SOLUTIONS (UNIT -1)

SOLUTION is the homogeneous mixture of two or more than two components. Most of the solutions are
binary i.e. consists of two components out of which that is present in the largest quantity iscalled solvent & one
which is present in smaller quantity called solute.
EXPRESSING CONCENTRATIONS OF SOLUTIONS
 Mass percentage: Mass of solute per100gof solution Mass % = (mass of solute / total mass of solution) X 100
 Volume percentage : volume of soluteper100mlof solution Volume % = (volume of solute/ total volume of solution) X 100
 Parts per million:parts of a component per million (106)parts of the solution.
ppm = no. of parts of the component / total no. of parts of all components of the sol. X 106
 Mole fraction(x): It is the ratio of no. of moles of one component to the total no. of all the components
present in the solution . For binary solution :‐ the no. of moles of A and B are nA and nB respectively
so, xA= nA/nA+nB ; xB= nB/ nA+nB
In binary solution x A+ x B= 1
 Molarity: No. of moles of solute dissolved in one litre of solution .
Molarity(M) = moles of solute/ vol. of solution in litre
 Molality(m) : No. of moles of solute per kg of the solvent.
Molality(m) = moles of solute/mass of solvent in kg
Molality is independent of temp. whereas molarity is a function of temp. because vol. depends on temp. and
mass does not.
HENRY'S LAW
It states that at a constant temp. the solubility of the gas in liquid is directly proportional to the pressure of the
gas above the surface of the liquid .
It also states that the partial pressure (p) of a gas in vapour phase is proportional to the mole fraction of the
gas (x) in the solution.
P=KHX
KH is Henry's law constant .
APPLICATION OF HENRY'S LAW
 To increase the solubility of CO2 in soda water and soft drinks the bottle is sealed under high
pressure.
 To avoid bends , toxic effects of high concentration of nitrogen in the blood the tanks used
Used by scuba divers are filled with air diluted with He.
RAOULT'S LAW :‐ it states that :
1) For a solution of volatile liquid , the partial vapour pressure of each component of the solution is
directly proportional to its mole fraction present in solution.
PA= P0 A XA PB= P0 B XB
The total pressure is equal to sum of partial pressure. Ptotal = PA + PB
2) For a solution containing non‐volatile solute the vapour pressure of the solution is directly proportional
to the mole fraction of the solvent.
P Aα X A P A = P 0 A X A
IDEALSOLUTION
The solution which obeys Raoult's law over the entire range of concentration when enthalpy of mixing and vol.
of mixing of pure component to form solution is zero.
CONDITIONS
I. P A
=PA0 XA PB= P0 B XB
II. ∆Hmix= 0
III. ∆Vmix= 0
This is only possible if A‐B interaction is nearly equal to those between A‐A and B‐B interactions. Ex:‐
solution of n‐hexane and n‐heptane.
NON IDEAL SOLUTIONThe solution which do not obey Raoult's law over the entire range of concentrations.
CONDITIONS :
I. PA ≠P0 A XA and PB≠ P0BXB
II. ∆H mix≠0
III. ∆Vmix ≠0
The vapour pressure of such solutions is either higher or lower than that predicted for Raoult's law .
I. If vapour pressure is higher, the solutions shows positive deviation (A‐B interactions are weaker than those
between A‐A and B‐B ).
Ex: mixture of ethanol and acetone .
PA>PA 0 XA ; PB > PB0 XB
∆Hmix = Positive ; ∆Vmix= Positive
II. If vapour pressure is lower, the solution shows negative deviation (A‐B interactions are stronger than those
between A‐A and B-B).
III. Ex: mixture of chloroform and acetone .
IV. PA <PA0 XA ; PB< P0 B XB
V. ∆H mix= negative ∆Vmix= negative
AZEOTROPE
Mixture of liquid having the same composition in liquid and vapour phase and boil at constant temp.
Azeotropes are of two types :‐
a) Minimum boiling azeotrope :‐ the solution which shows a large positive deviation from Raoult's law . Ex‐
ethanol – water mixture.
b) Maximum boiling azeotrope :‐ the solution which shows large negative deviation from Raoult's law. Ex‐
nitric acid – water mixture.
COLLIGATIVE PROPERTIES Properties of ideal solution which depends upon no. of particles of solute but
independent of the nature of the particles are called colligative properties.
1. RELATIVE LOWERING OF VAPOUR PRESSURE
P0A– PA/P0A= XB
XB= nB/ nA+nB
For dilute solution, nB<< nA, hence nB is neglected in the denominator.
P0A– PA/P0A= nB/nA
P0A– PA/ P0A= W B MA/MB W A
2. ELEVATION OF BOILING POINT
∆Tb= kbm Where , ∆Tb= Tb– T0b
Kb= molal elevation constant / Ebullioscopic constant
m = molality
M= kb 1000 WB/∆Tb WA
3. DEPRESSION IN FREEZING POINT
∆Tf= Kfm where, ∆Tf= T0f– Tf
Kf= molal depression constant / Cryoscopic constant
m = molality
M= kf 1000 W B/∆Tf W A
4. OSMOTIC PRESSURE
The excess pressure that must be applied to a solution side to prevent osmosis i.e. to stop the passage of
solvent molecules into it through semi‐permeable membrane is called osmotic pressure.
Π= CRT
Π= n/VRT ( n= no.of moles; V= volume of solution(L)
R= 0.0821 L atm mol–1 ; T= temperature in kelvin
ISOTONIC SOLUTION
Two solutions having same osmotic pressure and same concentration are called isotonic solutions.
Hypertonic solution have higher osmotic pressure and hypotonic solution have lower
osmotic pressure than the other solution.
0.91% of sodium chloride is isotonic with fluid present inside blood cell.
VAN'T HOFF FACTOR (i)
Ratio of normal molecular mass to the observed molecular mass of the solute.
i = normal molecular mass/ observed molecular mass
= observed colligative properties / calculated value of colligative properties
i<1 (for association ) i>1 (for dissociation)
MODIFIED FORMS OF COLLIGATIVE PROPERTIES
1) P0A-PA / P0A = i n B / n A
3) ∆Tb = iKb m
4) ∆Tf = iKf m
5) Π= iCRT
FREQUENTLY ASKED QUESTIONS (1 MARK QUESTIONS)

