Class XII Chemistry Ch. 2: Solutions 9888737919 1 WHAT IS A SOLUTION? Homogeneous mixtures of two or more chemically non reacting substances. Homogeneous means (single phase) Solute + solvent = solution e.g. sugar +water , salt +water, urea + water, liquid medicine. TYPES OF SOLUTION ON BASIS OF NUMBER OF COMPONENTS: Binary solution: A solution having two components is called a binary solution. Components of a binary solution: (a) Solute (b) solvent. SOLUTE: Component of a solution which is generally in small quantity. SOLVENT: Component of a solution which is in large quantity. If two component: binary solution If three component: tertiary solution TYPES OF SOLUTION ON THE BASIS OF SOLVENT: AQUEOUS SOLUTION: A solution where solvent is water. NON- AQUEOUS SOLUTION: A solution where solvent is not water like ethanol (C 2 H 5 OH), chloroform (CHCl 3 ). TYPES OF SOLUTION ON THE BASIS OF PHYSICAL STATE: a. Solid solution: When solvent is in solid state. b. Liquid solution: When solvent is in: liquid state. c. Gaseous solution: When solvent is in gaseous state. TYPES OF SOLUTION IN LIQUID LIQUID SOLUTIONS: a. completely miscible liquids: when both solute and solvent are polar or non polar. E.g. ethyl alcohol + water, e.g. benzene + hexane. b. completely immiscible liquids: when one liquid is polar and other is non polar. E.g. benzene + water, cyclohexane + water. c. partially miscible liquids: when two liquids are not exactly similar in nature and also not completely dissimilar in nature. E.g. ether and water, phenol +water, nitrobenzene + hexane. Q1 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage. Q2. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example. Q3. Give an example of a solid solution in which the solute is a gas.? ADSORPTION
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Class XII Chemistry Ch. 2: Solutions
9888737919 1
WHAT IS A SOLUTION?
Homogeneous mixtures of two or more chemically non reacting substances. Homogeneous means (single phase)
Solute + solvent = solution
e.g. sugar +water , salt +water, urea + water, liquid medicine. TYPES OF SOLUTION ON BASIS OF NUMBER OF COMPONENTS:
Binary solution: A solution having two components is called a binary solution.
Components of a binary solution: (a) Solute (b) solvent.
SOLUTE: Component of a solution which is generally in small quantity.
SOLVENT: Component of a solution which is in large quantity.
If two component: binary solution If three component: tertiary solution
TYPES OF SOLUTION ON THE BASIS OF SOLVENT:
AQUEOUS SOLUTION: A solution where solvent is water.
NON- AQUEOUS SOLUTION: A solution where solvent is not water like ethanol (C2H5OH), chloroform (CHCl3).
TYPES OF SOLUTION ON THE BASIS OF PHYSICAL STATE:
a. Solid solution: When solvent is in solid state.
b. Liquid solution: When solvent is in: liquid state. c. Gaseous solution: When solvent is in gaseous state.
TYPES OF SOLUTION IN LIQUID LIQUID SOLUTIONS:
a. completely miscible liquids: when both solute and solvent are polar or non polar. E.g. ethyl alcohol + water, e.g. benzene + hexane. b. completely immiscible liquids: when one liquid is polar and other is non polar. E.g.
benzene + water, cyclohexane + water. c. partially miscible liquids:
when two liquids are not exactly similar in nature and also not completely dissimilar in nature. E.g. ether and water, phenol +water, nitrobenzene + hexane.
Q1 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.
Q2. Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Q3. Give an example of a solid solution in which the solute is a gas.? ADSORPTION
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Solute Solvent Types of solution
Examples
Solid Gas Aerosol or Solid in gas
Iodine vapours in air, Smoke, dust, fumes, camphor in water.
Solid Liquid Solid in liquid NaCl + H2O, Salt or glucose (C6H12O6) in water, sugar (C12H22O11) solution, urea. Paints, starch,
proteins, gold solution, glue, Indian ink, muddy water, milk of magnesia, white of an egg.
