissolve non-polar solutes.

For ionic solutes, the solubility, in general, is related to the magnitude of hydration (or salvation) energy and lattice energy. In general if the magnitude of hydration energy is greater than lattice energy, the solute is soluble otherwise it is insoluble.

  • Liquid in liquid solutions : When two liquids are mixed, the mixture may be of the following types :
  • The two components may be almost immiscible : In this case, one of the liquids is polar while the other is of non-polar For example, benzene and water.
  • The miscibility of the component may be partial : If the intermolecular attraction of one liquid is different from intramolecular attraction of the other, there may be a partial miscibility of the two liquids. For example, ether and
  • The two components may be completely miscible : In this case, the two liquids are of the same nature, e., they are either polar (like alcohol and water) or non-polar (like benzene and hexane).

Cause of Miscibility of Liquids

  • Chemically alike liquids dissolve in one another more freely as compared to others, for example, alkanes are miscible in all proportions with one another. Alkanes are however, not miscible with water because they cannot form H-bonds with water
  • DipoleDipole interactions also play an important role in forming liquid
  • Molecular sizes of liquids which are mutually soluble, are also approximately
  • Gas in Liquid solutions : Most of the gases are soluble in water to some The solubility of gas in water generally depends upon the following factors
  • Nature of the gas : In general the gases which are easily liquefiable are more soluble in
  • Temperature : The dissolution of gas in water is exothermic process. Hence, the solubility of gas decreases with rise in
  • Pressure : Effect of pressure on the solubility of gas is explained by “Henry’s law”. According to Henry’s law “m

Solubility µ Pressure; S = KP where K is Henry’s constant.

Higher the value of K at given pressure, the lower is the solubility of the gas in the liquid. Value of K increases with increase in temperature while solubility of gas decreases. It is due to this reason that aquatic species are more comfortable in cold water rather than hot water.

Applications of Henry’s law : Henry’s law finds several applications in industry and explains some biological phenomena. Notable among these are :

  • To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.
  • To minimise the painful effects accompanying the decompression of deep sea divers, oxygen diluted with less soluble helium gas is used as breathing
  • In lungs, where oxygen is present in air with high partial pressure of oxygen to form In tissues where partial pressure of oxygen is low. Oxohaemoglobin releases oxygen for utilization in cellular activities.

 

 

 

 

Concentration of solution is the amount of solute dissolved in a known amount of the solvent or solution. The concentration of solution can be expressed in various ways as discussed below.

  • Percentage : It refers to the amount of the solute per 100 parts of the It can also be called as parts per hundred (pph). It can be expressed by any of following four methods :

 

(i)  Weight to weight percent (% w/w)

= Wt. of solute Wt. of solution

´ 100

 

e.g., 10%

Na2 CO3

solution w/w means

10g

of  Na2 CO3

is dissolved in

100g

of the solution. (It means

 

10g

Na2 CO3

is dissolved in 90g

of H 2 O )

 

(ii)   Weight to volume percent (% w/v) =

Wt. of solute Volume of solution

´ 100

 

e.g., 10%

Na2 CO3

(w/v) means 10g

Na2 CO3

is dissolved in 100 cc

of solution.

 

(iii) Volume to volume percent (% v/v)

= Vol. of solute Vol. of solution

´ 100

 

e.g., 10% ethanol (v/v) means 10 cc

of ethanol dissolved in 100 cc

of solution.

 

 

(iv)  Volume to weight percent (% v/w)

= Vol. of solute ´ 100 Wt. of solution

 

e.g., 10% ethanol (v/w) means 10 cc

of ethanol dissolved in 100g

of solution.

 

  • Parts per million (ppm) and parts per billion (ppb) : When a solute is present in trace quantities, it is convenient to express the concentration in parts per million and parts per billion. It is the number of parts of solute per million (106 ) or per billion (109 ) parts of the It is independent of the temperature.

 

ppm mass of solute component ´ 106 ;

Total mass of solution

ppb mass of solute component ´ 109

Total mass of solution

 

 

Example:1          15 g of methyl alcohol is dissolved in 35 g of water. The weight percentage of methyl alcohol in solution is (a) 30%                                       (b) 50%                                   (c) 70%                           (d) 75%

æ                                           Weight of solute            ö

 

Solution : (a)

ç Weight percentage =                                         ´ 100÷

Weight of solution

 

è                                                                                        ø

Total weight of solution = (15 + 35) g = 50 g

Weight percentage of methyl alcohol = Weight of methyl alcohol ´ 100 = 15 ´ 100 = 30%

 

Weight of solution                   50

 

Example: 2        Sea water contains

million is

5.8 ´ 10 3 g

of dissolved oxygen per kilogram. The concentration of oxygen in parts per

 

(a)

5.8 ppm

(b)

58.5 ppm

(c)

0.58 ppm

0.05 ppm

 

Solution: (a)     Part per million = Mass of solute ´ 106 = 5.8 ´ 10-3 g ´ 106 =

 

Mass of solution

103 g

5.8 ppm

 

Example: 3         A

500gm

toothpaste sample has 0.2g

fluoride concentration. The concentration of fluoride ions in terms of

 

ppm level is                                                                                                                          [AIIMS 1994]

 

 

 

(a) 250 ppm                           (b) 200 ppm                            (c) 400 ppm                        (d) 1000 ppm

 

 

Solution: (c)      ppm of

F ions =

Mass of solute Mass of solution

´ 106

= 0.2 ´ 106

500

= 400 ppm

 

  • Strength : The strength of solution is defined as the amount of solute in grams present in one litre (or

dm3 ) of the solution. It is expressed in g/litre or (g / dm3 ).

 

 

Strength =

Mass of solute in grams Volume of solution in litres

 

  • Normality (N) : It is defined as the number of gram equivalents (equivalent weight in grams) of a solute present per litre of the Unit of normality is gram equivalents litre–1. Normality changes with temperature since it involves volume. When a solution is diluted x times, its normality also decreases by x times. Solutions in term of normality generally expressed as,

 

N = Normal solution; 5 N =

Penta normal, 10 N =

Deca normal;

N / 2 =

semi normal

 

N / 10 = Deci normal;

N / 5 =

Penti normal

 

N / 100

or 0.01 N =

centinormal,

N / 1000

or 0.001= millinormal

 

Mathematically normality can be calculated by following formulas,

 

 

(i)

 

(ii)

Normality (N) = Number of g.eq. of solute =

Volume of solution (l)

N Wt. of solute per litre of solution ,

g eq. wt. of solute

Weight of solute in g.

