SOLUTIONS

Types of solutions : - R.B.

● The homogeneous mixture of two or more components is called solution.
● A binary solution contains one solute and one solvent.
● The component which is present in large quantity is called solvent while the component
which is present in small quantity is called solute.
Solute + solvent Solution
● Based on the physical state of solute and solvent, solutions are 9 types.
Type of solution Solute Solvent Example
Gaseous solutions Gas Gas Air
Liquid Gas CHCl3 in N2
Solid Gas Camphor in N2
Liquid solutions Gas Liquid O2 in water
Liquid Liquid Alcohol in water
Solid Liquid NaCl in water
Solid solutions Gas Solid H2 in palladium
Liquid Solid Hg in Na
Solid Solid Alloys like brass, bronze etc.,
Concentration terms :
a) Mass percentage (w/w) :
The of a component is given by,

Mass of the component

Mass % of a component = 100
Total mass of solution
Eg. 10% sucrose (w/w) means, 10g of sucrose is dissolved in 90g of water where the
total mass of the solution is 100g.
b) Volume percentage (V/V) ;

Volume of the component

Volume % of a component =  100
Total volume of solution
Eg. 10% (V/V) ethanol means, 10ml of ethanol is dissolved in 90ml of water resulting
in 100ml of solution.
c) Mass/volume percentage :
10% (w/v) glucose means 10g of glucose is present in 100ml of solution.
d) Parts per million (ppm) :

Number of parts of component

Parts per milliion = 106
Total number of parts of all components of solution
e) Molarity (M) :
* The number of moles of solute dissolved in one litre of solution at a given temperature
is called molarity.

1
 1
* Units of molarity : Mol. lit
* Molarity depends upon the temperature because the molarity expression contains
volume term.
* Molarity (M) = n/V
n Number of moles of solute
V Volume of solution in lit.

w 1000
Molarity (M) =
* GMW  V
W weight of the solute in grams
GMW gram molecular weight of the solute
V Volume of the solution in ml.
* For dilution :
M1V1 = M2V2
M1 Molarity of concentrated solution
V1 Volume of concentrated solution
M2 Molarity of dilute solution
V2 Volume of dilute solution
Volume of water to be added V2 – V1.
f) Molality (m) :
* The number of moles of the solute dissolved in one kg of the solvent is called molality.
1
* Units of molality : Mol. kg
* Molality is independent of temperature because the expression of molality does not
contain volume term.

Number of moles of solute

Molality (m) =
* Mass of solvent in kg

W1  1000

GMw  W2
W1 weight of solute in grams
W2 weight of solvent in grams
GMw gram molecular weight of solute.
g) Mole fraction :
* The ratio of the number of moles of a component to the total number of moles of all
the components of solution is called mole fraction of that component.
Let n1 is the number of moles of solute and n2 is the number of mole of solvent.
Then,

n1
 X1  
mole fraction of solute n1  n2

2
 and,

n2
 X2  
mole fraction of solvent n1  n2
* Sum of the mole fractions of solute and solvent is equal to one.
* Mole fraction has no units.
* Mole fraction is independent of temperature because its expression contains only weight
terms.
Solubility :
* The maximum amount of substance that can be dissolved in a specified amount of
solvent at a specified temperature is called solubility.
Solubility of a solid in a liquid :
1) Polar compounds are soluble in polar solvents while non-polar solutes are soluble
in non-polar solvents.
2) The solution which cannot dissolve some more solute at a given temperature is
called saturated solution.
3) The solution which is capable of dissolving some more solute at a given temperature
is called unsaturated solution.
4) The solution which contains more solute than that required for saturated at a given
temperature is called supersaturated solution.
Supersaturated solutions are meta stable.
5) (i) If the lattice energy of a substance is higher than its hydration energy, then the
dissolution process is endothermic and the solubility of that compound increases with raise in
temperature.
(ii) If the lattice energy of a substance is lower than its hydration energy, then the
dissolution process is exothermic and the solubility of that compound decreases with raise in
temperature.
6) Pressure does not affect the solubility of a solid in a liquid.
Solubility of a gas in a liquid :
1) The solubility of gases increases with increase of pressure.
2) According to Henry’s law, at a constant temperature the solubility of a gas in a
liquid is directly proportional to the pressure of the gas.
(OR)
The partial pressure of the gas in vapour phase (P) is directly proportional to the
mole fraction of the gas (X) in the solution.
P ∝X
(OR) P = KH.X
Where KH is called Henry’s law constant.

