Solutin in chemistry

SOLUTION
[1]
When two or more chemically non-reacting substances are mixed, they form mixtures. Mixtures are of two types, Homogeneous
and Heterogeneous mixture.
A solution is a homogeneous mixture of two or more substances in variable proportion. The substances that make up the solution
are called the components of a solution. We will be concerned here of binary solutions only where our components are named as
solute and solvent.
A solute is a substance dissolved into other substance and a solvent is a substance in which dissolution takes place. Generally in a
solution the component present in smaller amount is called as solute and the component present in larger amount is called solvent.
But if solution is such that the state of aggregation of the solution is same as that of a component present in smaller amount, the
latter is called the solvent. Therefore solvent in a solution is its constituent substance which has the same state of aggregation as
that of the solution.
A homogeneous mixture is where the components that make up the mixture are uniformly distributed throughout the volume and
its components cannot be separated by any physical process. Eg –sugar-water.
A heterogeneous mixture is where the components of a mixture are not uniform throughout but are in different proportion and its
components can be separated by simple physical process. Eg : pizza.
Type of solution Solute Solvent Examples
Gaseous solution Gas
Liquid
Solid
Gas
Gas
Gas
Mixture of O2 and N2( Air).
Chloroform mixed with N2.
Camphor in nitrogen gas, smoke.
Liquid solution Gas
Liquid
Solid
Liquid
Liquid
Liquid
Oxygen dissolved in water, Soft Drinks.
Ethanol dissolved in water.
Glucose dissolved in water.
Solid Solution Gas
Liquid
Solid
Solid
Solid
Solid
Hydrogen in Palladium.(As Catalyst)
Amalgam of mercury with sodium, Liquid
additive on solid ice cream
Copper dissolved in gold.(Alloys)
The solution in which water is the solvent is known as aqueous solution; and where water is not the solvent is known as non-
aqueous solution.
CONCENTRATION OF SOLUTION:
The concentration of a solution is a macroscopic property. It is used to describe about the composition of a solution.
Concentration can be expressed qualitatively or quantitatively. By former we say about quality of the solution whether it is dilute
or concentrated. By latter method we say about the concentration of solution by measurement:
1. Per cent by mass: It is defined as amount of solute in gram present in 100g of a solution.
10% by mass of solution of glucose means 10 gram of glucose is present in 100 gram of solution
2. Per cent by volume: It is defined as volume of solute in ml present in 100ml of a solution.
10 % by volume means 10 ml of a component is dissolved in 100 ml of solution or 90 ml of solvent.
3. Per cent by mass by volume: It is defined as the mass of solute in gram present in 100 ml of the solution.
4. Parts per million: When solute particles is present in very minute amount i.e. traces we express the concentration in
parts per million parts of the solution ( ppm). It is defined as quantity of solute present in 106
gram of solution. It is
commonly used as measure of small levels of pollutants in air, water etc.
5. Molarity: It is number of moles of solute dissolved per litre of the solution. It is denoted by symbol “M”. 1 molar
solution means a solution which contains 1mole of solute dissolved per litre of solution.
6. Molality: It is number of moles of solute dissolved per 1000gm of solvent. It is denoted by “m.” 1 molal solution means
that 1 mol of solute is dissolved in 1000 g of solvent.

SOLUTION
[2]
TYPES FORMULA APPLICATION
MASS % Mass of the component
100
Total mass of the solution

In industrial applications like
commercial bleaching solution
contains 3.62 mass percentage of
sodium hypochlorite.
VOLUME % Volume of the component
100
Total Volume of the solution

Solution containing liquids are
commonly expressed in this unit.
Eg: 35% (v/v ) sol of ethylene
glycol = An antifreeze
MASS BY VOL
%
Mass of the solute dissolved
100
Volume of the solution

It is commonly used in medicine
and pharmacy.
PARTS PER
MILLION 6Number of parts of the component
10
Total number of parts of all components in the solution

It is used when solute is present in
traces. The concentration of
pollutants in water or air is
expressed in ppm.
MOLE
FRACTION
(X)
Number of moles of one component
Total number of moles of all components.
It is used to relate physical
properties of solutions.
MOLARITY(M)
Moles of solute
Volume of the solution in L
solute
solute
= 1000
m (in ml)solution
w
V


Temperature dependent.
MOLALITY(m)
Moles of solute
Mass of the solvent in Kg
solute
solute
= 1000
m solvent
w
w


Temperature independent.
NORMALITY
(N)
No. of gram equivalent of soulte
Volume of solution
N 
Temperature Dependent
FORMALITY
F =
Mass of solute
Formula Mass of solute x Volume of solution in L
Temperature dependent
Molality is considered better for expressing the concentration as compared to molarity because of expansion and contraction of
liquid with temperature. Volume of solution increases with rise in temperature and as a result molarity also decreases. Molality
has no volume term rather mass term therefore it remains same all where ; It does not change with temperature. Molality and
solubility is related by following relation:
Molality (m) =
Solubility 10
Molecular mass of solute

where
Mass of solute in gram
Solubility
Mass of solvent in gram
100 
Therefore molarity of solution at higher temperature will be less than that of solution at lower temperature.
If a solution having molarity M1 and volume V1 is diluted to volume V2 so that the new molarity is M2, then as total number of
moles of solution remains same, we have
If V1 ml of solution of molarity M1 is mixed with V2 ml of solution of molarity M2 then overall molarity of the solution will be -
For a solution containing x% solute by mass and solution having density “d”
solute
M =
x 10
m
d 
where msolute
is molecular mass of solute.
7. Mole fraction: The mole fraction, X, of a component in a solution is the ratio of the number of moles of that component to
the total number of moles of all components in the solution.
M1 x V1 = M2 x V2
2 2 1 1
3
1 2
M M
M
V V
V V