Mole fraction of water can also be calculated as: Or Xethylene glycol + Xwater = 1 Xwater = 1- 0.068 = 0.932
: 1 – 0.068 = 0.932 Xethylene glycol = 0.068
Xwater = 0.932 7. MASS FRACTION:
Mass fraction of a component (x) = Mass of a component (g) Total mass of all components (g)
If a and b are two components: xa = wa xb = wb wa+ wb wa+ wb
8. PARTS PER MILLION: (ppm)
ppm = Number of parts of component x 106 Total number of parts of all components of solution. Molality, Mole fraction, Mass fraction etc are preferred to molarity, normality
etc. Because molality, mole fraction, mass fraction involves mass of solute and solvent while normality, molarity involves volume of solution. Temperature has no effect on
mass but it has significant effect on volume. Hence molality, mole fraction, mass fraction do not change with temperature whereas molarity, normality has significant effect on volume hence preferred.
SOLUBILITY:
Maximum amount of solute in g that can be dissolved in fixed amount of solvent at a specified temperature.
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FACTORS ON WHICH THE SOLUBILITY OF A SOLID DEPENDS:
(a) Size of solute particles :
solubility surface area of solute. smaller is the size , more is the surface area
exposed to the solvent, and therfore gretaer is the solublility of that solute. E.g. powder sugar easily
dissolves while crystal sugar takes more time comparatively.
(b) Stirring or mixing: solubility mixing.
stirring increases the contact of solute with the solvent thus increases solubility. We use mixer for making the solution very fast.
(c) Temperature:
(i) Solids solubility temperature
Mostly all solids in water solubility increases with temperature but it is not necessary in all cases. In Endothermic process(ΔHsolution= +ve): solubility increases with the increase in
temperature and vice versa. e.g. solubility of potassium nitrate KNO3 increases with the increase in temperature.
In exothermic process (ΔHsolution= -ve): Solubility decrease with the increase in temperature. e.g. solubility of calcium oxide (CaO) decreases with the increase in temperature.
(d) NATURE OF SOLUTE AND SOLVENT Solubility rule: like dissolves like:
A polar solute is dissolved in polar solvent and vice versa. e.g. water (polar) cannot be dissolved in oil which is non polar
NaCl (polar) is easy dissolve in H2O (polar) Naphthalene (non polar) and anthracene (non polar) dissolve readily in benzene (non polar) but sodium chloride (polar) and sugar do not.
DISSOLUTION: process of increase in concentration of solution due to dissolving of
solute in a solvent.
Crystallisation: process of separating out of solute in form of crystal from solution. State of dynamic equilibrium is reached when
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rate of dissolution = rate of crystallisation . Solute + Solvent Solution
At this stage the concentration of solute in solution will remain constant under the given conditions, i.e., temperature and pressure.
(e) PRESSURE: no effect on solubility of solid.
FACTORS ON WHICH THE SOLUBILITY OF A GAS DEPENDS:
For gases: solubility of solute decreases with temperature.
solubility of gas 1 Temperature
Gas+ solvent solution + heat Dissolution is an exothemic process, thus solubility decreases with increase of
temperature (Lechatlier principle). e.g. hot water taste flat because dissolved O2 decreses with temperature as solubility of O2 decreases with temperature.
Henry’s law:
For an ideal gas , at constant temperature, mass of gas dissolved in a fixed volume of liquid is directly proprtional to the presure of gas on the surface of the liquid.
Mathematically: m P also m = K P where K is henry’s constant which depends on the gas to be dissolves, temperature, type of the solvent used for dissolvoing gas.
Higher the KH value lower is the solubility. With the increase in temperature, KH value increases therefore solubility decreases.
Gas Temperature (K) KH (Kbar-1) O2 293 34.86 O2 303 46.82
Henry’s law can also be stated as :
The partial pressure of gas in vapour phase (p) is proportional to the mole fraction of the gas in the solution.
pA = KH xA where xA is mol fraction of gas in solution, KH is Henry’s constant. e.g. for dissolving CO2 in coca cola soft drinks, high presure is used as CO2 is having
poor solubility in aqueous solvent.