3
 * A plot of partial pressure of the gas versus mole fraction of the gas in solution gives a
straight line passing through the origin. Slope is equal to Henry’s law constant.
* Higher the value of KH at a given pressure, the lower is the solubility of the gas in the
liquid.
* As temperature increases, the KH value increases, ie, the solubility of a gas decreases
with increase in temperature.
Applications of Henry’s law :
1. The soda water bottles are sealed under high pressure to increase the solubility of CO2
in soda water.
2. When scuba divers go into the deep sea, due to increase in pressure the solubility of
the gases in blood increases. When divers come out towards surface, the pressure
gradually decreases and dissolved gases release from blood by forming bubbles of
nitrogen in blood. This blocks capillaries and creates a medical condition called bends.
Because of this divers feel more pain and dangerous to life.
To avoid bends, the tanks used by scuba divers are filled with air and helium because helium
is less soluble in blood.
3. At higher altitudes, the partial pressure of oxygen is low. Due to this the people living
there will have low concentration of oxygen in blood. Because of this they suffer with
anoxia by which they cannot think and work properly.
* Vapour pressure of liquid-liquid solutions :
1. The pressure exerted by the vapour over the surface of the liquid when the liquid is
in equilibrium with its vapour at a given temperature is called vapour pressure of the

liquid P .
o

2. The vapour pressure of a liquid increases with in temperature.

* For the mixture of two volatile liquids :
Let A and B are the two volatile liquids, XA and XB are the mole fractions of A and B liquids
and PA and PB are the partial pressures of liquid A and B in the solution.
According to Raoult’s law, for a solution of two volatile liquids, the partial vapour pressure
of a component liquid is directly proportional to its mole fraction in the solution.
ie., PA ∝ XA

(or) PA  PAo . X A o
where PA is the vapour pressure of the pure liquid ‘A’.
similarly,
PB ∝ XA

(or) PB  PBo . X B o
where PB is the vapour pressure of the pure liquid ‘B’.

PTotal  PA  PB

(or) PT  PAo . X A  PBo . X B

4
 PA

* Mole fraction of ‘A’ in vapour phase PT

PB

Mole fraction ‘B’ in vapour phase PT .
* The solution which obeys Raoult’s law over the entire range of concentrations is called ideal
solution.
i) If the intermolecular attractive forces between the A – A and B – B are nearly equal to those
between A – B, then an ideal solution forms.
ii) For an ideal solution,

Vmixing  0
iii) For an ideal solution,

H mix  0
ie., no heat is absorbed or liberated during the formation of solution.
iv) This type of solution boils at an expected temperature.
Eg. (1) n-hexane + n-haptane
(2) Benzene + toluene
(3) Bromo ethane and chloroethane
Non-ideal solutions :
* The solution which does not follow over the entire range of concentration is called
non-deal solution.
For the non-ideal solution with +ve deviation from Roult’s law :
1) It the A-B interactions are weaker than those between A-A and B-B, then this type of
solutions results.
2) For this solution,

Vmix  ve
3) For this solution,

H mix  ve
i.e., the solution is formed with the absorption of heat.
4) The solution shows more vapour pressure than expected and boils at a lower temperature
than expected.