SOLUTION
[3]
The mole fraction of A, XA, in a solution consisting of A, B, C, ... is calculated using the equation:
and
Normality (N): It is the number of gram equivalent of solute dissolved per litre of solution.(Unit : gm equiv/L)
No. of gram equivalent of soulte
Volume of solution
N 
where
Gm equiv of solute =
Mass of soulte
Equivalent weight
RELATION OF
MOLARITY, MOLALITY AND NORMALITY
1
1 2 2
10001000
1000
XM
Molality
d Mm X m
 

N = M x Basicity (for acids)
N = M x Acidity (For bases)
Acidity = Number of replaceable OH—
ions in a solution.
Basicity = Number of replaceable H+
ions in a solution.
Molarity x Molecular Mass = Strength of the solution (gram per litre)
Normality x Equivalent Mass = Strength of the solution (gram per litre)
Hence, Molarity x Molecular mass = Normality x Equivalent mass
EQUIVALENT WEIGHT
In all chemical reaction the reactants and products obey law of equivalent proportion i.e. their equivalents are same.
aA + bB cC +dD ; a,b,c,d, are the stoichiometric coefficients. By law of equivalence
CA B D
A B C D
Constant
WW W W
E E E E
   
(I) Equivalent weight of Base Line Elements : Al (OH)3
EH = 1, EO = 8, ECl =35.5
2H2 + O2 = 2H2O
4g H2 is equivalent to 32 g O2
Therefore 1g is equivalent to 8g = EO.
(II) Equivalent weight of Radicals : ER =
Ionic weight
Total +ve/-ve charge
ENH4+ = (14+4x1/ 1) = 18
(III) Equivalent weight of Acids : EA = +
Molecular weight Molecular weight
Basicity of AcidNo. of replaceable H ion

E H2SO4 = 98/2 = 49
(IV) Equivalent weight of Bases : EB = -
Molecular weight Molecular weight
Acidity of BaseNo. of replaceable OH ion

E Al (OH)3 = M/3
Q.1 Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass.
Ans : Xglycol = 0.068
Q.2 Calculate the molarity of a solution containing 5g of NaOH in 450 mL solution.
Ans: M = 0.278 mol/L

SOLUTION
[4]
Q.3 Calculate Molality of 2.5 g of ethanoic acid (CH3 COOH) in 75 g of benzene (C6H6). Ans: 0.556 mol/kg
Q.4 Calculate the mass percentage of benzene (C6H6 )and carbon tetra chloride (CCl4 ) if 22g of benzene is dissolved in 122g of
carbon tetrachloride.
Q.5 Calculate the mole fraction of benzene in the solution containing 30% by mass in carbon tetrachloride.
Q.6 Calculate the molarity of each of the following solutions: (a) 30g of Co(NO3)2.6H2O in 4.3 L of the solution (b) 30 mL of 0.5
M H2SO4 diluted to 500mL.
Q.7 Calculate the mass of urea () required in making 2.5 kg of 0.25 molal aqueous solution.
Q.8 Calculate (a) Molality (b) Molarity (c) Mole fraction of KI if the density of 20% (w/w) aqueous KI is 1.202 g/mL.
Q.9 Sucrose solution which is 4% by mass is heated till it becomes 5% by mass. What is the water lost from 100g of the solution.
Q.10 Calculate the Molality of 1M solution of NaNO3 . The density of the solution is 1.25 g/cm3
. Ans: 0.86 mol/Kg.
Q.11 What is the Molality of NH3 in a solution containing 0.85g of NH3 in a 100 cm3
of liquid having density 0.85 g/cm3
.
Ans: 0.59 mol/kg
Q.12 0.75g of NaHCO3 is dissolved in 250 ml of a solution. Calculate its molarity and Normality?
Ans: M = 0.0357 mol/L , N = 0.0357 equiv /L
Q.13 Calculate the normality and molarity of H2SO4 in a solution containing 9.8 g of per dm3
of the solution.
Ans: M = 0.1 mol/L, N = 0.2 equiv / L
Q.14 Calculate the molarity of pure water. (density = 1 g/ml).
Q.15 Calculate the moles of methanol in 5 L of its 2m solution if density of the solution is 0.981 kg/L.
Ans: 9.22 mol
Q.16 Concentrated nitric acid used in the laboratory work is 68% nitric acid bys mass in the aqueous solution. What should be the
molarity of such sample of the acid if the density of the solution is 1.504 g/ mL.
Q.17 A solution of glucose in water is labelled as 10% (w/w) , what would be the Molality, Molarity of the solution and mole
fraction of each component in the solution? (Density of solution = 1.2 g/mL).
Q.18 How many mL of 0.1 M HCl are required to react completely with 1 g mixture of NaCO and NaHCO containing equimolar
amounts of both ?
Q.19 A solution is obtained by mixing 300g of 25% solution and 400g of 40% solution by mass. Calculate the mass percentage of
the resulting solution.
Q.20 An antifreeze solution is prepared from 222.6 g of ethylene glycol and 200 g of water. Calculate the Molality of the solution.
If the density of the solution is 1.072 g/mL then what shall be the molarity of the solution.
Q.21 A sample of drinking water was found to be severely contaminated with chloroform supposed to be carcinogen. The level of
contamination was 15 ppm (by mass).
(1) express this in percent by mass.
(2) determine the molality of chloroform in the water sample.
Q.22 If the density of some lake water is 1.25 g/ml and contains 92g of Na+
ions per kg of water, calculate the Molality of Na+
ions in the lake.
Q.23 Calculate the mass percentage of aspirin () in acetonitrile () when 6.5g of aspirin is dissolved in 450g of acetonitrile.
Q.24 Nalorphene similar to morphine is used to combat withdrawal symptoms in narcotic users. Dose of Nalorphene generally
given is 1.5mg. Calculate the mass of 1.5x 10-3
m aqueous solution required for the above dose.
Q.25 Calculate the amount of benzoic acid required for preparing 250mL of 0.15 M aqueous solution in methanol.