 

  1. eq. weight of solute ´ Volume of solution (l)

 

 

(iii)

N     Wt. of solute    ´           1000         

 

g.eq. wt. of solute    Vol. of solution in ml

 

 

(iv)

 

 

N Percent of solute ´ 10 ,

g eq. wt. of solute

N = Strength in g l 1 of solution g eq. wt. of solute

 

 

 

  • If volume V1

and normality

N1 is so changed that new normality and volume

N1 V1 = N 2 V2

(Normality equation)

N 2 and V2

then,

 

  • When two solutions of the same solute are mixed then normality of mixture (N) is

 

N =N1 V1 + N 2 V2
V1 + V2

 

 

  • of water to be added i.e., (V2V1 ) to get a solution of normality

æ N1 – N 2 ö

 

N 2 from V1

ml of normality N1

 

V2 – V1 = ç

è     N 2

÷ V1

ø

 

 

 

  • If Wg of an acid is completely neutralised by V ml of base of normality N

    Wt. of acid    = VN ; Similarly,    Wt. of base     = Vol. of acid ´ N of acid

 

g eq. wt. of acid

1000

g eq. wt. of base

1000

 

  • When Va ml

of acid of normality

Na  is mixed with Vb ml

of base of normality Nb

 

  • If Va Na= Vb Nb (Solution neutral)

 

  • If Va Na > Vb Nb
  • If Vb Nb  > Va Na

(Solution is acidic) (Solution is basic)

 

  • Normality of the acidic mixture = Va NaVb Nb

(Va + Vb )

  • Normality of the basic mixture = Vb Nb +Va Na

(Va + Vb )

 

 

N = No. of meq * of solute

 

(* 1 equivalent = 1000 milliequivalent or meq.)

 

Vol. of solution in ml

 

 

Example: 4        Normality of a solution containing 9.8 g of

HSO4

in 250 cm3 of the solution is

[MP PMT 1995, 2003; CMC Vellore 1991; JIPMER 1991]

 

  • 8 N

 

Solution: (a)     Eq. wt. of H  SO

  • 1N

Mol. mass of H 2SO4

= 98 = 49

(c) 0.08 N

  • 8 N

 

2     4      Basicity of H  SO               2

2       4

 

\ Number of g equivalent of H SO

= Weight in g = 9.8 = 0.2

 

 

2     4       Eq. mass         49

 

250 cm3

of solution contain H 2 SO4 = 0.2g

equivalent

 

\ 1000 cm3

of the solution contain H

2SO4

= 0.2 ´ 1000g 250

equivalent = 0.8g

equivalent

 

Hence normality of the solution = 0.8 N

Example: 5         Amount of NaOH present in 200 ml of 0.5 N

solution is                                                        [AIIMS 1992]

 

(a) 40 g                                               (b) 4 g                                (c) 0.4 g                                       (d) 4.4 g

Solution: (b)     Wt. of solute = N ´ V ´ g eq. wt. = 0.5 ´ 200 ´ 40 = 4 g

1000                   1000

 

Example: 6

50 ml

of 10N H 2SO4 , 25ml

of 12N HCl

and 40 ml

of 5 N HNO3

were mixed together and the volume of

 

the mixture was made 1000ml

by adding water. The normality of the resulting solution will be

[MP PMT 1998, 2002, Kerala CET 2003]

 

  • 1N
  • 2N
  • 3 N
  • 4 N

 

Solution: (a)

N1 V1 + N 2 V2 + N 3 V3 = N 4 V4

50 ´ 10 + 25 ´ 12 + 40 ´ 5 = N 4 ´ 1000 or

N 4 = 1N

 

Example: 7

100ml

of 0.3N HCl

is mixed with 200ml

of 0.6 N H 2SO4 . The final normality of the resulting solution will

 

be                                                                                                                                       [DPMT 1994]

 

  • 1 N
  • 2 N
  • 3N
  • 5 N

 

Soltuion: (d)

100ml

of 0.3N HCl

contains HCl = 0.03g eq.,

 

200ml

of 0.6 N H

2SO4

contains H

2SO4

= 0.6

1000

´ 200 = 0.12g

eq.

 

Total g eq. = 0.15 , Total volume = 300ml

Finally normality = 0.15 ´ 1000 = 0.5

300

 

Alternatively

N1 V1 + N 2 V2 = N 3 V3 i.e., 0.3 ´ 100 + 0.6 ´ 200 = N 3 ´ 300 or 0.3 + 1.2 = 3N 3

 

or N 3 = 1.5 / 3 = 0.5

 

Example: 8         An aqueous solution of

6.3g

oxalic acid dihydrate is made up to

250ml . The volume of

  • N NaOH

 

required to completely neutralize 10ml of this solution is                                                     [IIT 2001; CPMT 1986]

 

  • 40 ml
  • 20 ml
  • 10 ml
  • 4 ml

 

Solution: (a)     Normality of oxalic acid solution = 6.3 ´ 1000 = 0.4

 

63     250

N1 V1 = N 2 V2

0.1 ´ V1 = 0.4 ´ 10 or V1 = 40 ml

 

Example: 9

10.6 g of

Na2 CO3

was exactly neutralised by 100 ml

of HSO4

solution. Its normality is

 

(a) 1 N                                       (b) 2 N                                     (c) 1.5 N                              (d) 0.5 N

Solution: (b)     Weight of base (w) = 10.6g ; g eq. wt. of base = 53; Vol. of acid (V) = 100 ml ; Normality of acid (N) = ?

 

       w        V ´ N ;

 

10.6100 ´ N ;

N 1000 ´ 10.6 = 2

 

g eq. wt.

1000       53

1000

100 ´ 53

 

  • Molarity (M) : Molarity of a solution is the number of moles of the solute per litre of solution (or number of millimoles per of solution). Unit of molarity is mol/litre or mol/dm3 For example, a molar (1M) solution of

 

sugar means a solution containing 1 mole of sugar (i.e., 342 g or solution. Solutions in term of molarity generally expressed as,

6.02 ´ 1023

molecules of it) per litre of the

 

1M = Molar solution,

2M = Molarity is two,

M or 0.5 M = Semimolar solution,

2

 

M or 0.1 M = Decimolar solution,

10

   or 0.01 M = Centimolar solution

100

 

M      or 0.001 M = Millimolar solution

1000

  • Molarity is most common way of representing the concentration of solution.

 

  • Molarity is depend on temperature as,

Molarity µ

1

 

Temperature

 

  • When a solution is diluted (x times), its molarity also decreases (by x times) Mathematically molarity can be calculated by following formulas,

 

(i)

M No. of moles of solute (n) ,

Vol. of solution in litres

 

 

 

 

(ii)

M = Wt. of solute (in gm) per litre of solution

Mol. wt. of solute

 

(iii)

M = Wt. of solute (in gm) ´

Mol. wt. of solute

1000

 

Vol. of solution in ml.

 

(iv)

 

(v)

 

(vi)

 

(vii)

M = No. of millimoles of solute Vol. of solution in ml

M Percent of solute ´ 10 Mol. wt. of solute

M = Strength in gl 1 of solution

Mol. wt. of solute

M 10 ´ Sp. gr. of the solution ´ Wt. % of the solute

Mol. wt. of the solute

 

  • If molarity and volume of solution are changed from

M1 , V1 to

M 2 , V2 . Then,

 

MV1 = M 2 V2 (Molarity equation)

  • In balanced chemical equation, if n1 moles of reactant one react with n2

moles of reactant two. Then,

 

MV1

n1

M 2 V2

n2

 

  • If two solutions of the same solute are mixed then molarity (M) of resulting

M M1 V1 + M 2 V2

(V1 + V2 )

 

  • Volume of water added to get a solution of molarity

æ M1 – M 2 ö

 

M 2 from Vml

of molarity M1 is

 

V2 – V1 = ç

è     M 2

÷ V1

ø

 

Relation between molarity and normality

 

Normality of solution = molarity ´

Molecular mass Equivalent mass

 

Normality ´ equivalent mass = molarity ´ molecular mass

 

For an acid,

Molecular mass = basicity

Equivalent mass

 

So, Normality of acid = molarity ´ basicity.