Effect of pressure on the solubility of a gas. The concentration of dissolved gas is proportional to the pressure on the gas above the
solution. Clausius Clapeyron equation: d lnC =ΔH
dT RT2
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lnC2 = ΔH (1 - 1) C1 RT2 (T1 T2)
LIMITATIONS OF HENRY’S LAW: (a) Gas must be ideal in nature i.e. low pressure and high temperature.
(b) Gas should not undergo compound formation with the solvent or association or dissociation in the solvent.
APPLICATIONS OF HENRY’S LAW: (a) Production of carbonated drinks like coke etc to increase the solubility of CO2 in cold
drinks at high pressure. (b) In deep sea diving e.g. He is used in artificial respiration to see divers (scuba divers).
11.7% He + 56.2% N2 + 32.1% O2 (c) For climbers or high altitude living people as. As pressure decreases concentration of gas decreases.
At constant temperature: density of gas pressure of gas (Boyles law PV =constant). i.e. Lesser amount of O2 is available in mountains for breathing due to low pressure,
therefore mountain trackers take O2 cylinder with them. Q. If N2 gas 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. The solubility of gas is related to the mole fraction in aqueous solution. By Henry’s law. pN2 = K xn2
x (Nitrogen) = p (nitrogen) = 0.987 = 1.29x10-5 K 76.48 x 1000
Voume of water = 1 litre Moles of water = 1000 = 55.5 mol of it,
18 if n represents number of moles of N2 in solution,
x (Nitrogen) = n = 1.29x 10–5
n+ 55.5 mol (n in denominator is neglected as it is < < 55.5) Thus n = 1.29 10-5 x 55.5 mol = 7.16 x 10-4 mol = 7.16×10-4 mol × 1000 m mol =
0.716 mmol
TYPES OF SOLUTION ON THE BASIS OF SOLUBILITY:
UNSATURATED SOLUTION: Solution which contains less amount of solute than is required to saturate it at fixed
temperature. i.e. we can add more solute to the unsaturated solution at the same T.
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Check for Unsaturated solution: when we add solute to the unsaturated solution and
stirr, it will dissolve more solute which means solution is unsaturated solution. SATURATED SOLUTION:
When no more of solute can be dissolved in a given solvent at the same temperature and pressure.
Or When we add crystal of solute and stirr it will not dissolve and remains as crystals. Solution which is in dynamic equilibrium with undissolved solute is the saturated solution and contains the maximum amount of solute dissolved in a given amount of solvent.
Thus, the concentration of solute in such a solution is its solubility.
SUPER SATURATED SOLUTION: When we heat the saturated solution excess of solute can be dissolved in a given
amount of solvent. A super saturated solution at a particular temperature is one that is more concentrated (contains more solute) than its saturated solution at that temperature.
CHECK: If a crystal of solute is added to this solution, the excess of solute crystallizes. VAPOR PRESSURE (P0):
Pressure exerted by vapour present in equilibrium with the solution at a particular temperature.
Factors on which vapour pressure depends: (a) Nature of liquid: lesser is the intermolecular forces, more is the vapour pressure.
(b) Temperature: higher the temperature, higher is the vapour pressure. log (P2/P1) = ΔHvap (T2-T1) 2.303R (T1T2)
Where P1, P2 are the vapour pressure at temperature T1 and T2 respectively.
How vapour pressure varies If solute is added to pure solvent? When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of solution is less than that of pure solvent.
P10-Ps = n2 = x2 n2 is no of moles of solute, n1 is no. of moles of solvent , x2 is
P10 (n2+n1)
mole fraction of solute.