PA  PAo . X A , PB  PBo . X B

PT   PAo . X A  PBo . X B 
and
Eg. (1) Ethanol + Acetone
(2) Water + Ethanol etc.,
For the non-ideal solution with –ve deviation from Raoult’s law :

5
1) If the A – B interactions are stronger than those between A – A and B – B, then this type of
solution result’s.
2) For this type of solution,

Vmix  ve
3) For this type of solution,

H mix  ve
i.e., the solution is formed with the liberation of heat.
4) The solution shows less vapour pressure than expected and boils at an higher temperature
than expected.
o o
i.e., PA  PA . X A ; PB  PB . X B
o o
and PT  PA . X A  PB . X B
Eg. (i) Chloroform + acetone
(ii) Phenol + Aniline
(iii) Water + HNO3
Azeotropes :
* The binary mixture of two liquids having same composition in liquid and vapour phase and
boil at a constant temperature without change in the composition is called azeotrope (or)
constant boiling mixture.
* The solution of two liquids which show a large positive deviation form Raoult’s law form
minimum boiling azeotrope at a specific composition.
Eg. Ethanol + Water [95.6% ethanol + 4.4% water]
with boiling point
* The solution of two liquids which show a large negative deviation from Raoult’s law form
maximum boiling azeotrope at a specific composition.
Eg. HNO3 + H2O [68% HNO3 + 32% water]
with boiling point 393.5 K.
Colligative properties:

The property of a dilute solution which depends upon the number of particles of the
solute, irrestpective of their nature, present in the solution is called colligative property.

Lowering of vapour pressure, elevation of boiling point, depression of freezing point and
Osmatic pressure are the colligative properties.

Lowering of vapour pressure:

1) When a non volatile solute is dissolved in a pure liquid, then the vapour pressure of the
liquid decreases. This decrease in vapour pressure is called lowering of vapour pressure

2) According to Raoult’s law, vapour pressure of the solution depends upon the mole fraction
of the solvent

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 PA X A (or ) PA  PA0 . X A
P  PA0  PA  PA0  PA0 . X A
 PA0 1  X A   PA0 . X B
P
 XB
PA0
PA0  PA
 or  0  X B
PA

0
Here PA  PA = lowering of the vapour pressure

PA0  PA
PA0 = Relative lowering of vapour pressure

Relative lowering of vapour pressure is equal to the mole fraction of the non-volatile
solute present in the solution

PA0  PA nB
0
 XB 
PA nA  nB

For a dilute solution nA  nB

PA0  PA nB
 
PA0 nA
PA0  PA wM
 or  
PA0 mW

PA0  vapour pressure of pure liquid

PA  vapour pressure of solution

w  weight of solute in gram

W  weight of solvent in gram

m  molar mass of solute in gram

M  molar mass of solvent in gram

To calculate the molar mass of given unknown solute (m) accurately following
expression can be used

PA0  PA wM

PA mW

Elevation of boiling point:

When a non-volatile solute is added, then the boiling point of the liquid increases. This
increase in boiling point is termed as elevation of boiling point.
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 Reason: When a non-volatile solute is added, the vapour pressure of the liquid
decreases. To make the vapour pressure of the solution as 1 atm, higher temperature is
required where the solution boils. Hence boiling point elevates.

Tb0  B.P of pure liquid

Tb  B.P of solution

Tb  Tb0

Tb  Tb  Tb0

According to experimental studies, the elevation of boiling point is directly proportional

to the molality of the solution.

i.e., Tb molality

 or  Tb  Kb . molality
where Kb is the molal boiling point elevation constant (or) Ebullioscopic constant.

This expression is valid for non-volatile non-electrolyte solutes only.

The elevation in the boiling point observed by dissolving 1 mole of non-volatile

non-electrolyte solute in 1kg solvent is called Kb.

Kb is given by
2
R.Tb0
Kb 
1000.lv

0
Where Tb is the boiling point of pure liquid and lv is the latent heat of vapourisation

For water,

2  373  373
Kb 
1000  540
 0.515 K .mol 1.kg
0.52 K .mol 1.kg

1
* units of Kb= K .mol .kg

Depression of freezing point

* When a non-volatile solute is added, then the freezing point of the solvent decreases. This
decrease in freezing point is termed as depression of freezing point.

Reason: When anon-volatile solute is added then the vapour pressure of the solvent
decreases and now it would become equal to that of solid solvent at lower temperature. Thus
the freezing point of the solvent decreases.