SOLUTION
[5]
SOLUBILITY
Solubility of a substance at a given temperature is defined as amount of solid that dissolves in 100g of the solvent at a given
temperature to form a saturated solution. The solubility of one substance in another depends on the attractive forces between its
particles as well as those between the molecules of a solvent.
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent.
Such a solution in which no more solute can be further dissolved at a given condition of temperature and pressure is called
saturated solution. A saturated solution is those in which dissolved solute is in equilibrium with un-dissolved solute.
A solution in which more solute can be dissolved at the same temperature and pressure is called unsaturated solution.
Solubility of a solid in liquid depends upon
1. Nature of solvent and solute: As we know that like dissolves in like therefore polar compounds dissolve readily in polar
solvent like water, methanol etc and non polar compounds an readily dissolve in non-polar solvents like benzene,
toluene, carbon tetrachloride etc. ( Polar solvents are molecules whose electric charges are unequally distributed leaving
one end of each molecule more positive than the other) .
2. Effect of temperature: Variation of solubility of a solid with temperature depends upon enthalpy of solution. (Enthalpy is
energy required for a substance to change itself from one state/ phase to another state / phase.)
(a) the solubility of solute increases with increase in temperature because the dissolution process is endothermic.
Solute + solvent + Heat = Solution ∆H = +ve
(b) The solubility of solids decreases with increase in temperature because the dissolution process is exothermic.
Solute + Solvent = Solution + Heat ∆H = -ve
3. Effect of pressure: The effect of pressure on solubility of solids in liquids is generally very small or insignificant because
solids and liquids are highly incompressible.
Solubility of Gases in Liquids
1. Nature of the Gas and Solvent: Generally. The gases which can be easily liquefied are more soluble in common
solvents. For example CO2 is more soluble in water than O2 or H2. The gases which react with the solvent possess high
solubility. For example HCl and NH3 are highly soluble in water.
2. Effect of temperature: As the temperature increases the solubility of a gas in liquid decreases as the dissolution is an
exothermic process. More gas is present in a solution at lower temperature as compared to the solution at higher
temperature. The reason of this gas solubility relationship is that vapour pressure increases with temperature. Increased
temperature causes increase in K.E of gaseous molecules as a result of which molecules break intermolecular bonds and
escape from the solution.
3. Effect of Pressure: The solubility of gaseous particles increases with increase of pressure.
Henry’s Law: The solubility of a gas in a liquid is directly proportional to the pressure of gas over the solution at a
definite temperature. The solubility (i.e. mass of gas dissolved per unit volume of the solvent) at constant temperature is
directly proportional to the pressure of gas over the solution.
m α P
m = KP
where K is proportionality constant. The magnitude of K depends upon the nature of gas. Nature of solvent, temperature
and the units of pressure.
If solubility of the gas is known at one particular pressure, then it can be calculated at other pressures using the following relation :
1 1
2 2
m P
m P

Where m1 is solubility of gas at pressure P1 and m2 is the solubility of the gas at pressure P2 .
Henry’s Law restated: The partial pressure of a gas in vapour phase is proportional the mole fraction of gas (X) in the solution.
Mathematically
P = KHX
Where KH is Henry’s law constant .

SOLUTION
[6]
 KH is a function of nature of gaseous molecule.
 Higher the value of KH at a given pressure lower the solubility.
 For any gas the value of KH increases with temperature.
Limitations of Henry’s Law: This law is valid if
 Pressure is low. (To avoid deviation)
 Temperature is not very low.
 Gas does not combines chemically with solvent i.e. gas do not either associate or dissociates in the solvent.
Application of Henry’s Law:
 In production of carbonated beverages: Soda water which is solution of CO2 gas in water is sealed under high pressure to
increase the solubility of CO2 in the soft drink.
 In deep sea diving: The phenomenon of CO2 in soda water is same as is of concern in scuba diving. As divers descend, they
are under increasing pressure due to the increasing mass of water above them. Each 33 ft of water exert a pressure of 1 atm.
As the diver descends, the pressure on the air in the lungs increases rapidly, and the blood dissolves more than normal
amounts of both nitrogen and oxygen. As the diver comes up, the pressure decreases and the gases come out of solution. If
the change in pressure is too swift, the bubbles of gas are released into the blood stream and tissues, instead of into the lungs.
The bubbles cause sharp pain wherever they occur, most notably in the joints. This dangerous and sometimes fatal condition
is called bends, the name deriving from the temporary deformities caused by the affected diver's being unable to straighten
his or her joints. Because helium, at all pressures, is less soluble than nitrogen, divers often use a mixture of breathing gas
(11.7% He, 56.2 % N2 & 32.1 % O2 ) to decrease chances of bends.
 At high altitudes: concentration of oxygen at high altitudes is less as a result of which people living at high altitudes have
less oxygen in their tissues and blood. Deficiency or low concentration of oxygen causes ANOXIA i.e. inability to think and
act properly.
Q. 26 If N2 is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water. Assume that N2
exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293 K is 76.48 kbar. Ans: 0.716mmol
Q.27 H2S a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is
0.195m, calculate Henry’s law constant.
Q.28 Henry’s law constant for carbon dioxide in water is 1.67x108
Pa at 298 K. Calculate the quantity of CO2 in 500ml of soda
water when packed under 2.5 atm CO2 pressure at 298 K. CBSE 2008 Comptt.
Q.29 What role does the molecular interaction play in the solution of alcohol and water?
Ans: The intermolecular interaction between alcohol and water leads to
 Complete miscibility of alcohol and water.
 Positive deviation from Raoult’s law.
 Formation of a minimum boiling Azeotropes.
Q.30 Why do gases nearly always tend to be less soluble in liquids as the temperature is raised?
Ans: When temperature is raised, the molecules of dissolved gas prevent in a liquid gain kinetic energy. Higher the kinetic energy
of the gas molecules make them to escape from its solution. That is why, gases tend to be less soluble in liquids at higher
temperature.
Q.31 The partial pressure of ethane over a saturated solution containing 6.56 x 10-2
g of ethane is 1 bar. If the solution contains
5.00 x 10-2
g of ethane, then what shall be the partial pressure of the gas.
Ans: Partial pressure of ethane over a saturated solution = 1bar.
Mass of ethane in the saturated solution at 1 bar = 6.56 x10-2
g
Mass of ethane in the solution at 1 bar = 5.00 x10-2
g
Partial pressure of ethane gas = ?
According to Henry’s Law,