P0 is vapour pressure of solvent Ps is vapour pressure of solution
P10-Ps
= i n2 n2 no. of moles of solute, n1 moles of solvent P1
0 n1 n2<<<<<n1 (for dilute solutions)
P10-Ps
= i w2/ M2 w2 is mass of solute, M2 is molar mass of solute P1
0 w1/M1 w1 mass of solvent, M1 is molar mass of solvent w2<<<<<w1 (for dilute solutions)
P0 is vapour pressure of solvent Ps is vapour pressure of solution
RAOULT’S LAW: According to Raoult’s law:
For a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
PA = p0A xA PB = po
BxB
Using Dalton’s law of partial pressure the total pressure of solution is calculated. Ptotal = PA+ PB = PA
0 xA+ PB0 xB we know XA+ XB = 1
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= PA0(1-xB)+ PB
0 xB
saturated vapour pressure : closed system :
Equilibrium when number of particles breaking away from the surface is exactly the
same as the number sticking on to the surface again.
At equilibrium, more particles in the vapour Fewer particles in the vapour
Saturated vapour pressure is more Saturated vapour pressure is lower. Q. Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 298 K are
200 mm Hg and 415 mm Hg respectively. (i) Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at 298 K and, (ii) mole
fractions of each component in vapour phase. (i) Molar mass of CH2Cl2 = 12x1 + 1x2 + 35.5x2 = 85 g mol-1 Molar mass of CHCl3 = 12x1 + 1x1 + 35.5x3 = 119.5 g mol-1
Moles of CH2Cl2 = 40 g = 0.47 mol 85 g mol-1
Moles of CHCl3 = 25.5 g = 0.213 mol 119.5 g mol-1 Total number of moles = 0.47 + 0.213 = 0.683 mol
x CH2Cl2 = 0.47 mol = 0.688
0.683 mol x CHCl3 = 1.00 – 0.688 = 0.312
Using equation Ptotal = PCH2Cl2+ PCHCl3 = P CH2Cl20 x CH2Cl2+ P CHCl3
0 x CHCl3 = (415) 0.688 + 200 x 0.312
= 285.5 + 62.4 = 347.9 mm Hg (ii) Using the relation , yi = pi
ptotal, mol fraction of the components in gas phase (yi).
p CH2Cl2 = 0.688 x 415 mm Hg = 285.5 mm Hg p CHCl3 = 0.312 x 200 mm Hg = 62.4 mm Hg
y CH2Cl2 = 285.5 mm Hg/347.9 mm Hg = 0.82 y CHCl3 = 62.4 mm Hg/347.9 mm Hg = 0.18
Note: Since, CH2Cl2 is a more volatile component than CHCl3, [p0
CH2Cl2 = 415 mm Hg
and p0CHCl3 = 200 mm Hg] and the vapour phase is also richer in CH2Cl2 [y CH2Cl2 = 0.82
and y CHCl3 = 0.18], it may thus be concluded that at equilibrium, vapour phase will be always rich in the component which is more volatile.
COMPARISION OF RAOULT’ LAW AND HENRY’S LAW, It is observed that
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(a) The partial pressure of volatile component or gas is directly proportional to its mole fraction in solution.
(b) both apply to the volatile component of the solutions. DISSIMILARITIES:
In proportionality constant: In case of Henry’s Law the proportionality constant is KH and it is different from p1
0 which is partial pressure of pure component.
Raoult’s Law becomes a special case of Henry’s Law when KH becomes equal to p10
in Henry’s law.
IDEAL AND NON IDEAL SOLUTIONS: Liquid –liquid solutions can be classified into ideal and non-ideal solutions on basis of
Raoult’s Law.
Ideal solutions Non- ideal solutions
The solutions that obey Raoult’s Law over the entire range of concentrations
are known as ideal solutions.
When a solution does not obey Raoult’s Law over the entire range of concentration,
then it is called non-ideal solution.
ΔHmix =0 and ΔVmix=0 ΔHmix ≠0 and ΔVmix ≠0
The intermolecular attractive forces between solute molecules and solvent
molecules are nearly equal to those present between solute and solvent molecules i.e. A-A and B-B interactions
are nearly equal to those between A-B
The intermolecular attractive forces between solute molecules and solvent
molecules are not equal to those present between solute and solvent molecules i.e. A-A and B-B interactions are not equal to
those between A-B
Benzene +toluene,
n- hexane + n heptanes, chlorobenzene + bromobenzene.