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 T f0  F.P of pure solvent

T f  F.P of solution

T f  T f0  T f

* For a dilute solution, the depression of freezing point is directly proportional to the
molality of the solution

T f  molality

(or) T f  K f .molality

where Kf is called molal freezing point depression constant (or) cryoscopic constant.

w  1000
T f  K f .
m W

Where w  weight of solute in grams

W  weight of solvent in grams

m  molar mass of solute

 This expression is valid for non-volatile non-electrolyte solute only

The depression in the freezing point observed when one mole of non-volatile non-
electrolyte solute is dissolved in one kg solvent is called Kf.

Kf is given by

2
R.T f0
Kf 
1000.l f

Where
T f0 is freezing point of pure solvent

l f is latent heat of fusion

2  273  273
Kf 
For water, 1000  80

 1.86 K .mol 1.kg

units of
K f  K .mol 1.kg

Osmotic pressure:

The spontaneous flow of solvent molecules from solvent to solution through a

semi permeable membrane is called Osmosis.

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 The minimum pressure that must be applied on solution to stop the osmotic flow of
solvent molecules is called osmotic pressure.

 V  nRT
n
 or    RT
V
 or    CRT
w  1000
 or     R T
m V
Where c  molarity of the solution

n  number of moles of solute

V  Volume of the solution

w  weight of solute in grams

m  molecular weight of solute in grams

R  Molar gas constant

T  Absolute temperature

The solutions having same osmotic pressure at a given temperature are called isotonic
solutions.

At a given temperature, isotonic solutions will have same molarity.

 1  c1.RT
& 2  c2 RT
If  1   2 then c1  c2

Among two solutions, the solution with higher osmotic pressure is called hypertonic
while the solution with lower osmotic pressure is called hopotonic.

When blood cells are placed in 0.9% (w/v) aq.NaCl solution, then blood cells
collapse due to loss of water by osmosis. People taking a lot of salt experience water retention
in tissue cells and intercellular spaces because of osmosis. The resulting puffiness or swelling
is called edema.

When a pressure higher than osmotic pressure is applied, then solvent

molecules flow from solution to solvent through semipermeable membrane. This phenomenon
is called reverse osmosis.

From sea water, pure water is obtained by reverse osmosis.

Abnormal molar masses – Vant Hoff factor:

When the solute undergoes dissociation then the experimentally determined

molar mass is always lower than the true value.

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 When the solute undergoes association then the experimentally determined
molar mass is always higher than the true value.

This is called abnormal molar mass

To account for the degree of association or degree of dissociation, van’t Hoff introduced
a factor (i), known as Vant Hoff factor.

Vant Hoff factor (i) is given by

Number of particles of the solute present in the solution

i
Number of molecules of the soluteinitially taken
Observed colligative property

Calculate colligative property
Normal molar mass

Experimental molar mass

For dissociating solutes,

i>1

Normal molar mass > Experimental molar mass and

Observed colligative property > Calculated colligative property

Degree of dissociation (α) is given by

i 1

n 1
Where n is the number of ions per molecule obtained due to dissociation.

For associating solutes,

i<1; Normal molar mass < experimental molar mass and observed colligative property <
normal colligative property.

Degree of association (α) is givne by

1 i

1  1/ n
Where ‘n’ is the number of molecules undergoing association.

By considering ‘i’ value,

i) Relative lowering of vapour pressure

PA0  PA n 
0
 i B 
PA  nA 

ii) Elevation of boiling point

Tb  Kb .molality.i

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 iii) Depression of freezing point

T f  K f .molality.i

iv) Osmotic pressure:

  C.RT .i
As dilution increases, i value increases i.e., the degree of dissociation increases.

NUMERICALS

1. Calculate the mass percentage of benzene and carbontetrachloride of 22g of benzene is

dissolved in 122g of carbontetrachloride.

