SOLUTIONS
GIST:
HOW TO CALCULATE THE CONCENTRATION OF A SOLUTION:
1. MOLARITY CAN BE CALCULATED AS :
No. of moles of the solute
M =
Volume of solution in liters
2. MOLALITY CAN BE CALCULATED AS :
No. of moles of the solute
m =
Volume of the solvent in Kilo grams.
3.
NORMALITY CAN BE CALCULATED AS :
No. of gram equivalents of solute
N =
Volume of solution in liters
4.
MOLE FRACTION OF EACH COMPONENT :
No. of moles of the components
XA (or) XB =
Total Number of moles
HOW TO CALCULATE THE NUMBER OF MOLES:
Weight of the component
Number of moles =
Molecular weight of the component
1. .
XA =
nA
n A  nB
XB =
nB
n A  nB
Where XA and XB are the mole fractions of the components of A and B
4. XA + XB = 1 . If you calculate XA, you can calculate XB = 1 - XA
Henry’s law:. The law states that at a constant temperature, the solubility of a gas in
a liquid is directly proportional to the pressure of the gas.
The solubility of a gas in a liquid solution is a function of partial pressure
of the gas. If we use the mole fraction of a gas in the solution as a
measure of its solubility, then it can be said that
“The mole fraction of gas in the solution is proportional to the partial
pressure of the gas over the solution.”
p = KH x
Here KH is the Henry’s law constant.
X is molefraction of gas in the solution.
P partial pressure of the gas.
Henry’s law & its applications:
1) Solubility of a gas increases with decrease of temperature. It is due to this
reason that aquatic species are more comfortable in cold waters rather than in
warm waters.
2) To increase the solubility of CO2 in soft drinks and soda water, the bottle is
sealed under high pressure.
3. Scuba divers must cope with high concentrations of dissolved gases while
breathing air at high pressure underwater. Increased pressure increases the
solubility of atmospheric gases in blood. When the divers come towards
surface, the pressure gradually decreases. This releases the dissolved gases
and leads to the formation of bubbles of nitrogen in the blood. This blocks
capillaries and creates a medical condition known as bends, which are painful
and dangerous to life.
To avoid bends, as well as, the toxic effects of high concentrations of
nitrogen in the blood, the tanks used by scuba divers are filled with air
diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
4) At high altitudes the partial pressure of oxygen is less than that at the
ground level. This leads to low concentrations of oxygen in the blood and
tissues of people living at high altitudes or climbers. Low blood oxygen causes
climbers to become weak and unable to think clearly, symptoms of a condition
known as anoxia.
Ideal and Non-Ideal Solutions :
a) Ideal Solutions : The solution which obeys Raoult’s law exactly at all concentration
and all temperatures.
The ideal solution also have following characteristics:
i) It should obey the Raoult’s law (Total pressure, P = p0A xA + p0B xB )
ii) Heat exchange on mixing is zero ( Δ H mix = 0)
iii) Volume exchange on mixing is zero ( Δ Vmix = 0 )
b) Non ideal solutions:
The solution which do not obey Raoult’s law are called non ideal solutions. For
these solutions
i) pA # p0A xA
mixing # 0
p0B xB
and pB #
, ii) Δ H mixing # 0, iii) Δ V
Types of non ideal Solutions :
A) Non ideal solution showing positive deviation from Raoult’s law :
→ The Mixture contains two components namely A and B, If the interaction
between A-B molecules is weaker than the interactions between A-A or B-B or
both, then the solution deviate from the ideal behavior and Each component of
solution has a partial vapour pressure greater than expected on the basis of
Raoult’s law .
→The total vap. Pressure will be greater than corresponding Vap. Pressure
expecterd in case of ideal solution of the same composition
pA
>
p 0A
xA
and
pB >
p0B
xB
Total Vapour pressure P , P = (pA + pB) > ( p0A xA + p0B xB )
→ Example: Mixture of Ethyal alcohol and Cyclohexane.