Properties of solution which depends on only the number of solute particles in fixed amount of solvent but not on the nature of solute are called colligative properties.
There are four colligative properties: a. Relative lowering of vapour pressure
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(p1o - p1 ) = x2 p1
o vapour pressure of pure solvent
p10 p1 vapour pressure of solution.
x2 mol fraction of solute b. Elevation of boiling point.
bTb- Tb0
bKb x m m is molality Kb is molal elevation constant or ebullioscpic constant.
c. Depression of freezing point: bb
ΔTf = Kf x m Kf is molal depression constant or cryoscopic constant M is molality.
d. Osmotic pressure: CRT
Osmotic pressure () Molarity (C) of the solution R = 0.082atm L /K .gmol T is temperature in Kelvin
USES OF COLLIGATIVE PROPERTY: Colligative properties help in calculation of molar mass of solutes.
Colligative property no. of particles
no. of molecules in case of non electrolyte.
no. of ions (in case of electrolytes)
no. of moles of solute
mol fraction of solute Relative lowering of vapour pressure:
Difference in the vapour pressure of pure solvent (P10) and solution (ps) represents
lowering in vapour pressure (p1o - ps ).
Dividing lowering in vapour pressure by vapour pressure of pure solvent is called relative lowering of vapour pressure (p1
o - ps )
p10
Relative lowering of vapour pressure is directly proportional to mole fraction of solute (x2). Hence it is a colligative property.
Q. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar
mass 78 g mol-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?
Ans. The various quantities known to us are as follows: w2 = 0.5 g; w1 = 39 g p1
0 = 0.850 bar; ps = 0.845 bar; M1 = 78 g mol-1 ,
Let M is the moles of solute= M g/ gmol moles of solute = 0.5/ M
moles of solvent = 39/ 78 p1
o - ps = 0.850 bar – 0.845 bar = n solute = 0.5 g × 78 g mol
Po 0.850 bar n solvent M2 × 39 g
Therefore, M2 = 170 g mol–1 Elevation of boiling point:
The difference in boiling points of solution ( ΔTb ) and pure solvent (Tb0 ) is called
elevation in boiling point
bTb- Tb0
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For a dilute solution elevation of boiling point is directly proportional to molal concentration of the solute in solution. Hence it is a colligative property.
b bb
ΔTb =Kb x 1000x w2 w2 is wt of solute, w1 is wt of solvent , M2 mol wt of solute
M2w1 Kb is molal elevation constant or ebullioscpic constant , Kb T2
Q. 18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? Kb for water is 0.52 K kg mol-1.
Sol. Kb for water is 0.52 K kg mol-1 Moles of glucose = 18 g/ 180 g mol-1 = 0.1 mol
Number of kilograms of solvent = 1 kg =1000g Thus molality of glucose solution = 0.1 mol kg-1
For water, change in boiling point ΔTb = Kb m = 0.52 K kg mol-1x 0.1 mol kg-1 = 0.052 K Since water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of
solution will be 373.15 + 0.052 = 373.202 K. The boiling point of benzene is 353.23 K.
Q. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is
2.53 K kg mol-1? Solution
W2 = 1.80 W1 = 90 M2: molar mass of solute
Kb (benzene) = 2.53 K kg mol-1 The elevation (ΔTb) in the boiling point = 354.11 K – 353. 23 K = 0.88 K ΔTb =Kb x 1000x w2 w2 is wt of solute, w1 is wt of solvent , M2 mol wt of solute
M2w1 Kb is molal elevation constant ΔTb = Kb m
0.88 = 2.53 x (1.8/M2) x1000
90 we get M2 = 2.53 K kg mol-1 × 1.8 g × 1000 g kg–1 = 58 g mol-1
0.88 K × 90 g Therefore, molar mass of the solute, M2 = 58 g mol-1
Depression of freezing point: The lowering of vapour pressure of solution causes a lowering of freezing point compared
to that of pure solvent.The difference in freezing point of the pure solvent (Tf0) and
solution ( Tf ) is called the depression in freezing point.
ff
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For a dilute solution depression in freezing point is a colligative property because it is directly proportional to molal concentration of solute.