Mass of component
Mass percentage of benzene = 100
* Total mass
22
 100
22  122
= 15.28%.
122
Mass percentage of CCl4 = 100
144
= 84.72%.
2. Calculate the mole fraction of benzene in solution containing 30% by mass in CCl4.
* Mass of benzene = 30g
Mass of CCl4 = 100 – 30 = 70g

Mass
Number of moles of benzene =
gram molar mass
30

 0.3846
78
Mass
Number of moles of CCl4 
gram molar mass
= 70/154 = 0.4545.
n Benzene
Mole fraction of benzene =
Total moles
0.3846

0.3846  0.4545
= 0.4583.

Mole fraction of CCl4 = 1 – 0.4583 = 0.5417.

3. Calculate the molarity.
(a) 30g of Co(NO3)2.6H2O in 4.3L of solution.
(b) 30ml of 0.5 M H2SO4 diluted to 500ml

w 1000
M
(a) m  V (ml )
w = 30g, m = 291, V = 4300ml
30 1000
Molarity =   0.024M
291 4300
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 (b) M 1V1  M 2V 2

0.5  30  M 2  500
(or) M 2  0.03M
4. Calculate the mass of urea [MW = 60] required in making 2.5 kg of 0.25 molal aqueous
solution.

w 1000
Molality (m) =
* GMW  W
w = weight of solute = ?
GMW = 60, W = weight of solvate = 2500 g
Molality = 0.25 m

m  GMW  W 0.25  60  2500

w 
1000 1000
= 37.5 g.
5. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of the 20% (w/w)
aqueous KI solution is 1.202 g/ml.
(a) w = 20g, GMW = 39 + 127 = 166
W = 80g

w  1000 20  1000
molality (m) = 
GMW  W 166  80
 1.506 mol.kg 1
Mass 100
w  20 g , GMW  166, V  
(b) density 1.202
w 1000
molarity (M) =
GMW  V (ml )
20 1000
 1.202
166 100
= 1.448 M.

Mass of KI
Number of moles of KI =
(c) Molar mass of KI
20
  0.12
166
80
Number of moles of H 2O   4.444
18
Number of moles of KI
Mole fraction of KI =
Total moles
0.12
  0.0263
4.444  0.12 .
6. Concentrated nitric acid is 68% by mass. Density of the solution is 1.504 g/ml. Calculate
the molarity of the acid.
* w = 68g, GMW = 63,

Mass of solution 100

Volume of solution (v) = 
density 1.504

13
 w  1000
Molarity (M) =
GMW  V (ml )
68 1000
 1.504
63 100
= 16.234 M.
7. A glucose solution in water is 10% (w/w). Its density is 1.2 g/ml. Calculate (a) molality and
(b) mole fraction of each component.
* (a) w = 10, W = 90, GMW = 180

w  1000 10 1000
Molality (m) = 
GMW × W 180  90
= 0.617 m.
Mass of glucose
Number of moles of glucose =
(b) Molar mass
10
  0.0555
180
Mass of water
Number of moles of water =
Molar mass
90
 5
18
Number of moles of glucose
Mole fraction of glucose =
Total moles
0.0555

5.0555
= 0.011.

Number of moles of water

Mole fraction of water =
Total moles
5

5.0555
= 0.989.

(c) w  10, GMW  180

weight 100
Volume of solution = 
density 1.2
w 1000 10 1000
Molarity(M) =  1.2
GMW  V 180 100
= 0.667 M.
8. If N2 gas is bubbled through water at 293K, how many milli moles of N2 gas would dissolve
in 1 litre of water? Partial pressure of N2 is 0.987bar and Henry’s law constant of N2 at
293K is 76.48 k bar.
3
* P = 0.987 bar, K H  76.48 10 bar

X N2  ?
P  K H .X
P 0.987
X N2  
(or) K H 76.48 103
 1.28 105
14
 nN2 nN 2
X N2 
nN2  nH 2O nH 2O
n N 2  X N 2  nH 2 O 
1000
 1.29  105   7.17 104
18
= 0.717 milli mole.
9. If the solubility of H2S in water at STP is 0.195m, calculate Henry’s law constant.
* 0.195 m = 0.195 mole H2S in 1 kg water.