SOLUTION
[7]
Partial pressure of the gas = KH x Mole fraction of gas in the solution
So, 1 bar α KH x 6.56 x10-2
g (i)
And p bar α KH x 5.00 x 10-2
g (ii)
Solving both equations we have p = 0.76 bar.
Q.32 Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 105 mm Hg. Calculate the solubility of
methane in benzene at 298 K under 760 mm Hg.
Q.33 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of
20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law
constants for oxygen and nitrogen at 298 K are 3.30 107 mm and
6.51 107 mm respectively, calculate the composition of these gases in water.
VAPOUR PRESSURE OF LIQUIDS
TO VACUUM PUMP
CLOSED VESSEL (TEMPERATURE FIXED)
When a volatile liquid is allowed to evaporate in a closed vessel, due to evaporation liquid changes into vapour and level of liquid
decreases. As the evaporation proceeds number of gaseous particles in vapour phase increases . As a result of which molecules
strikes the surfaces of liquid. Slowly an equilibrium is established between vapour and liquid phase. At this point the maximum
pressure exerted by vapours above the liquid surface at a given temperature is called vapour pressure.
The partial vapour pressure is a measure of the tendency of molecule of the components to escape from the solution into the
vapour state.
Vapour pressure of a liquid depends upon
1. Nature of Liquid: The liquids which have weaker intermolecular forces tend to escape readily into vapour phase and
therefore have greater vapour pressure.
Q. Suppose there are two liquids A and B which are volatile in nature. If liquid A is more volatile than liquid B the What
is relation between their vapour pressure values?
2. Temperature : The vapour pressure of a liquid increases with increase in temperature.
3. Presence of impurities: The presence of non-volatile impurities lowers the vapour pressure.
A: VAPOUR PRESSURE OF A LIQUID – LIQUID SOLUTION
Raoult’s Law: At a given temperature for the solution of volatile liquids the partial vapour pressure of a component is equal to the
product of vapour pressure of pure component and its mole fraction. Mathematically PA = P0
AXA
PB = P0
BXB
According to Dalton’s Law of partial pressure PT = PA + PB = P0
AXA + P0
BXB
PT = P0
A+ (P0
B – PA
0
)XB
PT = PB
0
+ (P0
A – PB
0
)XA
We can see from above relation that vapour pressure of the components are linear function of their mole fraction
 When XA =1, then XB = 0 and PA = PA
0
and PT = PA
0

When XA =0, then XB = 1 and PB = PB
0
and PT = PB
0
O O O O
O O O VAPOUR O
O O O O
………LIQUID………..………………
……………………………………………
………………………………..

SOLUTION
[8]
From the curve it is clear that when there is component A or volatile liquid A in the container then the vapour pressure of the
liquid is PA
0
.As soon as component B or liquid B is mixed with liquid A vapour pressure of liquid A decreases till it becomes
zero.
COMPOSITION OF VAPOUR PHASE IN EQUILIBRIUM WITH SOLUTION
If YA and YB are mole fraction of components in vapour phase then according to Dalton’s law of partial pressure
PA = YA PTotal
PB = YB PTotal
B. VAPOUR PRESSURE OF SOLUTION OF SOLID IN LIQUID
In a pure liquid the entire surface is occupied by the molecules of liquid. When a non-volatile solute is added to the volatile liquid
solvent to make a solution the vapour pressure of the solution is solely due to solvent atoms. There is a decrease in vapour
pressure of solvent due to addition of non-volatile solute .The decrease in vapour
Mathematically: (Vapour pressure of solvent) PA α XA (mole fraction of solvent)
PA = P0
XA
Generalised form of Raoult’s Law: For any solution the partial vapour pressure of each volatile component in the solution is
directly proportional to its mole fraction.
Q.32 Heptane and Octane form ideal solution. At 373K, the vapour pressure of the two liquid components are 105.2Pa and 46.8
kPa, respectively. What will be the vapour pressure, in bar of a mixture of 25.0 g of heptane and 35.0 g of Octane?
Ans: PHeptane
0
= 105.2 kPa POctane
0
= 46.8 kPa
WHeptane = 25.0 g Woctane = 35.0 g
Mheptane = 100 g/mol Moctane = 114 g/mol
No. mole of heptanes = 25.0
100 /
g
g mol
= 0.25 mol No. of mole of octane = 35.0
114 /
g
g mol
= 0.31 mol
Mole fraction of Heptane =
tan
tan tan
hep e
hep e oc e
n
n n
mole fraction of Octane = an
an n
oct e
oct e hept e
n
n n
= 0.446 = 0.554
Vapour pressure of mixture , P = Xh x Ph + Xo x Po
= 0.446 x 105.2 Pa + 0.554 x 46.8 kPa
P = 72.8 kPa = 0.728 bar
Q.33 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 molecular mass of the solute?
Q.34 The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molar solution of a non volatile solute
in it.
Q.35 Vapour pressure of chloroform and Dichloromethane at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate the
vapour pressure of the solution prepared by mixing 25.5 g of chloroform and 40g of Dichloromethane at 298 K and Mole fractions
of each components in vapour phase.
Q.36 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively at 350 K. Find out the composition of the
liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour pressure.
Q.37 The vapour pressure of two liquids P and Q are 80 and 60 torr respectively. What will be the total vapour pressure of the
solution obtained by mixing 3 mol of P and 2 mol of Q?