→Explanation for positive deviation:
→In ethyl alcohol the molecules are held together by hydrogen bonding.
H
|
C2H5 − O ……. H – O……….H – O………
|
|
C2H5
C2H5
→When Cyclohexane is added to ethyl alcohol, the Cyclohexane
molecules occupy spaces between ethyl alcohol molecules. As a result
, some hydrogen bonds in alcohol molecules break. The escaping
tendency of alcohol and Cyclohexane molecules from solution
increases. So there is increase in vapour pressure. In such case
i)
ΔH
mixing
> 0,
ii)
ΔV
mixing
>0
B) Non ideal solution showing negative deviations from Raoult’s law:
→ The Mixture contains two components namely A and B, If the interaction
between A-B molecules is stronger than the interactions between A-A or
B-B or both, then the solution deviate from the ideal behavior and Each
component of the solution has a partial vapour pressure less than from
pure liquids. As a result, each component of solution has a partial vapour
pressure less than expected on the basis of Raoult’s law. Hence total
vap.pressure becomes less than the corresponding vap. pressure expected
in case of ideal solution
pA < p0A
xA and
pB < p0B
xB
Total Vapour pressure P , P = (pA + pB) < ( p0A xA + p0B xB )
→ Example: Mixture of Acetone and Chloroform
→Explanation for negative deviation:
→By Mixing Acetone and chloroform, a new attractive forces are formed due to
in intermolecular hydrogen bonding. Hence the attractive forces become
stronger and the escaping tendency of each liquid from the solution
decreases.
CH3
Cl
|
|
C = O-------------H - C - Cl
|
CH3
|
Cl
→So, the Vapour pressure of the solution is less than expected for an ideal solution
. In such case
(i) Δ H mixing < 0
(ii) Δ V mixing < 0
Azeotropes:
The binary mixtures having the same composition in liquid and vapour
phase and boil at a constant temperature.
Types of Azeotropes:
i) Minimum boiling azeotropes:
.
The solutions which show a large positive deviation from Raoult’s law
form minimum boiling azeotrope at a specific composition.
Example: Ethanol-water mixture (obtained by fermentation of
sugars) on fractional distillation gives a solution containing
approximately 95% by volume of ethanol. Once this composition,
known as azeotrope composition, has been achieved, the liquid and
vapour have the same composition, and no further separation occurs.
ii) Maximum boiling azeotropes:
The solutions that show large negative deviation from Raoult’s law form
maximum boiling azeotrope at a specific composition.
Example: Nitric acid and water is an example of this class of
azeotrope. This azeotrope has the approximate composition, 68% nitric
acid and 32% water by mass, with a boiling point of 393.5 K.
PROBLEMS ON COLLIGATIVE PROPERTIES:
1.
Relative Lowering of Vapour pressure:
Xsolute =
P o solvent  Psolution
P o solvent
P o solvent  Psolution
w  M solvent
=
P o solvent
M solute  W
(or)
(Xsolute =
w  M solvent
)
M solute  W
Let us identify each term in it :
Posolvent = Vapour pressure of the pure Solvent
Psolution = Vapour pressure of the solution.
XB = Mole fraction of the solute
w = Weight of Solute
W = Weight of Solvent
Msolvent = Molecular mass of solvent
Msolute = Molecular mass of solute
NOTE: From the value of mole fraction (XB), you can find out the molecular weight of the Non –
Volatile solute(MB).
XB =
nB
n A  nB
where nB=
WB
MB
Where WB = Weight of the Non – Volatile solute
MB = Molecular weight of the Non –Volatile solute.
2. Problems on Elevation of boiling point:
Msolute =
K b 1000  w
Tb  W
Let us identify the each term in it :
Msolute = Molecular weight of the non-volatile solute
Kb
= Molal Elevation constant
w
= Weight of the non – volatile solute
W = Weight of the solvent
Tb = Elevation of boiling point.