ΔTf = Kf x1000x w2 M2w1 w2 is mass of solute,
M2 mol wt of solute, w1 mass of solvent.
Kf is molal depression constant or
cryoscopic constant Kf T2 ΔTf = Kf x1000x w2
M2w1
APPLICATIONS OF DEPRESSION IN FREEZING POINT: (a) In making antifreeze solutions (ethylene glycol in water) for car radiators in hilly areas.
(b) melting of ice/ snow on roads by adding NaCl or CaCl2
Q. What are anti-freeze solutions? Give one example.
Ans. Substance which depress the freezing point of the solution/ water. i.e. after adding depressant water would not freeze at 273K and can handle sub zero temperature without solidification/ freezing. e.g. ethylene glycol is added in car radiators or NaCl,
CaCl2 are scattered on hilly roads to melt the ice.
Q. 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate (a) the freezing point depression and (b) the freezing point of the solution. Depression in freezing point is related to the molality, therefore, the molality of the
solution with respect to ethylene glycol. Moles of ethylene glycol = 45 g = 0.73 mol
62 g mol-1 Mass of water in g = 600g
Hence molality of ethylene glycol = 0.73 x 1000 = 0.73 mol = 1.2 mol kg-1 600 0.60 kg Therefore freezing point depression, ΔTf = Kf x m = 1.86 K kg mol-1x1.2 mol kg-1 =2.2 K
Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K
Q. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K
kg mol–1. Find the molar mass of the solute. Ans. ΔTf = Kf x1000x w2
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M2w1
M2 = Kf x 1000x w1
w2 M2 = 5.12 K kg mol-1× 1000 x1.00 g = 256 g mol-1 0.40 × 50 g
Thus, molar mass of the solute = 256 g mol-1
OSMOSIS: The phenomenon of flow of solvent molecules through a semi permeable membrane from pure solvent to solution is called osmosis.
REVERSE OSMOSIS:
The process of movement of solvent through a semi permeable membrane from the solution to the pure solvent by applying excess pressure on the solution side is called
reverse osmosis.
REVERESE OSMOSIS
EXCESS PRESSURE APPLIED MUST BE APPLIED ON THE SOLUTION TO PREVENT
OSMOSIS. OSMOTIC PRESSURE:
The excess pressure that must be applied to solution to prevent the passage of solvent into solution through a semi
permeable membrane is called osmotic pressure. Osmotic pressure is a colligative property as it depends on the Number of solute particles and not on their identity.
For a dilute solution,
osmotic pressure () is directly proportional to the molarity (C)
of the solution i.e. CRT R = 0.082atm L /K .mol
Osmotic pressure can also be used to determine the molar mass of solute using the equation M2 =iw2 RT w2 is mass pof solute, M2 mol wt of solute, V volume of solution.
V
i n2RT V ISOTONIC SOLUTION.
Two solutions having same osmotic pressure at a given temperature are called isotonic solution.
n1 = n2 or w1 = w2
V1 V2 m1V1 m2V2
Human blood is isotonic 0.9% solution of NaCl
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HYPERTONIC SOLUTION: If a solution has more osmotic pressure than other solution it is called hypertonic
solution. A solution of NaCl with concentration > 0.9% is hypertonic solution.
HYPOTONIC SOLUTION:
If a solution has less osmotic pressure than other solution it is called hypotonic solution. A solution of NaCl with concentration < 0.9% is hypotonic solution.
RBCs will shrink in this condition called plasmolysis.
Q. 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57x10-3 bar. Calculate the
molar mass of the protein.
Ans. The various quantities known to us are as follows: = 2.57x10-3 bar, V = 200 cm3 = 0.200 litre (1lt 1000cm3)
T = 300 K R = 0.083 L bar mol-1 K-1
CRT
Let M2 is the molar mass of solute
wRT M2
M2 = wRT = 1.26 g × 0.083 L bar K-1 mol -1× 300 K = 61,022 g mol-1
2.57×10-3 bar × 0.200 L
ABNORMAL MOLAR MASS: Molar mass that is either lower or higher than expected or normal molar mass is called as abnormal molar mass.