1000
nH 2 S  0.195, nH 2O   55.55
18
P  kH .X H2S
1
k H  P / X H2S 
0.195(0.195  55.55)
= 285.87 atm.

8
10. Henry’s law constant for CO2 in water is 1.67 10 Pa at 298K. Calculate the quantity of
CO2 in 500 ml of soda water when packed under 2.5atm CO2 pressure at 298K.

* K H  1.67 108 Pa
PCO2  2.5  1.013  105  2.5325  105
X CO2  ?
PCO2  k H . X CO2
PCO2 2.5325  105
X CO2  
(OR) kH 1.67 108
 1.5164 103
nCO2
X CO2 
nCO2  nH 2O
nCO2
1.5164 103 
(500 /18)
nCO2  1.5164  103  500 /18
= 0.042

weight of CO2 in 500ml water  0.042  44  1.853g .

3
11. The partial pressure of ethane over a solution containing 6.56 10 g of ethane is 1bar. If
2
the solution contains 5 10 g of ethane, then what shall be the partial pressure of the gas?
* According to Henry’s law,
Solubility (s) α partial pressure.

S1 P1

S2 P2
15
 P1.S 2 1 5 102
P2  
(or) S1 6.56 103
= 7.62 bar.
12. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the
normal boiling point of the solvent. What is the molar mass of the solute?

P o  Ps w  M

* Pso m W
1.013  1.004 2 18
 
1.004 m  98
m = 40.98 g/mol.
13. Heptane and octane form ideal solution. At 372K, the vapour pressure of the two liquids
are 104.2k Pa and 46.8k Pa respectively. What will be the vapour pressure of a mixture of
26 g of heptane and 35 g of octane.

Mass
No. of moles of heptane =
* Molar mass
26
  0.260
100
35
Number of moles of octane =  0.307
114
No. of moles of heptane
Mole fraction of heptane =
Total moles
0.26
  0.458
0.26  0.307
No. of moles of octane
Mole fraction of octane =
Total moles
0.307
  0.542
0.26  0.507
Vapour pressure of the solution (PS )  PAo . X A  PBo . X B
= 105.2 (0.458) + 46.8(0.542)
= 48.287 + 25.319
= 73.606k Pa.
14. The vapour pressure of two liquids A and B are 450mm of Hg and 700 mm of Hg
respectively at 350K. Find out the composition of the liquid mixture if total vapour
pressure is 600 mm of Hg. Also find the composition in the vapour phase.
o o
* PS  PA ( X A )  PB ( X B )
but X A  X B  1
(or) X B  1  X A
600  450( X A )  700(1  X A )
 450 X A  700  700 X A
(or) 250 X A  700  600  100
100
XA   0.4
250
16
 and X B  1  0.4  0.6
In liquid phase A and B are in 4 : 6 mole ratio.

PA  PAo . X A  450  0.4  180 mm of Hg

PB  PBo . X B  700  0.6  420 mm of Hg
P 180
YA  A   0.3
PS 600
P 420
YB  B   0.7
PS 600
In vapour phase A and B are in 3 : 7 mole ratio.
15. The vapour pressure of pure benzene at a certain temperature is 0.85 bar. A non-volatile
non electrolyte solid weighing 0.5g when added to 39 g of benzene [Molar mass = 78 g/mol].
Vapour pressure of the solution is 0.845 bar. What is the molar mass of the solid
substance?
* w = 0.5 g, M = 78 g/mol
m = ? W = 39 g

P o  0.85 bar , PS  0.845 bar

Po  PS w  M

Po m W
w  M  Po
m o
( P  PS )  W
0.5  78  0.850

0.005  39
= 170 g/mol.
16. 18g of glucose is dissolved in 1kg of water. At what temperature will water boil at
1
1.013bar? Kb for water is 0.52 K . Kg. mol .
* w = 18g, m = 180 g/mol

W  1kg  1000 g , K b  0.52 K . Kg. mol 1

Tb  ?
Tb  Kb .molality. i
w 1000
 0.52  1
m W
18 1000
 0.52  1
180  1000
= 0.052

Tb  Tbo  Tb
= 373 + 0.052 = 373.052 K.
17. The boiling point of benzene is 353.23 K. When 1.8g of a non-volatile solute was dissolved
in 90g of benzene, the boiling point is raised to 354.11K. Calculate the molar mass of
1
solute. Kb for benzene is 2.53K. Kg. mol .
1
* w = 1.8 g, W = 90 g, K b  2.53 K. Kg. mol

Tb  354.11  353.23  0.88K , m  ?