Students, it is observed that you are doing a mistake while calculating the Tb value. Its
value always should be in Kelvin only.
Tb  T2 K  T1K
Where T1K = Boiling point of the pure Solvent, T2K = Boiling point of the solution.
3. Problem on Depression in Freezing Point:
K f 1000  w
Msolute =
T f  W
Let us identify the each term in it :
Msolute = Molecular weight of the non-volatile solute
Kf
= Molal Elevation constant
w
= Weight of the non – volatile solute
W = Weight of the solvent
T f = Depression in Freezing point.
NOTE :Students it is observed frequently that you are doing mistake while calculating T f
value. Its value always should be in Kelvin only. Some times, T f value is given in the negative sign.
But, you should take only the positive value only.
Tf  T o f  T o f
Where Tof = Freezing point of pure solvent, Tf = Freezing point of the solution.
4. Problems on Osmotic Pressure :
  MRT
(or)

n
RT
V
(or)
Msolute =
wRT
V
Let us identify the each term in it :
 = Osmatic pressure in atmospheres only.
R = Gas constant = 0.0821 Ltrs. Atms.
V = Volume of the solution in Litres only
T = Temperature in Kelvin only.
n = No. of moles of the Non – Volatile solute.
M = Concentration of the solution in molarity.
NOTE :1. Students, the common mistake committed by most of you is substituting the ‘R’
value. It is observed that the value of ‘R’ is taken as 8.314 instead of 0.0821.
2. As the value of ‘R is in Ltrs Atms, care should be taken that the volume of the
solution should be in Litres and also the value of Osmotic Pressure in Atmospheres
inly.
3. Some times, Osmatic pressure value will be given in mm mercury. Hence, it is to be
converted into atmospheres. We know that 760mm Hg = 1 atmosphere.
4. Isotonic solutions were given to find out the molecular weight of the non – volatile
solute, Then equate both the Osimatic pressure values and use this formula:
n1 = n2
Where n1 , n2 are the number of moles of the two solutes.
HOW TO CALCULATE THE VAN’T HOFF FACTOR (i) , DEGREE OF ASSOCIATION AND DEGREE OF
DISSOCIATION.
Normal Moleculat Mass
i =
Observed Molecular mass
NOTE :In case of dissociation the value of i < 1 and in case of association the value of i > 1.
To calculate degree of dissociation and association:
1.For Dissociation:
i =
1
1
(or) i = 1 +

Let us Know :
i = Van’t Hoff factor,
 = Degree of dissociation
% of Dissociation = Degree of dissociation x 100
2.For Association:
i =1-

2
Let us Know : i = Van’t Hoff factor,
 = Degree of association
% of Association = Degree of Association x 100
Very Short Answr Type Questions (VSA) One Mark Questions:
1. What will be the mole fraction of water in C2H5OH solution containing equal number of
moles of water and C2H5OH?
2. Gases tend to be less soluble in liquids as the temperature is raised. Why?
3. What will happen to the boiling point of the solution formed on mixing two miscible liquids
showing negative deviation from Raoult’s law?
4. Which type of deviation is shown by the solution formed by mixing cyclohexane and
ethanol?
5. What is reverse osmosis? Give one large scale use of it.
6. What is the value of van’t Hoff factor (i) if solute molecules undergo dimerisation
SA (I) - TYPE QUESTIONS (2 - MARK QUESTIONS)
1. Show that the relative lowering of vapour pressure of a solvent is a colligative property.
2. Define azeotropes with one example of each type.
3. Conc. H2SO4 has a density 1.9g/ml and is 99% H2SO4 by weight. Find molarity of
solution.
4. State Raoult’s law for solutions of volatile liquid components. Taking a suitable
example, explain the meaning of positive deviation from Raoult’s law.
5. Calculate the mole fraction of ethanol in 40% of this solution by mass in water.
SA (II) TYPE QUESTIONS (3 - MARK QUESTIONS)
1. (a) State Henry’s Law.