SOLUTION
[9]
Q.38 Two liquids X and Y form an ideal solution at 300 K. Vapour pressure of the solution containing 1 mol of X and 3 mole of
Y is 550 mm of Hg. At the same temperature if 1mol of Y is further added to this solution vapour pressure of solution increases by
10 mm of Hg. What will be the vapour pressure of X and Y in their pure state.
Q.39 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour
pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the
solution if the total vapour pressure of the solution is 475 Torr.
IDEAL AND NON-IDEAL SOLUTION
IDEAL SOLUTION
Such solution which obey Raoult’s Law are known as Ideal Solution. i.e. PA = P0
XA
The total volume of an ideal solution will be equal to the sum of the volume of components mixed. i.e. Mixing
V 0 
When two pure components mix to form an ideal solution there is no evolution of heat or absorption of heat. i.e. enthalpy of
mixing is zero. i.e. Mixing
0H 
Example of ideal solutions are:- 1) Benzene and Toluene 2) n- Hexane and n- Heptane
3) Bromoethane and Iodoethane 4) Chlorobenzene and Bromobenzene
NON-IDEAL SOLUTION:
Such solutions which do not obey Raoult’s Law are known as Non –Ideal solutions. The deviation from Raoult’s law are due to
the differences in the molecular structure of the two components which result in the difference in the intermolecular forces.
Volume on mixing is not equal to zero. i.e. Mixing
V 0 
Enthalpy of mixing is not equal to zero. i.e. Mixing
0H 
There are two types of non- ideal solutions:
NON IDEAL SOLUTION SHOWING POSITIVE DEVIATION
If the cohesive forces between the unlike components (A-B) are weaker than those pure liquids (A-A, B-B) the escaping
tendencies of the components in the solution are higher than that predicted by Raoult’s Law. Cohesive forces are intermolecular
forces (like H-bonding , vander waal force) which causes tendency in liquids to resist separation. These attractive forces exist
between molecules of same substance. The total vapour pressure of the binary solution is greater than the sum individual partial
pressure of two components. Such solutions are said to boil at lower temperature than the boiling point of either of the pure
components. Eg: In a solution of ethyl alcohol and water the boiling point of C2H5OH is 78.40
C , that of water is 1000
C and that of
the solution is 78.20
C.
Mathematically: PA > P0
AXA Mixing
V 0 
PB > P0
BXB Mixing
0H 
Example: Ethyl Alcohol + Cyclo-hexane ; Acetone + Carbon Disulphide; Benzene + Acetone; Carbon Tetrachloride +
Chloroform ; Ethyl Alcohol + water ; Acetaldehyde + Carbon disulphide; Ethyl Alcohol + Chloroform ; water + Propyl alcohol
NON IDEAL SOLUTION SHOWING NEGATIVE DEVIATION
If the cohesive forces (attractive forces) between the unlike components (A-B) are stronger than those pure liquids (A-A, B-B)
the escaping tendencies of the components in the solution are lesser than that predicted by Raoult’s Law. The total vapour
pressure of the such binary solution is lower than the sum individual partial pressure of two components. Such solutions are said
to boil at higher temperature than the boiling point of either of the pure components. Eg: In a solution of HCl and H2O ,the boiling
point of HCl is -850
C and that of H2O is 1000
C but the boiling point of the solution is 1100
C.
Mathematically: PA < P0
AXA Mixing
V 0 

SOLUTION
[10]
PB < P0
BXB Mixing
0H 
Example: Acetone + Chloroform ; Chloroform + Diethyl ether; Chloroform + Nitric Acid; Acetone + Aniline; Water + Nitric
Acid; Water + HCl
AZEOTROPES (A means NO, Zeo means Boil, and Tropes means change i.e No change on boiling)
Azeotropes are binary mixtures which have same composition in liquid and vapour phase. i.e. X1 = Y1 & X2 = Y2. They show
larger deviation from Raoult’s law. Since their composition is same in both phase therefore they boil at constant temperature. It is
unable to separate such mixture using fractional distillation because for mixture to be separated by fractional distillation their
composition should be different in liquid and vapour phase.
The liquid mixture which boil at constant temperature and remain unchanged in composition are called Azeotropes. There are two
types of Azeotropes
1. Minimum Boiling Azeotropes: They have maximum values in vapour pressure and minimum in boiling point. Such
mixtures are obtained in case of positive deviation. Eg: Ethyl Alcohol and water
2. Maximum boiling Azeotropes: They have minimum value in vapour pressure and maximum in boiling point. Such
mixtures are obtained in case of negative deviation. Eg: HCl and water.
Q.40 Suggest the most important type of intermolecular attractive interaction in the following pairs.
i. n-Hexane and n- octane
ii. I2 and CCl4
iii. NaClO4 and H2O
iv. Methanol and acetone
v. Acetonitrile (CH3CN) and acetone
Ans: i. Dispersion forces. ii. Vander waal force iii. Ion- Dipole interaction iv. Dipole –Dipole interaction, H-bonding, v.
Dipole – dipole interaction.
Q.41 Based on solute solvent interactions, arrange the following in order of increasing solubility in n- octane and explain.
Cyclohexane , KCl, CH3OH, CH3CN.
Ans: The increasing order of polarity is : Cyclohexane < CH3CN < CH3OH < KCl
The order of increasing solubility in n-octane is
KCl < CH3OH < CH3CN < Cyclohexane
The n-octane is a nonpolar molecule. So non polar molecule like cyclohexane will have higher solubility in it and ionic salt like
KCl will have lowest solubility.
Q.42 Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(1) phenol (2) Toluene (3) Formic acid (4) Ethylene glycol (5) Chloroform (6) Pentanol