 

For a base,

Molecular mass = Acidity

Equivalent mass

 

So, Normality of base = Molarity ´ Acidity.

 

 

Example:10        The molarity of pure water (d = 1 g / l) is                                              [KCET 1993; CMC 1991, CPMT 1974, 88,90]

(a) 555 M                                  (b) 5.55 M                                (c) 55.5 M                             (d) None

Solution: (c)     Consider 1000ml of water

Mass of 1000ml of water = 1000 ´ 1 = 1000 g

 

 

 

Number of moles of water = 1000 = 55.5

18

Molarity = No. of moles of water = 55.5 = 55.5M

 

Volume in litre               1

 

Example:11        The number of iodine atoms present in 1cm3

of its 0.1M

solution is                                       [Pb. CET 1990]

 

(a) 6.02 ´ 10 23

(b) 6.02 ´ 10 22

(c) 6.02 ´ 1019

(d) 1.204 ´ 10 20

 

Solution: (d)

1 cc

of 0.1M I2 solution = 0.1 = 104 mol

1000

= 104 ´ 6.02 ´ 10 23 molecules

 

= 6.02 ´ 1019

molecules

= 2 ´ 6.02 ´ 1019 atoms = 1.204 ´ 10 20

 

Example:12        Equal volumes of

  • M AgNO3

and 0.2M NaCl

are mixed. The concentration of

NO

ions in the mixture

 

3

will be                                                                          [MP PMT 1996; CPMT 1991; Pb. PMT 1994; BHU 1994]

 

  • 1M

(b) 0.05M

  • 2 M

(d) 0.15M

 

Solution: (b)

AgNO3 + NaCl ¾¾® AgCl + NaNO3

 

  • M
  • M

 

0.1M AgNO3

reacts with 0.1M NaCl

to produce

 

0.1M AgCl

and 0.1M NaNO3

 

\ NO = 0.1M = 0.05M

3           2

[Q when equal volumes are mixed dilution occurs]

 

Example:13        The molarity of H 2 SO4

solution that has a density of 1.84 g / cc

at 35 o C

and contains 98% by weight is

 

[CPMT 1983, 2000; CBSE 1996, 2000; AIIMS 2001]

 

(a) 4.18M

Solution: (c)     Molarity = Wt. of solute ´

(b) 8.14M

1000

(c) 18.4 M

= 98 ´ 1000

 

 

= 18.4 M

  • 18 M

 

Mol. wt.

Vol. of solution (in ml.)

98    54.34

 

æ                                     mass      100                   ö

ç Vol. of solution =                   =            = 54.34 ml ÷

density    1.84

è                                                                                       ø

 

Example:14        Amount of oxalic acid ((COOH)2 .2H 2O) in grams that is required to obtain

250ml

of a semi-molar solution

 

is                                                                                                                                         [CBSE 1994]

(a) 17.25 g                                 (b) 17.00 g                                (c) 15.75 g                             (d) 15.00 g

Solution: (c)     Molecular mass of oxalic acid = 126

 

1000ml

of 1M

oxalic acid require oxalic acid = 126g

 

 

\ 250 ml

of 1M

oxalic acid will require oxalic acid =

126

 

1000

´ 250 = 31.5g

 

Hence 250ml of M oxalic acid will require oxalic acid = 31.5 ´ 1 = 15.75 g

 

2                                                                2

\ Mass of oxalic acid required = 15.75g .

 

Example:15        Volume of 10M HCl

should be diluted with water to prepare 2.00L of 5M HCl

is                   [AIEEE  2003]

 

  • 2L

1.5L

1.00L

0.5L

 

Solution: (c)     In dilution, the following equation is applicable :

M1 V1          =        M 2 V2

 

 

 

 

10M HCl

10 ´ V1 = 5 ´ 2.00

5M HCl

 

V = 5 ´ 2.00 L = 1.00L

1        10

 

Example:16        The volume of 95%

HSO4

(density = 1.85g cm3 ) needed to prepare 100cm3 of 15% solution of

H 2 SO4

 

(density = 1.10g cm3 ) will be                                                                                                    [CPMT 1983]

(a) 5 cc                                     (b) 7.5 cc                                 (c) 9.4 cc                             (d) 12.4cc

 

Solution: (c)     Molarity of 95% H

 

Molarity of 15% H

2SO4

 

2SO4

= 95 ´

98

= 15 ´

98

1

 

100 / 1.85

1

 

100 / 1.10

´ 1000

 

´ 1000

= 17.93M

 

= 1.68 M

 

M1 V1       =    M 2 V2

(95% H 2 SO4 )         (15% H 2 SO4 )

17.93 ´ V1 = 1.68 ´ 100 or V1 = 9.4cm3

 

Example:17        The molarity of a 0.2N Na2 CO3

solution will be                                                                   [MP PMT 1987]

 

(a)

0.05M

(b)

0.2M

(c)

0.1M

(d)

0.4 M

 

 

Solution: (c)

N M ´ molecular mass (M 2 )

Equivalent mass (E)

 

 

So,

M N ´ Equivalent mass (E)

Molecular mass (M 2 )

 

 

Equivalent mass of salt =

Molecular mass Total positive valency

 

Equivalent mass of

Na2 CO3

M 2

2

 

M = 0.2 ´ M 2 / 2 ;

M 2

M 0.2

2

= 0.1M

 

Example:18        If 20ml

of 0.4 N NaOH

solution completely neutralises 40ml

of dibasic acid. The molarity of acid solution is

[EAMCET 1992; DPMT 1994; JIPMER 1994]

 

(a)

  • M

(b)

  • M

(b)

  • M

(d)

0.4 M

 

Solution: (a)     We know N1 V1 = N 2 V2

0.4 ´ 20 = N 2 ´ 40 ; N 2 = 0.2N

 

Molarity = Normality of acid ; Here acid solution is dibasic, so

basicity

M = 0.2 = 0.1M

2

 

  • Molality (m) : It is the number of moles or gram molecules of the solute per 1000 g of the solvent. Unit of

 

molality is

mol / kg . For example, a

0.2

molal (0.2m)

solution of glucose means a solution obtained by dissolving

 

  • mole of glucose in

1000gm

of water. Molality (m) does not depend on temperature since it involves

 

measurement of weight of liquids. Molal solutions are less concentrated than molar solution. Mathematically molality can be calculated by following formulas,

  • m =Number of moles of the solute ´ 1000 = Strength per 1000 grams of solvent

Weight of the solvent in grams                      Molecular mass of solute

 

 

 

 

m = No. of gm moles of solute Wt. of solvent in kg

 

m =     Wt. of solute     ´

Mol. wt. of solute

1000

 

Wt. of solvent in g

 

 

 

 

m = No. of millimoles of solute Wt. of solvent in g

m = 10 ´ solubility Mol. wt. of solute

m = 1000 ´ wt. % of solute (x) (100 – x) ´ mol. wt. of solute

m =                      1000 ´ Molarity                                       (1000 ´ sp. gravity) – (Molarity ´ Mol. wt. of solute)

 

 

Relation between molarity (M) and molality (m)

 

 

Molality (m) =

Molarity

 

Density – Molarity ´ molecular mass

1000

 

 

Molarity (M)

=         Molality ´ density                  1 + Molality ´ molecular mass

1000

 

Example:19

HSO4

solution whose specific gravity is 1.98g ml 1

and

HSO4

by volume is 95%. The molality of the

 

solution will be

(a) 7.412                         (b) 8.412                         (c) 9.412                      (d) 10.412