17
 Tb  Kb . molality. i
w 1000
=K b  i
m W
1.8 1000
0.88  2.53  1
m  90
2.53  1.8  1000
m  58 g/mol
0.88  90
18. Calculate the weight of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75g of acetic
1
acid to lower its melting point by 1.5 C . K f  3.9 K . Kg . mol
o

* w  ?, W  75 g , m  12  6  1 8  16  6  176
T f  1.5, K f  3.9 K. Kg. mol1
T f  K f . molality
w 1000
=K f .
m W
T f  m  W 1.5  176  75
w 
(or) K f 1000 3.9  1000
= 5.0769g.
19. Calculate the osmotic pressure in Pascals exerted by a solution prepared by dissolving 1.0
o
g of polymer of molar mass 185000 in 450ml of water at 37 C .

*   CRT

w 1000
  R T
m  V (ml )
11000
  0.0821 310
18500  450
 0.003057 atm
=0.003057 1.013 105 Pa
=309.69 Pa .
20. At 300K, 36g of glucose is present in a litre of its solution has an osmotic pressure of 4.98
bar. If the osmotic pressure of the solution is 1.52bar at the same temperature, what
would be its concentration?

*  1  C1 RT
 2  C2 RT
 1 C1

 2 C2
4.98 36 1000
 C2
1.52 180 1000
1 4.98 180

C2 1.52  36
1.52  36
C2   0.061 mol. lit 1
(or) 4.98  180

18
21. 19.5 g of CH2(F)COOH is dissolved in 500 g of water. The depression in the freezing point
o
of water observed is 1.0 C . Calculate the of fluoroactic acid.

* T f  K f  molality  i
w 1000
T f  K f  i
m W
19.5 1000
1.0  1.86  i
(or) 78  500
i  1.0753
i  1 1.0753  1
   0.0753
n 1 2 1
K a  C. 2 /(1   )
19.5 1000 (0.0753) 2
 
78  500 (1  0.0753)
 3.066 103 .
22. Calculate the depression in the freezing point of water when 10g of CH3CH2CHClCOOH is
3 1
added to 250 g of water. K a  1.4  10 & K f  1.86 K. Kg. mol .
2
* K a  C.
10 1000
1.4  103   2
122.5  250
 2  0.0042875
(or)   0.0655
i 1 i 1
 (or) 0.0655 =
n 1 2 1
(or) i  1.0655
T f  K f . molality. i
10  1000
 1.86   1.0655
122.5  250
 0.647 0.65K .
23. Determine the amount of CaCl2 (i = 2.47) dissolved din 2.5 lit of water such that its
0
osmotic pressure is 0.75 atm at 27 C .
*   0.75 atm, T = 27 + 273 = 300 K
Molar mass (m) = 40 + 71 = 111
V = 2.5 lit, R = 0.0821 lit. atm. K 1. mol1
  CRT . i
w 1000
  R T. i
m V
 .mV
.
w
1000  R  T  i
0.75 111 2500

1000  0.0821 300  2.47
 3.421g .

19
24. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4
o
in 2 lit of water at 25 C . Assume that K2SO4 dissociates completely.

*   CRT  i

w 1000 25 103  1000

C 
m  V (ml ) 175  2000
 7.143 105 Mol. lit 1
R  0.0821 lit. atm. K 1. mol1
T  25  273  298K
  ?, i  3
  7.143 105  0.0821 298  3
 1.7476 103 atm  3
 5.243 103 atm

****

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