SOLUTION
[11]
Ans: (1) Phenol has polar group –OH and non-polar group C6H5. Thus is partially soluble
(2) Toluene has no polar groups. Thus it is insoluble in water.
(3) Formic acid has the polar group – OH and can form H-bond with water. Thus it is Highly soluble.
(4) Ethylene glycol has polar OH group and can form H- bond. Thus it is highly soluble in water.
(5) Chloroform is insoluble in water.
(6) Pentanol has polar group –OH, but it also contains a very bulky non-polar group C5H11. Therefore it is partially
soluble.
COLLIGATIVE PROPERTIES
Properties of the solution which depends only upon the number of solute particles present in the solution and independent from the
nature of solute particles are known as colligative properties. These are also called Democratic properties as in democratic country
there is preference to the number of people.
Types of Colligative Properties
1. Relative lowering of Vapour Pressure
2. Elevation in Boiling Point
3. Depression in Freezing Point
4. Osmotic pressure
All colligative properties are interrelated with each other . i.e. by knowing the value of any one of the colligative property others
can be known. All colligative properties depends on the Molality of the solution.
Condition of Colligative Property
Solution should be dilute, solute should be non-volatile. Sugar, NaCl , KNO3 are examples of non volatile solute, and the solute
does not dissociate or associate in the solution.
RELATIVE LOWERING IN VAPOUR PRESSURE:
At any given temperature the vapour pressure of a solution containing non-volatile solute is less than that of pure solvent.
FIGURE: The vapour pressure of pure water is shown as a solid line; the vapour pressure of an aqueous solution is shown as a dashed line. Note
the differences between the solution and the pure substance in melting point and boiling point.

SOLUTION
[12]
FIGURE Vapour pressure lowering: (a) the vapour pressure of a pure liquid; (b) the vapour pressure of a solution.
The surface of a pure solvent (Figure) is populated only by solvent molecules. Some of these molecules are escaping from the
surface, and others are returning to the liquid state (see Section 10.3A). The surface of a solution is populated by two kinds of
molecules; some are solvent molecules others are solute molecules. Only the solvent molecules are volatile. They alone can escape
to build up the vapour pressure of the solution. There are fewer solvent molecules on the surface of the solution than on the
surface of the pure liquid. Fewer will vaporize and, as a consequence, the vapour pressure of the solution will be less than that of
the pure liquid at the same temperature.
Experimentally for a dilute solution: Lowering in Vapour Pressure is given by
∆P1 = P1
0
– P1
∆P1 = P1
0
- P1
0
X1 = P1
0
(1-X1) = P1
0
X2
0
1 1 2 2
0
1 2 1 1
0
1 1 2 1
0
1 1 2
p p n n
p n n n
p p w M
p w M


 
 


Q.43 Calculate the mass of a non volatile solute (molecular mass = 40) which should be dissolved in 114 g octane to reduce its
vapour pressure to 80%.
Q.44 A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further,
18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the
solute (ii) vapour pressure of water at 298 K.
Q.45 Vapour pressure of water at 200
C is 17.5 mm of Hg and lowering of vapour pressure of a sugar solution is 0.061 m of Hg.
Calculate the (i) Reltive lowering of vapour pressure
(ii) Vapour pressure of the solution
(iii) Mole fraction of sugar and water.
ELEVATION IN BOILING POINT
The boiling point of a liquid is the temperature at which the vapour pressure of the liquid equals atmospheric pressure. The vapour
pressure of the liquid is lowered when a non-volatile solute is added to it. To boil the solution the reduced vapour pressure of the
solution must be raised to 1 atmosphere. To increase the vapour pressure more heat is required. Thus, the boiling point of a
solution containing a non-volatile solute is higher than that of the pure solvent (see Figure) This effect is called boiling point
elevation.
Elevation in Boiling point (∆Tb) = Boiling point of the solution(Tb) – Boiling point of the pure solvent(∆Tb
0
)
∆P1/ P1
0
= X2

SOLUTION
[13]
Experimentally for a dilute solution it was found that ∆Tb α m
∆Tb = Kb m
Where Kb is called Boiling point elevation constant, Molal elevation constant or Ebullioscopic constant. Kb is defined as molal
elevation in boiling point for a 1 Molal solution. Unit = K.Kg/mol
DETERMINATION OF MOLECULAR WEIGHT OF NON-VOLATILE SOLUTE
b b
2
b b
1
2
b b
2 1
b 2
2
b 1
T = K m
T = K 1000
T = K 1000
K
= 1000
T
n
w
w
M w
w
M
w

 
 


Q.46 The boiling point of benzene is 353.23 K. When 1.8 g of a non volatile solute was dissolved in 90g of benzene the boiling
point is raised to 354.11 K. Calculate the molar mass of the solute. (KB of benzene = 2.53 K.Kg/mol)
Ans: 57.5 g/mol
Q.47 18 g of glucose is dissolved in 1000 g of water in a saucepan. At what temperature will water boil at 1.013 bar? (KB of water
is 0.52 K. Kg/ mol) Ans: 373.098 K
DEPRESSION IN FREEZING POINT
Freezing and melting point are two terms that describe the same temperature, at which the vapour pressure of the solid equals the
vapour pressure of the liquid and at which the solid and the liquid are in dynamic equilibrium. Remember, too, that vapour
pressure decreases as the temperature decreases. The vapour pressure of a solution is lower than that of the solvent, so the vapour
pressure of a solution will equal that of the solid at a lower temperature than in the case of the pure solvent. Thus, the freezing
point will be lower for a solution than for the pure solvent (see Figure 11.6). This effect is called freezing point depression.
Remember that, just as it is the solvent that vaporises when a solution boils, it is the solvent, not the solution , that becomes solid
when a solution freezes. When a salt solution freezes, the ice is pure water (solid); the remaining solution contains all the salt.
Depression in Boiling point (∆Tf) = Freezing point of the pure solvent (Tf
0
) – Freezing point of the pure solvent(∆Tf).
Experimentally for a dilute solution it was found that ∆Tf α m
∆Tf = Kf m
Where Kf is called Freezing point depression constant, Molal depression constant or Cryoscopic constant. Kf is defined as
Molal depression in freezing point when 1 mole of solute is dissolved in 1000g of solvent. Unit = K.Kg/mol.
DETERMINATION OF MOLECULAR WEIGHT OF NON-VOLATILE SOLUTE
f f
2
f f
1
2
f f
2 1
f 2
2
f 1
T = K m
T = K 1000
T = K 1000
K
= 1000
T
n
w
w
M w
w
M
w