Q 1. Two liquids X and Y boil at 380 K and 400K respectively, which of them is more volatile?
Ans. X is more volatile since it has low boiling point.
Q2. How does the molarity of a solution change with temperature?
Ans. Molarity decreases with increase in temperature as volume of solution increases with increase in
temperature.
Q3. Under what condition do non ideal solutions show negative deviation ?
Ans. When the new forces of interaction between the components are stronger than those in the pure
components, then non ideal solutions show negative deviation.
Q4. What are minimum boiling azeotropes? Give one example.
Ans. Minimum boiling azeotropes are those which boil at lower temperature than boiling point of each
component in pure state, e.g., 95.5% ethyl alcohol and 4.5% water by mass.
Q5. What do you understand by the term that Kf for water is 1.86 K kg/mol?
Ans. It means that the freezing point of water is lowered by 1.86 K when 1 mol of non volatile solute is
dissolved in 1 kg of water.
Q6. Why is osmotic pressure of 1 M KCl higher than 1 M urea solution ?
Ans. This is because KCl dissociates to give K+ and Cl– ions while urea being a molecular solid does not
dissociate into ions in the solution.
Q 7. What is the value of van’t Hoff factor for a dilute solution of
(i) K2SO4 in water
(ii) acetic acid in benzene.
Ans .(i) 3 (ii) 1/2

ASSERTION -REASON TYPE

A statement of assertion is followed by a statement of reason. Mark the correct
choice from the options given below:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
1. Assertion : In an ideal solution, ∆mix H is zero.
Reason : In an ideal solution, A - B interactions are lower than A-A and B-B interactions.
( Ans -c)
2. Assertion : Osmosis does not take place in two isotonic solutions separated by semi-permeable
membrane.
Reason : Isotonic solutions have same osmotic pressure. ( Ans - a)
3. Assertion : Lowering of vapour pressure is not dependent on the number of species present in
the solution.
Reason : Lowering of vapour pressure and relative lowering of vapour pressure are colligative
properties. ( Ans - d)
4. Assertion : 1 M solution of KCl has greater osmotic pressure than 1 M solution of glucose at
same temperature.
Reason : In solution KCl dissociates to produce more number of particles. ( Ans - a)
5. Assertion : Two liquids nitric acid and water form a maximum boiling azeotrope when mixed in
the ratio of 68% and 32% respectively.
Reason : Interaction between nitric acid and water are stronger than nitric acid - nitric acid
interactions and water - water interactions. ( Ans - a)
One - word answers
1. Liquid ‘Y’ has higher vapour pressure than liquid ’ X’ . Which of them will have higher boiling
point. ( Ans - X)
2. Liquids A and B on mixing produce a warm solution. Which type of deviation from Raoult’s law is
shown? ( Ans - Negative deviation)
3. Under what condition Van’t Hoff factor is less than 1? ( Ans - Association)