 

Solution: (c)

HSO4

is 95% by volume

 

Wt. of H 2SO4 = 95g

Vol. of solution = 100ml

 

\ moles of H

2SO4

= 95

98

and weight of solution = 100 ´ 1.98 = 198g

 

Weight of water = 198 – 95 = 103 g

Molality = 95 ´ 1000 = 9.412

98 ´ 103

 

Hence molality of

HSO4

solution is 9.412

 

Example:20        The density of

HSO4

solution is

1.84 gm ml 1 . In 1 litre solution

HSO4

is 93% by volume then, the

 

molality of solution is                                                                                                          [UPSEAT 2000]

(a) 9.42                           (b) 10.42                         (c) 11.42                      (d) 12.42

 

Solution: (b)     Given  H 2 SO4

Wt. of H 2SO4

is 93% by volume

= 93 g

 

 

Volume of solution =100ml

Q Density =

mass volume

\ mass = d ´ volume

 

\ weight of solution = 100 ´ 1.84 g = 184 g

 

 

 

  1. of water = 184 – 93 = 91 g

Molality =          Moles        = 93 ´ 1000 = 10.42

  1. of water in kg 98 ´ 91
  • Formality (F) : Formality of a solution may be defined as the number of gram formula masses of the ionic solute dissolved per litre of the solution. It is represented by F . Commonly, the term formality is used to express the concentration of the ionic solids which do not exist as molecules but exist as network of ions. A solution containing one gram formula mass of solute per litre of the solution has formality equal to one and is called formal It may be mentioned here that the formality of a solution changes with change in temperature.

 

Formality (F)= Number of gram formula masses of solute =

Volume of solution in litres

Mass of ionic solute (g)

 

(gm. formula mass of solute) ´ (Volume of solution (l))

 

 

Thus,

F =     WB (g)

GFM ´ V(l)

or WB (g) ´ 1000

GFM ´ V(ml)

 

  • Mole fraction (X) : Mole fraction may be defined as the ratio of number of moles of one component to the total number of moles of all the components (solvent and solute) present in the solution. It is denoted by the letter X . It may be noted that the mole fraction is independent of the Mole fraction is dimensionless.

 

Let us suppose that a solution contains the components A and B and suppose that WA g

present in it.

of A and WB g

of B are

 

M
A

Number of moles of A is given by, n = WA

A

and the number of moles of B is given by, n   = WB

M
B

B

 

where

M A and M B

are molecular masses of A and B respectively.

 

Total number of moles of A and

B = nA  + nB

 

 

Mole fraction of A ,

XA =

nA

nA   + nB

; Mole fraction of B ,

XB =

nB

nA + nB

 

The sum of mole fractions of all the components in the solution is always one.

 

XA + XB

=     nA

nA  + nB

  • B nA + nB

= 1 .

 

Thus, if we know the mole fraction of one component of a binary solution, the mole fraction of the other can be calculated.

Relation between molality of solution (m) and mole fraction of the solute (XA).

X     =         m      

A          55.5 + m

Example: 21      A solution contains 16gm of methanol and 10gm of water, mole fraction of methanol is

[BHU 1981, 87; EAMCET 2003]

(a) 0.90                           (b) 0.090                         (c) 0.1                          (d) 1.9

Solution: (b)     Mass of methanol = 16g , Mol. mass of CH 3OH = 32

\ No. of moles of methanol = 16 = 0.5 moles

32

 

 

 

No. of moles of water = 90 = 5 moles

18

 

\ Mole fraction of methanol =

0.5

 

5 + 0.5

= 0.090

 

Example: 22      A solution has 25% of water, 25% ethanol and 50% acetic acid by mass. The mole fraction of each component will be                                                                                                              [EAMCET 1993]

(a) 0.50, 0.3, 0.19             (b) 0.19, 0.3, 0.50             (c) 0.3, 0.19, 0.5           (d) 0.50, 0.19, 0.3

Solution: (d)     Since 18 g of water = 1mole

 

 

25g

of water =

25 = 1.38

18

mole

 

Similarly, 46g of ethanol = 1 mole

 

 

25g

of ethanol = 25 = 0.55 moles

46

 

Again, 60g of acetic acid = 1 mole

 

 

50g

of acetic acid = 50 = 0.83 mole

60

 

\ Mole fraction of water =

1.38

 

1.38 + 0.55 + 0.83

= 0.50

 

Similarly, Mole fraction of ethanol =

0.55

 

1.38 + 0.55 + 0.83

= 0.19

 

Mole fraction of acetic acid =

0.83

 

1.38 + 0.55 + 0.83

= 0.3

 

  • Mass fraction : Mass fraction of a component in a solution is the mass of that component divided by the

 

total mass of the solution. For a solution containing w A gm of A and wB gm

of B

 

 

Mass fraction of A =

w A

w A   + wB

; Mass fraction of B =

wB

w A + wB

 

Note : ® It may be noted that molality, mole fraction, mass fraction etc. are preferred to molarity, normality, etc. because the former involve the weights of the solute and solvent where as later involve volumes of solutions. Temperature has no effect on weights but it has significant effect on volumes.

  • Demal unit (D) : The concentrations are also expressed in “Demal unit”. One demal unit represents one mole of solute present in one litre of solution at 0o C .

 

Certain properties of dilute solutions containing non-volatile solute do not depend upon the nature of the solute dissolved but depend only upon the concentration i.e., the number of particles of the solute present in the solution. Such properties are called colligative properties. The four well known examples of the colligative properties are

  • Lowering of vapour pressure of the
  • Osmotic pressure of the

 

 

 

  • Elevation in boiling point of the
  • Depression in freezing point of the solvent.

Since colligative properties depend upon the number of solute particles present in the solution, the simple case will be that when the solute is a non-electrolyte. In case the solute is an electrolyte, it may split to a number of ions each of which acts as a particle and thus will affect the value of the colligative property.

Each colligative property is exactly related to any other and thus if one property is measured, the other can be calculated. The study of colligative properties is very useful in the calculation of molecular weights of the solutes.

The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure of the liquid. The vapour pressure of a liquid depends on

  • Nature of liquid : Liquids, which have weak intermolecular forces, are volatile and have greater vapour pressure. For example, dimethyl ether has greater vapour pressure than ethyl
  • Temperature : Vapour pressure increases with increase in temperature. This is due to the reason that with increase in temperature more molecules of the liquid can go into vapour
  • Purity of liquid : Pure liquid always has a vapour pressure greater than its solution.

Raoult’s law : When a non-volatile substance is dissolved in a liquid, the vapour pressure of the liquid (solvent) is lowered. According to Raoult’s law (1887), at any given temperature the partial vapour pressure (pA) of any component of a solution is equal to its mole fraction (XA) multiplied by the vapour pressure of this component in

A
A
A
A

the pure state (p0 ) . That is,     p   = p0 ´ X

 

 

The vapour pressure of the solution (Ptotal )

is the sum of the parital pressures of the components, i.e., for the

 

solution of two volatile liquids with vapour pressures

p A  and

pB .

 

Ptotal

= p A

  • pB

= (p0 ´ X

) + (p0 ´ X )

 

A
B
A
B

Alternatively, Raoult’s law may be stated as “the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.”