 
 



SOLUTION
[14]
Application of Depression in Freezing point
 Sea water remains liquid at temperature below 00
C which is freezing point of water, because non volatile solute i.e. NaCl is
dissolved.
 Road salting takes advantage of this effect to lower the freezing point of Ice on which it is placed. Lowering the freezing
point allows the street ice to melt at lower temperature.
 Radiator fluid is in automobile and cars is mixture of water and ethylene glycol (antifreeze) . We thus lower the freezing
point of the solvent (water), and the solution remains a liquid even at lower temperatures.
# Equimolal quantities of different substances dissolved in the same quantity of solvent bring out the same depression in freezing
point of the solvent under identical conditions.
Q.48 A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in
water if freezing point of pure water is 273.15 K.
Q.49 Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of
AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1
K kg mol-1
. Calculate atomic masses of A and B.
Q.50 At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the
solution is 1.52 bars at the same temperature, what would be its concentration?
Q.51 The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and
trifluoroacetic acid increases in the order given above. Explain briefly.
OSMOSIS AND OSMOTIC PRESSURE
OSMOSIS (OSMOS means to push): When a semi permeable membrane is placed between a dilute solution and a concentrated
solution, it is observed that solvent molecules from a solution of lower concentration moves towards a solution of higher
concentration. This phenomenon was first observed by Abbe Nollet in 1748. Osmosis is a phenomenon of spontaneous flow of
solvent molecules through semi permeable membrane form pure solvent to the solution i.e. from low concentration region to high
concentration region.
We can understand the phenomenon of osmosis by following illustrations:
Two eggs of same size are taken and their outer hard shell is removed. They are dissolved in HCl solution. Now one is placed in
distilled water and other is placed in saturated salt solution. After sufficient time it was observed that egg placed in water swells
up and that placed in salt solution shrinks.
OSMOSIS DIFFUSION
SPM is required. No SPM is required.
The process involves movement of solvent molecules. Here both solute and solvent are involved.
Solvent molecules move from a region of low
concentration to high concentration.
Solvent molecules move from a region of high
concentration to low concentration.
It is limited to solution only. It is observed in gases as well as liquids.
A membrane which allows the passage of solvent molecules and prevents the passage of solute molecules through them is called
semi permeable. Ex: parchment paper, Egg membrane, goat’s bladder and cell membrane. Artificial SPM used is copper
ferrocyanide(Cu2[Fe(CN)6]).
Observation: Take a porous pot and deposit the SPM made of copper ferrocyanide on its walls. A long glass tube is fitted in the
pot with the help of rubber stopper. The pot filled with concentrated aqueous sugar solution is placed in distilled water. As time
goes osmosis is observed and level of solution in glass tube rises until equilibrium is established. At equilibrium osmosis stops.
The external pressure which must be applied on the solution in order to prevent the flow of the solvent molecules into the solution
through SPM is called osmotic pressure.
It is also a colligative property and depends upon the number of solute particles.

SOLUTION
[15]
The pressure exerted by a fluid at equilibrium
at a given point within the fluid, due to the force
of gravity.
Experimentally it was found for a dilute solution that
2
2
2
2
2
2
CT
CRT
n
RT
V
V n RT
w
V RT
M
w
M RT
V











When two or more solution have equal osmotic pressure i.e. π1= π2 then they are called isotonic solutions.
A solution having higher osmotic pressure than the other side is said to be hypertonic solution. When placed in a hypertonic
solution cells contract in size, a case of plasmolysis.
A solution having lower osmotic pressure than the other side is said to be hypotonic solution . when placed in a hypotonic solution
cells swell and burst, a case of haemolysis.
When isotonic solution is separated by SPM no osmosis occurs between them.
BIOLOGICAL EXAMPLE OF OSMOSIS
 In animals circulation of water to all parts of body takes place due to osmosis.
 Plant roots absorb water from soil due to osmosis.
 Water absorbed by plant rots is circulated in entire plant body and up to the top of tree due to osmosis.
 Red blood cells burst when placed in water due to endosmosis.
 When dried fruits and vegetables are placed under water they swell slowly to original form due to osmosis.
REVERSE OSMOSIS & WATER PURIFICAITION
When a solution is separated from pure water by a SPM, water moves
towards the solution side due to osmosis until osmotic pressure equals
hydrostatic pressure. If external pressure greater osmotic pressure is
applied on solution side the flow of solvent molecules from solution
towards solvent side can be initiated i.e. a direction reverse of osmosis.
This phenomenon is known as reverse osmosis.
--------------------------------
--------------------------------
--------------------------------
--------------------------------
--------------------------------
--------------------------------
----------------- -----------
--------------------------------
-----
………
………
………
………
………
………
………
…..
…
…
…
…
…
…
… SPM
SUGAR SOLUTION
WATER
Maximum
level
Initial level