(b) If O2 is bubbled through water at 393 K, how many millimoles of O2 gas would be dissolved
in 1L of water? Assume that O2 exerts a pressure of 0.95 bar.
(Given KH for O2 = 46.82 bar at 393K).
2. Given reason for the following :–
(a) Aquatic species are more comfortable in cold waters than in warm waters.
(b) To avoid bends scuba divers use air diluted with helium.
(c) Cold drinks bottles are sealed under high pressure of CO2.
3. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water,
the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is
much lower than the concentration of solvent, determine the vapour pressure (mm of Hg) of
the solution.
[Given : Kb for water = 0.76 kg mol–1] [Ans.: 724 mm of Hg] [Hint: ∆Tb = Kb.m ]
4. (a) Define osmotic pressure.
(b) Why osmotic pressure is preferred over other colligative properties for the determination of
molecular masses of macromolecules?
(c) What is the molar concentration of particles in human blood if the osmotic pressure is 7.2
atm at normal body temperature of 37°C? [Ans. : 0.283 M]
5. Describe a method of determining molar mass of a non-volatile solute from vapour pressure
lowering.
(b) How much urea (mol. mass 60 g mol–1) must be dissolved in 50g of water so that the
vapour pressure at the room temperature is reduced by 25%? Also calculate the molality of
the solution obtained.
[Ans. : 55.55 g and 18.5 m]
6. The vapour pressure of pure liquids A and B are 450 and 750 mm Hg respectively, at 350K.
Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find
the composition of the vapour phase. [Ans. : XA = 0.4, XB = 0.6, YA = 0.3, YB = 0.7]
Higher Order Thinking Skills
1.
Under what condition molality and molarity of a solution are identical. Explain with
suitable reason.
2. Addition of HgI2 to KI (aq.) shows decrease in vapour pressure. Why?
3. Account for the following :–
(a) CaCl2 is used to clear snow from roads in hill stations.
(b) Ethylene glycol is used as antifreeze solution in radiators of vehicles in cold countries.
(c) The freezing point depression of 0.01 m NaCl is nearly twice that of 0.01 m glucose
solution.
4. Give reasons for the following :–
(a) RBC swell up and finally burst when placed in 0.1% NaCl solution.
(b) When fruits and vegetables that have been dried are placed in water, they slowly
swell and return to original form.
(c ) A person suffering from high blood pressure is advised to take less amount of table
salt.
5. Glycerine, ethylene glycol and methanol are sold at the same price per kg. Which would be
cheaper for preparing an antifreeze solution for the radiator of an automobile?
6. Determine the correct order of the property mentioned against them :
(a) 10% glucose (p1), 10% urea (p2), 10% sucrose (p3) [Osmotic pressure]
(b) 0.1 m NaCl, 0.1 m urea, 0.1 m MgCl2
[Elevation in b.pt.]
(c) 0.1 m CaCl2, 0.1 m sucrose, 0.1 m NaCl [Depression in f.pt.]
7. Two liquids X and Y on mixing form an ideal solution. The vapour pressure of the solution
containing 2 mol of X and 1 mol of Y is 550 mm Hg. But when 4 mol of X and 1 mole of Y are
mixed, the vapour pressure of solution thus formed is 560 mm Hg. What will be the vapour
pressure of pure X and pure Y at this temperature? [Ans. : X = 600 mm Hg; Y = 400 mm Hg]
8. 2g of C6H5COOH dissolved in 25g of benzene shows depression in freezing point equal to
1.62K. Molar freezing point depression constant for benzene is 4.9 K kg mol–1. What is the
percentage association of acid if it forms a dimer in solution?
[Ans. : 99.2%]
9. Calculate the amount of NaCl which must added to one kg of water so that the freezing point
is depressed by 3K. Given Kf = 1.86 K kg mol–1, Atomic mass : Na = 23, Cl = 35.5).
[Ans. :
0.81 mol NaCl]