Relative lowering of vapour pressure is defined as the ratio of lowering of vapour pressure to the vapour pressure of the pure solvent. It is determined by Ostwald-Walker method.

Mole fraction of the solute is defined as the ratio of the number of moles of solute to the total numbr of moles in solution.

Thus according to Raoult’s law,

 

 

p0 – p =

p0

n      =

n + N

w m

w W

 

m      M

 

where,

p = Vapour pressure of the solution;

n = Number of moles of the solute;

p0 = Vapour pressure of the pure solvent

N = Number of moles of the solvent

 

 

 

 

w and m =

weight and mol. wt. of solute; W and

M = weight and mol. wt. of the solvent.

 

Limitations of Raoult’s law

  • Raoult’s law is applicable only to very dilute solutions.
  • Raoult’s law is applicable to solutions containing non-volatile solute
  • Raoult’s law is not applicable to solutes which dissociate or associate in the particular

 

 

 

Example: 23      34.2 g of canesugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be (a) 0.0099                              (b) 1.1597                       (c) 0.840                      (d) 0.9901

 

P 0P

W / M

34.2 / 342

0.1

 

Solution: (a)

     A             A   =                 B              A                  =

=            = 0.0099

 

P
A

0            WB / MB  + WA / M A

34.2 / 342 + 180 / 18

10.1

 

Example: 24       Lowering in vapour pressure is the highest for                                                             [Roorkee 1989; BHU 1997]

 

 

 

Solution: (d)

(a) 0.2m urea                   (b) 0.1 m glucose                (c) 0.1m MgSO4

P 0P

P
0

    A        A = Molality ´ (1 – ax + xa + gx)

A

(d) 0.1m BaCl 2

 

The value of

P 0PA

is maximum for BaCl 2 .

 

A

Example: 25        Vapour pressure of CCl 4

at 25 o C

is 143 mm Hg 0.5 g of a non-volatile solute (mol. wt. 65) is dissolved in

 

100 ml of CCl4 . Find the vapour pressure of the solution. (Density of CCl4

= 1.58 g / cm3 )

[CBSE PMT 1996]

 

(a) 141.93 mm                       (b) 94.39 mm                          (c) 199.34 mm                   (d) 143.99 mm

 

 

Solution: (a)

P 0P

= n   ;

143 – Ps  =

0.5 / 65 or        =

 

     A             A          B                                                                           

Ps          141.93 mm

 

P
A

0           n A

143

158 / 154

 

Example:26        The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour pressure in contact with equimolar solution of benzene and toluene is                         [Pb. CET 1988]

(a) 0.50                           (b) 0.6                            (c) 0.27                        (d) 0.73

 

Solution: (c)     For equimolar solutions,

XB  = XT

= 0.5

 

B

PB = XB ´ P 0 = 0.5 ´ 160 = 80 mm

T

PT = XT ´ P 0 = 0.5 ´ 60 = 30 mm

 

PTotal

= 80 + 30 = 110 mm

 

 

Mole fraction of toluene in vapour phase =

30 = 0.27

110

 

Example:27 The vapour pressure of a solvent decreases by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be if the decrease in vapour pressure is to be 20 mm of mercury then the mole fraction of the solvent is                                         [CBSE PMT 1998] (a) 0.8                                       (b) 0.6                            (c) 0.4                          (d) None

 

Solution: (a)

DP / P 0 = X 2

2

Hence DP / P0 = X2 / X1 i.e. 10 / 20 = 0.2 / XB

or XB = 0.4

 

Mole fraction of solvent = 1 – 0.4 = 0.6

 

 

 

B

Example:28 The vapour pressure of a solvent A is 0.80 atm. When a non-volatile substance B is added to this solvent its vapour pressure drops to 0.6 atm. the mole fraction of B in the solution is                                          [MP PMT 2000] (a) 0.25                   (b) 0.50                           (c) 0.75                        (d) 0.90

 

Solution: (a)

DP / P 0 = X

or XB = 0.2 / 0.8 = 0.25

 

 

 

  • Ideal solution : An ideal solution may be defined as the solution which obeys Raoult’s law over the entire range of concentration and temperature and during the formation of which no change in enthalpy and no change in volume takes So for ideal solutions the conditions are,

 

(i) It should obey Raoult’s law, i.e.,

PA   = P 0 XA

and

PB  = P 0 XB .

 

A
B

(ii)

(iii)

DH mixing = 0

DVmixing = 0

 

The solutions in which solvent-solvent and solute-solute interactions are almost of the same type as solvent- solute interactions, behave nearly as ideal solutions.

This type of solutions are possible if molecules of solute and solvent are almost of same size and have identical polarity. For example, solutions of following pairs almost behave as ideal solutions,

n-Heptane and n-hexane.; Chlorobenzene and bromobenzene. Benzene and toluene; Ethyl bromide and ethyl iodide.

Ethylene bromide and ethylene chloride; Carbon tetrachloride and silicon tetrachloride.

P

For such solutions the vapour pressure of the solution is always intermediate between the vapour pressures of

 

pure components A and B, i.e.,

0 and

P 0 .

 

B
A
  • Non-Ideal solution : The solutions which do not obey Raoult’s law and are accompanied by change in enthalpy and change in volume during their formation are called non-ideal solutions. so, for non-ideal solutions the conditions are :

 

  • It does not obey Raoult’s

PA  ¹ P 0 X

A ; PB

¹ P 0 XB

 

A
B

DH mixing ¹ 0

DVmixing ¹ 0

 

Either component of non-ideal binary solution do not follow Raoult’s law.

The non-ideal solutions are further divided into two types :

(a) Solutions showing positive deviations. (b) Solutions showing negative deviations.

  • Solutions showing positive deviation : In this type of deviations, the partial vapour pressure of each component (say A and B) of solution is

greater than the vapour pressure as expected according to Raoult’s law. This type of deviations are shown by the solutions in which solvent-solvent and solute-solute interactions are stronger than solvent-solute interactions.

 

 

 

For the non-ideal solutions exhibiting positive deviation.

 

A
A

P   > P 0 X

A , PB

  • P 0X ;

DHmixing = + ve ;

DVmixing = + ve

 

B
B

e.g. of solutions showing positive deviations (CH3 )2 CO + CS2  ;  (CH3 )2 CO + C2 H5 OH C6 H6 + (CH3 )2 CO ; CCl4 + C6 H6

CCl4  + CHCl3 ;  CCl4  + C6 H5 CH3

H2O + CH3 OH ;   H2O + C2 H5 OH

CH3 CHO + CS2 ; CHCl3  + C2 H5 OH

  • Solutions showing negative deviation : In this type of deviations the partial vapour pressure of each component of solution is less than the vapour pressure as expected according to Raoult’s law. This type of deviations are shown by the solutions in which solvent-solvent and solute-solute interactions are weaker than solvent-solute

For non-ideal solution showing negative deviation.