SOLUTION
[16]
Reverse osmosis is used in desalination of sea water or for
purifying water that contains higher salt concentrations. The
pressure applied on solution for reverse osmosis is quite high.
Q.53 Calculate the osmotic pressure and vapour pressure of 0.6% aqueous solution of a non-volatile, non-electrolyte solute, urea
(NH2CONH2) at 250
C. The vapour pressure of pure water at same temperature is 24 mm Hg. Take densities to be 1 g/ml and
assume ideal behaviour of the solution. (R = 0.0821 L atm mol-1
K-1
)
Q.54 Calculate the osmotic pressure at 250
C and freezing point of a 1.8% aqueous solution of glucose (C6H12O6). Assume ideal
behaviour of the solution. Take densities to be 1 g/ml and Kf for water to be 1.86 K kg mol-1
. (R = 0.0821 L atm mol-1
K-1
)
Q.55 5g of a non-volatile, non-electrolyte solute is dissolved in water and the solution was made up to 250 cm3
. The solution
exerted an osmotic pressure equal to 4x105
N m-2
at 298 K . Find the molar mass of the solute.
Q.56 A 5% solution of cane sugar is isotonic with 0.877% of solute A. Calculate the relative molar mass of A. Assume density of
solutions to be 1 g/cm3
.
ABNORMAL MOLECULAR MASS
Colligative properties are used for determining the molecular mass of solutes. The values of molar mass of non-electrolyte solute
obtained were found to be same as expected from their chemical formulae.
In certain cases i.e. for electrolytes the observed molar masses are found to be either higher or lower than the normal
molar masses. The molar mass values higher or lower than the normal values are termed as the abnormal molar masses. The
observed normal molar masses may arise due to:
 Molecular association.
 Molecular dissociation.
MOLECULAR ASSOCIATION – molecules of certain solutes undergo molecular association in the solution which leads to
formation of bigger molecules. The associated molecular species consisting of two normal molecules is called a dimer and so on.
2
3
2
3
2 (
3 (
Two molecules of A give one molecule of molecule of A )
Three molecules of A give one molecule of molecule of A )
Association
Association
A A
A A


6 5 6 5 2
Benzoic Acid Dimer of Benzoic Acid
3 3 2
Acetic Acid Dimer of Acetic Acid
2 ( )
2 ( )
Association
Association
C H COOH C H COOH
CH COOH CH COOH


As a result of molecular association no. of solute particles in the solution decreases. Since, the colligative property depends upon
the number of solute molecules in the solution hence the molecular association lowers the colligative property. The molecular
association of a solute in solution leads to a molar mass higher than the normal molar mass of the solute.
Acetic acid remains as normal molecule in their aqueous solutions. In benzene this acid forms dimer. So the number of solute
particles in the solution of acetic acid in benzene is half the number of normal molecules. As a result of which colligative property
of solutions of these acids will be half of the normal value. The molar mass is inversely related to magnitude of colligative
property so will double than the normal value.
MOLECULAR DISSOCIATION: Electrolytes in solution dissociate to give two or more ions. This splitting of a solute into its
constituent ions is called molecular dissociation. As a result of molecular dissociation the number of solute particles in solution
increases. Since, colligative property depends on number of solute particles so it is increased as a result of dissociation.

SOLUTION
[17]
+ -
2+ -
2 2
KCl K Cl )(Two ions are produced from one molecule of KCl
BaCl Ba + 2Cl (Three ions are produced from one molecule of BaCl )
 

As we know that the molar mass of any solute is related inversely to the magnitude of the colligative property so molecular
dissociation of a solute leads to molar mass which less than the normal value of the molar mass of solute.
When KCl is dissolved in water it dissociates to give two ions (K+
and Cl
-
). So the number of solute particles is almost double the
number of KCl molecules as a result of which colligative property is doubled and observed molar mass will be half of the normal
value.
Van’t Hoff Factor: It accounts for extent of association or dissociation
Normal Molecular Mass(Observed)
Abnormal Molecular Mass(Calculated)
i 
=
Observed Magnitude of colligative property
Calculated Magnitude of colligative property
soluteTotal No. of moles of particles after association or dissociation
Total No. of moles of solute particles before association or dissociation

KCl K+
+ Cl—
t=0 1 0 0 Total number of particles after dissociation = (1-x) + x +x = 1+x
t=t’
1 –x x x Total number of particles before dissociation = 1
therefore i
soluteTotal No. of moles of particles after association or dissociation
Total No. of moles of solute particles before association or dissociation
 = 1+x / 1 = 1+x
 When i = 1 , then the solute remains unaffected
 When i > 1 , then the solute undergoes dissociation in the solution.
 When i > 1 , then the solute undergoes association in the solution.
For solution which undergo association or dissociation in the solutions the equation for colligative property are modified.
COLLIGATIVE PROPERTY EQUATION FOR COLLIGATIVE PROPERTY
When solute is non-electrolyte When solute is electrolyte
Relative lowering of Vapour
Pressure
0
1 1
0
1
solute
p
p
p
X


0
1 1
0
1
solute
p
p
p
X


Elevation of Boiling Point
b b
T K m  b b
T iK m 
Depression of Freezing Point
f f
T K m  f f
T iK m 
Osmotic Pressure CRT  iCRT 
Q.52 What are the values of van’t hoff factor for
(i) KCl (ii) BaCl2 (iii) AlCl3 if they are 100% dissociated.
Q.53 A person suffering from high blood pressure should take less common salt in food. Why?

SOLUTION
[18]
Q.54 When a liquid A is mixed with liquid B, the resulting solution is found to be cooler. What do you conclude about the nature
of the solution?
Ans: The solution shows a positive deviation from Raoult’s law. Adsorption of heat takes place. A-B interaction are weaker than
A-A and B-B.
Q.55 When 50 cc of a liquid A are mixed with 50 cc of a liquid B the volume of resulting solution is found oto be 99 cc. what do
you conclude about the nature of the solution?
Q.57 Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27° C.
Q58 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is
completely dissociated.

Solutin in chemistry

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