VANT HOFF FACTOR:
van’t Hoff factor (i)accounts for the extent of dissociation or association .
= Total number of moles of particles after association/dissociation Total number of moles of particles before association/dissociation
For Association: i < 1 For dissociation : i> 1
Inclusion of van’t Hoff factor modifies the equations for colligative properties as: P1
0-P1 =i n2 P1
0 n1
ΔTb =i Kb m
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ΔTb =i Kb x 1000x w2 /M2w1 ΔTf = iKf m
ΔTf = i Kf x1000x w2 M2w1
i n RT V
For association: i = 1- (1-1/n)
= 1-i (1-1 )
n
for dissociation : i= (1- + n ) 1
= i-1 Where is degree of dissociation, n is no. of ions produced by complete (n-1) dissociation of 1 mole of substance.
Q. 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in
freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms dimer in solution? The given quantities are: w2 = 2 g; Kf = 4.9 K kg mol-1; w1 = 25 g,
ΔTf = 1.62 K ΔTf = Kf x1000x w2
M2w1 Substituting these values in above equation M2 = 4.9 K kg mol-1 × 2 g × 1000
25 g × 1.62 K Observed molar mass of benzoic acid M2 = 241.98 g mol-1
Thus, experimental molar mass of benzoic acid in benzene is = 241.98 g mol-1 Now consider the following equilibrium for the acid:
If represents the degree of association of the solute then we would have (1 – ) mol of benzoic acid left in unassociated form and correspondingly as associated moles of
benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium as below: 2C6H5COOH (C6H5COOH)2
T=0 1 0
T= equilibrium 1- /2
Therefore total no. of moles after association = 1- + /2 = 1- /2 Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.
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i = 1- /2 = 0.504 1
= (1-0.504) x 2 = 0.992 or %age dissociation = 99.2% Therefore, degree of association of benzoic acid in benzene is 99.2 %.
Q. 0.6 mL of acetic acid (CH3COOH), having density 1.06 g mL-1, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C.
Calculate the van’t Hoff factor and the dissociation constant of acid. Sol. Number of moles of acetic acid = 0.6 mL = 0.0106 mol = n
1.06 g mL x60 g mol
Molality = 0.0106 mol x 1000 mL = 0.0106 mol kg–1
1 g mL-1
Using equation ΔTf = Kf x m = 1.86 K kg mol-1x 0.0106 mol kg-1 = 0.0197 K
van’t Hoff Factor (i) = Observed freezing point = 0.0205 K = 1.041 Calculated freezing point 0.0197 K
Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen
ions per molecule of acetic acid. CH3COOH H+ + CH COO-
If is the degree of dissociation of acetic acid, then we would have C(1 – )
moles of undissociated acetic acid, C moles of CH3COO- and C moles of H+ ions, Inital moles 1 0 0
At equilibrium 1-
Thus total moles of particles are: C(1 – + + ) = C(1 + ) i = 1.041
Thus degree of dissociation of acetic acid = = 1.041– 1.000 = 0.041
Then [CH3COOH] = C(1 – ) = 0.0106 (1 – 0.041),
[CH3COO-] = C = 0.0106 x 0.041,
[H+] = C = 0.0106 x 0.041
Dissociation constant Ka = [C ][ C ] = [0.0106 x 0.041] x [0.0106 x 0.041]
[CH3COOH ] 0.0106 (1.00 - 0.041) Ka = 1.86 x 10-5
DIFFERENCE BETWEEN OSMOSIS AND DIFFUSION:
OSMOSIS DIFFUSION
Semi permeable membrane is used. No semi permeable membrane is used.
Flow of solvent from semi permeable membrane
Solvent and solute molecules move directly to each other.
Molecules move from lower concentration to higher concentration.
Molecules move from higher concentration to lower concentration.
Applies to liquid solutions only. It takes place in liquid and gases solution .