 

A
A

P   < P 0 X

A , PB

< P 0 X   ;

DHmixing

= – ve ;

DVmixing

= – ve

 

B
B

e.g. of solutions showing negative deviations

CH3 COOH + C5 H5 N (pyridine)

CHCl3  + (CH3 )2 CO ;  CHCl3   + C6 H6

CHCl3 + (C2 H5 )2 O ; H 2 O + HCl

H 2 O + HNO3 ; (CH3 )2 CO + C6 H5 NH 2

Differences between ideal and non-ideal solutions

 

Azeotropes are defined as the mixtures of liquids which boil at constant temperature like a pure liquid and possess same composition of components in liquid as well as in vapour phase. Azeotropes are also called constant boiling mixtures because whole of the azeotropes changes into vapour state at constant temperature and their components can not be separated by fractional distillation. Azeotropes are of two types as described below :

  • Minimum boiling azeotrope : For the solutions with positive deviation there is an intermediate composition for which the vapour pressure of the solution is maximum and hence, boiling point is At this

 

 

 

composition the solution distills at constant temperature without change in composition. This type of solutions are called minimum boiling azeotrope. e.g;

ComponentsMass % of B Boiling points (K)
AB ABAzeotrope
 H 2 O C2 H5 OH95.57373351.3351.1
 H 2 O C2 H5 CH2 OH71.69373370350.72
 CHCl 3 C2 H5 OH67334351.3332.3
 (CH3 )2 CO CS26.8329.25320312.2
  • Maximum boiling azeotrope : For the solutions with negative deviations there is an intermediate composition for which the vapour pressure of the solution is minimum and hence, boiling point is maximum. At this composition the solution distill`s at constant temperature without the change in composition. This type of solutions are called maximum boiling e.g
Components Mass % of B Boiling points (K)
AB ABAzeotrope
 H 2 OHCl20.3373188383
 H 2 O HNO358.0373359393.5
 H 2 O HClO471.6373383476

The flow of solvent from pure solvent or from solution of lower concentration into solution of higher concentration through a semi-permeable membrane is called Osmosis. Osmosis may be divided in following types,

  • Exo-Osmosis : The outward osmotic flow of water from a cell containing an aqueous solution through a semi-permeable membrane is called as Exo-osmosis. For example, egg (after removing hard shell) placed in NaCl solutions, will shrink due to exo-osmosis.
  • Endo-osmosis : The inward flow of water into the cell containing an aqueous solution through a semi- permeable membrane is called as endo-osmosis. g., an egg placed in water swells up due to endo-osmosis.
  • Reverse osmosis : If a pressure higher than osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through the semi-permeable Since here the flow of solvent is in the reverse direction to that observed in the usual osmosis, the process is called reverse osmosis.

Differences between osmosis and diffusion

 

OsmosisDiffusion
In osmosis movement of molecules takes place throughIn   diffusion   there    is    no  role   of   semi-permeable
a semi-permeable membrane.membrane.
It involves movement of only solvent molecules  fromIt  involves   passage   of   solvent   as   well   as   solute
one side to the other.molecules from one region to the other.
Osmosis is limited to solutions only.Diffusion    can    take  place    in    liquids,    gases    and
 solutions.
Osmosis  can   be   stopped   or   reversed   by   applyingDiffusion can neither be stopped nor reversed
additional pressure on the solution side. 

 

 

 

Osmotic pressure (p) : Osmotic pressure may be defined in following ways,

The osmotic pressure of a solution at a particular temperature may be defined as the excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semi-permeable membrane. It is denoted by p.                             Or

Osmotic pressure may be defined as the excess pressure which must be applied to a solution in order to prevent flow of solvent into the solution through the semi-permeable membrane.

Or

Osmotic pressure is the excess pressure which must be applied to a given solution in order to increase its vapour pressure until it becomes equal to that of the solution.

Or

Osmotic pressure is the negative pressure which must be applied to (i.e. the pressure which must be withdrawn from) the pure solvent in order to decrease its vapour pressure until it becomes equal to that of the solution.

Measurements of osmotic pressure : Following methods are used for the measurement of osmotic pressure : (i) Pfeffer’s method, (ii) Morse and Frazer’s method, (iii) Berkeley and Hartley’s method, (iv) Townsend’s negative pressure method, (v) De Vries plasmolytic method.

Determination of molecular mass of non-volatile solute from osmotic pressure (p) :

According to Van’t Hoff equation,

 

p = nB RT ;

V

where, WB

= known mass of solute in gm

 

This method is especially suitable for the determination of molecular masses of macromolecules such as proteins and polymers. This is due to the reason that for these substances the values of other colligative properties such as elevation in boiling point or depression in freezing point are too small to be measured. On the other hand, osmotic pressure of such substances are measurable. Conditions for getting accurate value of molecular mass are,

  • The solute must be non-volatile.
  • The solution must be
  • The solute should not undergo dissociation or association in the solution.

Van’t Hoff’s solution equation :

The osmotic pressure is a colligative property. For a given solvent the osmotic pressure depends only upon the molar concentration of solute but does not depend upon its nature. Osmotic pressure is related to the number of moles of the solute by the following relation :

 

pV = nRT

or p = n RT

V

Þ p = CRT ; here, C = concentration of solution in moles per litre

 

R = gas constant; T = temperature; n = number of moles of solute; V = volume of solution Above eq. is called Van’t Hoff’s solution equation

 

 

 

Relation of osmotic pressure with different colligative properties : Osmotic pressure is related to relative lowering of vapour pressure, elevation of boiling point and depression of freezing point according to the following relations.

 

æ PoP   ö

dRT

dRT

dRT

 

(1) p = ç   A             A ÷ ´

(2) p = DTb ´

(3) p = DTf ´

 

 

P
M

ç        o           ÷

è       A            ø        B

1000 ´ Kb

1000 ´ K f

 

In the above relations, p = Osmotic pressure; d = Density of solution at temperature T; R = Universal gas

 

constant; solvent

MB = Mol. Mass of solute;

Kb = Molal elevation constant of solvent; K f

= Molal depression constant of

 

Isotonic, Hypertonic and Hypotonic solutions

Isotonic or iso-osmotic solutions : Two solutions of different substances having same osmotic pressure at same temperature are known as isotonic solutions.

 

For isotonic solutions, p 1 = p 2

Primary Condition     …..(i)

 

Also,

C1  = C2

 

or                           n1

V1

n2 V2

Secondary Conditions

 

or                         w1

mV1

=   w2

m2 V2

…..(ii)

 

Eq. (ii) holds good only for those solutes which neither possess the tendency to get associate nor dissociate in solution, e.g.,

  • Urea and glucose are isotonic then, p 1 = p 2 and C1 = C2

 

  • Urea and

NaCl are isotonic then, p   = p

1         2

(dissociate)

but C1  ¹ C2

 

  • Urea and Benzoic acid are isotonic then,

(associate)

p 1 = p 2

but C1  ¹ C2

 

Hypertonic and Hypotonic Solution : The solution which has more osmotic pressure than the other solution is called as hypertonic solution and the solution which has lesser osmotic pressure than the other is called as hypotonic solution.

The flow of solvent is always from lower osmotic pressure to higher osmotic pressure i.e. from hypotonic to hypertonic solution.

 

 

Example: 29      Osmotic pressure is 0.0821 atm at a temperature of 300 K. find concentration in mole/litre                              [Roorkee 1990]

 

(a) 0.033                         (b) 0.066                         (c)

0.33 ´ 10 2

  • 3

 

 

Solution: (c)

C = P RT

=        0.0821      =

0.0821 ´ 300

1

 

300

= 0.33 ´ 10 2 mole / litre

 

Example: 30      The osmotic pressure of 5% (mass-volume) solution of cane sugar at 150 o C (mol. mass of sugar = 342) is

[BHU 1995]

(a) 4 atm                                 (b) 5.07 atm                              (c) 3.55 atm                          (d) 2.45 atm

 

 

Solution: (b)

C =   5

342

´   1

100

 

´ 1000 =

50 M ;

342

P = 50

342

 

´ 0.082 ´ 423 = 5.07 atm

 

 

 

 

Boiling point of a liquid may be defined as the temperature at which its vapour pressure becomes equal to atmospheric pressure, i.e., 760 mm. Since the addition of a non-volatile solute lowers the vapour pressure of the solvent, solution always has lower vapour pressure than the solvent and hence it must be heated to a higher temperature to make its vapour pressure equal to atmospheric pressure with the result the solution boils at a higher temperature than the pure solvent. Thus sea water boils at a higher temperature than distilled water. If Tb is the boiling point of the solvent and T is the boiling point of the solution, the difference in the boiling point (DT or D Tb) is called the elevation of boiling point.

T Tb = DTb or DT

Elevation in boiling point is determined by Landsberger’s method and Cottrell’s method. Study of elevation in boiling point of a liquid in which a non-volatile solute is dissolved is called as ebullioscopy.

Important relations concerning elevation in boiling point

  • The elevation of boiling point is directly proportional to the lowering of vapour pressure, e.,

 

DTb

µ p0 – p

 

  • DTb = Kb ´ m

 

where

Kb   =

molal elevation constant or ebullioscopic constant of the solvent;

m = Molality of the solution,

 

i.e., number of moles of solute per 1000g

of the solvent;

DTb =

Elevation in boiling point

 

 

  • DT

= 1000 ´ Kb ´ w

or m = 1000 ´ Kb ´ w

 

b                      m ´ W                                 DT ´ W

b

 

where,

Kb      is molal elevation constant and defined as the elevation in b.p. produced when 1 mole of the

 

solute is dissolved in 1 kg of the solvent. Sometimes the value of expression becomes

m = 100 ´ Kb ´ w

DTb ´ W

Kb     is given per

0.1kg

( 100g ), in such case the

 

Where w and W are the weights of solute and solvent and m is the molecular weight of the solute.

0.002(T )2

 

K   =           0    

b                        l

 

 

where

T0 =

V

 

Normal boiling point of the pure solvent;

lV  = Latent heat of evaporation in

cal / g

of pure

 

solvent; Kb for water is 0.52 deg- kg mol

-1

 

 

 

Example: 31      A solution containing 3.3 g of a substance in 125 g of benzene (b.p. 80 o C) boils at

80.66 o C . If

Kb    for one

 

litre of benzene is 3.28 o C , the molecular weight of the substance shall be                                              [CPMT 1992]

(a) 127.20                        (b) 131.20                       (c) 137.12                    (d) 142.72

 

Solution: (b)

M   = 1000 ´ Kb ´ w = 1000 ´ 3.28 ´ 3.3 = 131.2

 

 

2

W ´ DTb

125 ´ 0.66

 

 

 

Example: 32      The molal b.p. constant for water is 0.513o C kg mol 1 . When 0.1 mole of sugar is dissolved in 200 g of water, the solution boils under a pressure of 1 atm at                                                                                    [AIIMS 1991]

 

(a)

100.513o C

(b)

100.0513o C

(c)

100.256 o C

(d)

101.025 o C

 

Solution: (c)

DT   = K   ´ m = 0.513æ 0.1 ´ 1000ö = 0.2565 ; DT

 

= 100.2565o C

 

è
ø

b              b                                     ç 200            ÷                            b

 

Example: 33      An aqueous solution containing 1 g of urea boils at

100.25 o C . The aqueous solution containing 3 g of

 

glucose in the same volume will boil at                                                                          [BHU 1994, CBSE 2000]

 

(a)

100.75 o C

(b)

100.5 o C

(c)

100 o C

(d)

100.25 o C

 

Solution: (d)     1 g urea =

1 mol , 3 g glucose =

60

3 =

180

1 mol . Hence it will boil at the same temperature

60

 

 

Freezing point is the temperature at which the liquid and the solid states of a substance are in equilibrium with each other or it may be defined as the temperature at which the liquid and the solid states of a substance have the same vapour pressure. It is observed that the freezing point of a solution is always less than the freezing point of the pure solvent. Thus the freezing point of sea water is low than that of pure water. The depression in freezing point

 

(DT

or DTf ) of a solvent is the difference in the freezing point of the pure solvent (Ts ) and the solution (Tsol. ) .

 

Ts – Tsol

= DTf

or DT

 

NaCl  or

CaCl 2

(anhydrous) are used to clear snow on roads. They depress the freezing point of water and

 

thus reduce the temperature of the formation of ice.

Depression in freezing point is determined by Beckmann’s method and Rast’s camphor method. Study of depression in freezing point of a liquid in which a non-volatile solute is dissolved in it is called as cryoscopy.

Important relations concerning depression in freezing point.

  • Depression in freezing point is directly proportional to the lowering of vapour

 

 

DTf

DTf

= Kf  ´ m

µ p0 – p

 

where K f  =

molal depression constant or cryoscopic constant;

m = Molality of the solution (i.e., no. of moles

 

of solute per 1000g

of the solvent);

DTf

= Depression in freezing point

 

  • DT

= 1000 ´ Kf  ´ w or m = 1000 ´ Kf  ´ w

 

 

f                      m ´ W                                 DT ´ W

f

 

where K f

is molal depression constant and defined as the depression in freezing point produced when 1 mole

 

of the solute is dissolved in 1kg

the expression becomes

of the solvent. Sometimes the value of K f

is given per

0.1kg (100g) , in such case

 

m = 100 ´ Kf ´ w

DTf ´ W

where w and W are the weights of solute and solvent and m is the molecular weight of the solute.

R(T )2       0.002(T )2

 

  • K

=      0     =            0    

 

f

l f 1000          l f

 

where

T0 = Normal freezing point of the solvent;

l f  = Latent heat of fusion/g of solvent; K f

for water is

 

1.86 deg- kg mol -1

 

 

 

Relative lowering of vapour pressure, elevation in boiling point and depression in freezing point are directly proportional to osmotic pressure.

Example: 34      Solution of sucrose (Mol. Mass = 342) is prepared by dissolving 34.2 gm. of it in 1000 gm. of water. freezing

 

point of the solution is (K f

for water is 1.86 K kg mol -1 )

[AIEEE  2003]

 

(a) 272.814 K                         (b) 278.1 K                             (c) 273.15 K                       (d) 270 K

Solution: (a)     Molality of the solution = 34.2 = 0.1

342

DTf = K f ´ m = 1.86 ´ 0.1 = 0.186 K

Freezing point of solution = 273 – 0.186 = 272.814 K

Example: 35      An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7 g of water freezes at 272.817K. If the

value of K f  for water is 1.86 K kg mol 1 , the molecular mass of the acid is                                             [AMU 2002]

(a) 46.0                           (b) 48.6                           (c) 48.8                        (d) 46.8

Solution: (d)     Mass of solvent (WA ) = 21.7 g

Mass of solute (WB ) = 0.1 g

Depression in freezing point, (DTf ) = 273 – 272.817 = 0.183 K

 

 

DT  = K

  • m = K

. WB ´ 1000

 

 

f               f                           f    WA           MB

 

M     K f ´ WB ´ 1000 = 1.86 ´ 0.1 ´ 1000 = 46.8

 

B             WA ´ DTf

21.7 ´ 0.183

 

Example: 36      What is the molality of solution of a certain solute in a solvent if there is a freezing point depression of 0.184 o

and if the freezing point constant is 18.4                                                                             [Haryana CEET 1991]

(a) 0.01                           (b) 1                               (c) 0.001                      (d) 100

 

 

Solution: (a)

DTf            0.184

DT   = K   ´ m or m =             =               = 0.01

 

f                f                                        K f

18.4

 

Example: 37      A solution containing 6.8 g of a nonionic solute in 100 g of water was found to freeze at

  • 93o C . The

 

freezing point depression constant of water is 1.86. Calculate the molecular weight of the solute

[Pb. PMT 1994; ISM Dhanbad 1994]

(a) 13.6                           (b) 34                             (c)  68                           (d) 136

 

 

Solution: (d)

M     = 1000 ´ K f   ´ WB

1000 ´ 1.86 ´ 6.8 = 136

 

f

B            DT ´ W

A

100 ´ 0.93

 

Example: 38      The molar freezing point constant for water is 1.86 o C / mole . If 342 g of cane sugar (C12 H 22 O11 ) is dissolved in 1000 g of water, the solution will freeze at                                                                            [NCERT 1977; CPMT 1989]

 

 

 

Solution: (a)

(a)

 

DTf

– 1.86 o C                           (b)

ç         ÷

= 1.86æ 342 ö = 1.86

342

1.86 o C

(c)

– 3.92o C

(d)

2.42o C

 

 

DTf

è         ø

= -1.86 o C

 

 

 

 

The colligative properties of solutions, viz. lowering of vapour pressure, osmotic pressure, elevation in b.p. and depression in freezing point, depend solely on the total number of solute particles present in solution. Since the electrolytes ionise and give more than one particle per formula unit in solution, the colligative effect of an electrolyte solution is always greater than that of a non-electrolyte of the same molar concentration. All colligative properties are used for calculating molecular masses of non-volatile solutes. However osmotic pressure is the best colligative property for determining molecular mass of a non-volatile substance.

Points to remember

  • Colligative properties µ Number of particles

µ Number of molecules (in case of non-electrolytes)

µ Number of ions (In case of electrolytes)

µ Number of moles of solute

µ Mole fraction of solute

  • For different solutes of same molar concentration, the magnitude of the colligative properties is more for that solution which gives more number of particles on
  • For different solutions of same molar concentration of different non-electrolyte solutes, the magnitude of the colligative properties will be same for
  • For different molar concentrations of the same solute, the magnitude of colligative properties is more for the more concentrated
  • For solutions of different solutes but of same percent strength, the magnitude of colligative property is more for the solute with least molecular
  • For solutions of different solutes of the same percent strength, the magnitude of colligative property is more for that solute which gives more number of particles, which can be known by the knowledge of molecular weight and its ionisation

 

Molecular masses can be calculated by measuring any of the colligative properties. The relation between colligative properties and molecular mass of the solute is based on following assumptions.

  • The solution is dilute, so that Raoult’s law is
  • The solute neither undergoes dissociation nor association in

In case of solutions where above assumptions are not valid we find discrepencies between observed and calculated values of colligative properties. These anomalies are primarily due to

  • Association of solute
  • Dissociation of solute

 

 

 

  • Association of solute molecules : Certain solutes in solution are found to associate. This eventually leads to a decrease in the number of molecular particles in the solutions. Thus, it results in a decrease in the values of colligative As already discussed, the colligative properties are inversely related to the molecular mass,

 

colligative property µ

1

 

molecular mass of solute

, therefore, higher values are obtained for molecular masses

 

than normal values for unassociated molecules.

  • Dissociation of solute molecules : A number of electrolytes dissociate in solution to give two or more particles (ions). Therefore, the number of solute particles, in solutions of such substances, is more than the expected Accordingly, such solutions exhibit higher values of colligative properties. Since colligative properties are inversely proportional to molecular masses, therefore, molecular masses of such substances as calculated from colligative properties will be less than their normal values.

Van’t Hoff’s factor (i) : In 1886, Van’t Hoff introduced a factor ‘i’ called Van’t Hoff’s factor, to express the extent of association or dissociation of solutes in solution. It is ratio of the normal and observed molecular masses of the solute, i.e.,

i = Normal molecular mass Observed molecular mass

In case of association, observed molecular mass being more than the normal, the factor i has a value less than

  1. But in case of dissociation, the Van’t Hoff’s factor is more than 1 because the observed molecular mass has a lesser value than the normal molecular mass. In case there is no dissociation the value of ‘i’ becomes equal to one.

Since colligative properties are inversely proportional to molecular masses, the Van’t Hoff’s factor may also be written as :

i =                          Observed value of colligative property                                         Calculated value of colligative property assuming no association or dissociation

i = No. of particles after association or dissociation No. of particles before association or dissociation

Introduction of the Van’t Hoff factor modifies the equations for the colligative properties as follows :

P o P

P
o

Relative lowering of vapour pressure =  A               A = i XB

A

Elevation of boiling point, DTb = ikbm

 

Depression in freezing point,

Osmotic pressure, p = inRT ;

V

DTf  = ik f m

 

p = iCRT

 

From the value of “i”, it is possible to calculate degree of dissociation or degree of association of substance.

Degree of dissociation (a) : It is defined as the fraction of total molecules which dissociate into simpler molecules or ions.

a = i  1 ; m= number of particles in solution

m – 1

 

 

 

Degree of association (a) : It is defined as the fraction of the total number of molecules which associate or combine together resulting in the formation of a bigger molecules.

 

a =    i – 1

1 / m – 1

; m = number of particles in solution

 

Example: 39      The molal freezing point constant for water is 1.86 o C / m . Therefore, the freezing point of 0.1 M NaCl solution in water is expected to be                                                                                                                        [MLNR 1994]

 

(a)

– 1.86 o C

(b)

– 0.186 o C

(c)

– 0.372o C

(d)

+ 0.372o C

 

Solution: (c)

DTf

= iK f m = 2 ´ 1.86 ´ 0.1 = 0.372

 

Tf  = -0.372o C

Example: 40      The depression in freezing point of 0.01 M aqueous solutions of urea, sodium chloride and sodium sulphate is in the ratio of                                                                                                               [Roorkee 1990; DCE 1994]

(a) 1 : 1 : 1                       (b) 1 : 2 : 3                      (c) 1 : 2 : 4                   (d) 2 : 2 : 3

 

Solution: (b)     Concentration of particles of 0.01 M urea, NaCl and

Na2 SO4

= 0.01 M, 0.02 M, 0.03 respectively i.e., they

 

are in the ratio 1 : 2 : 3. Hence, depression in freezing point will be in the same ratio.

Example: 41      The Van’t Hoff factor for 0.1 M Ba(NO3 )2 solution is 2.74. The degree of dissociation is                               [IIT 1999]

(a) 91.3%                        (b) 87%                          (c) 100%                      (d) 74%

 

Solution: (b)

Ba(NO3 )2 ⇌

Ba ++ + 2NO

 

Initial      0.1

At. eq. (0.1- x)

3

0

xM                2xm

 

0

i = (0.1  x) + x + 2x = 0.1 + 2x = 2.74 = 0.1 + 2x = 0.274 or x = 0.174 = 0.087

0.1                  0.1                                                           2

%a = x ´ 100 = 0.087 ´ 100 = 87%

0.1               0.1

 

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