2 MARKS QUESTIONS
Q1. State Henry's law. What is the significance of KH ?
Ans. Henry's Law: It states that “the partial pressure of the gas in vapour phase (p) is directly proportional to
the mole fraction of the gas(x)in the solution” , and is expressed as : p=KH x where,K is the Henry's Law constant
H

Significance of KH : Higher the value of Henry's law constant KH ,the lower is the solubility of the
gas in the liquid .
Q2. How is that measurement of osmotic pressure is more widely used for determining molar masses of
macromolecules than the elevation in boiling point or depression in freezing point of their solutions?
Ans. The osmotic pressure method has the advantage over elevation in boiling point or depression in
freezing point for determining molar masses of macromolecules because
1. Osmotic pressure is measured at the room temperature and the molarity of solution is used instead of molality.
2. Compared to other colligative properties, its magnitude is large even for very dilute solutions.
Q3. Suggest the most important type of intermolecular interaction in the following pairs :
i) N-hexane and n-octane
ii) methanol and acetone
Ans. i) Dispersion or London forces as both are non-polar.
ii) Dipole-dipole interactions as both are polar molecules.
Q4. Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is
dissolved in 450 g of CH3CN .
Ans. Mass of solution = 6.5g + 450g = 456.5g
Mass of aspirin
Mass % of aspirin =
=6.5/456.5 X 100 = 1.424% Mass of solution x 100
3 MARK QUESTIONS
Q1. Non-ideal solution exhibit either positive or negative deviations from Raoult's law. What are these
deviation and why are they caused? Explain with one example for each type.
Ans. When the vapour pressure of a solution is either higher or lower than that predicted by Raoult's
law, then the solution exhibits deviation from Raoult's law. These deviation are caused when solute -solvent
molecular interactions A – B are either weak or stronger than solvent – solvent A – B or solute – solute B – B molecular
interactions. Positive deviations : When A – B molecular interactions are weaker than A – A and B – B molecular
interaction . For example, a mixture of ethanol and acetone.
Negative deviations: When A – B molecular interaction are stronger than A – A and B – B molecular interaction.
For example, a mixture of chloroform and acetone.
Q2 .a) Why is an increase in temperature observed on mixing chloroform and acetone?
b) Why does sodium chloride solution freeze at a lower temperature than water?
Ans: a) The bonds between chloroform molecules and molecules of acetone are dipole-dipole interactions
but on mixing, the chloroform and acetone molecules, they start forming hydrogen bonds which are stronger
bonds resulting in the release of energy. This gives rise to an increase in temperature.
b) When a non- volatile solute is dissolved in a solvent, the vapour pressure decreases. As a result, the solvent
freezes at a lower temperature.
Q3. A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500g of water. This solution has a
boiling point of 100.42C while pure water boils at 100-C. What mass of glycerol was dissolved to make the solution ? (Kb of
water = 0.512 K kg/mol)
Ans. ∆Tb = 100.42°C‐ 100°C = 0.42°C or 0.42K; WA = 500g ; Kb = 0.512 K kg / mol ;
MB = 92 g /mol Substituting these values in the expressions,
WB = ∆Tb x MB x WA

Kb x 1000

WB = 0.42 x 92 x 500 = 37.73 g

0.512 x 1000
Q4. 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 27⁰C.

Ans. π = i x WB x R x T

MB x V

Molar mass of CaCl ,M =40+22 X 35.5

B = 111 g mol –1

Therefore, Molar mass of CaCl2, WB = 0.75 atm x 111g/mol x 2.5 L

2.47 x 0.0821 x 300 K

= 3.42g
Q5. The molar freezing point depression constant for benzene is 4.90K kgmol–1. Selenium
exists as polymer SeX. When 3.26 gm of Se is dissolved in 226gm of benzene, the observed freezing point is 0.1120C
lower than for pure benzene. Decide the molecular formula of Selenium.(At.wt. of selenium is 78.8 g mol‐1)

1000 x Kf x WB
Ans ∆Tf =
W A X MB
0.112 K=1000x4.9x 3.26
226 x MB
MB= 1000X4.90X3.26/226X0.1112=63g/mol
No. of Se atoms in a molecule=631g / mol/78.8 g/mol=8
Therefore, molecular formula of Selenium = Se8
5 MARKS QUESTION
Q1. a) State Raoult's Law for a solution containing volatile components.
How does Raoult's law become a special case of Henry's Law?
b) 1.00 g of a non‐electrolyte solute dissolved in 50 g of benzene lowered the freezing point of a benzene by
0.40 K. Find the molar mass of the solute. (Kf for benzene = 5.12 K kg mol–1)
Ans. a) For a solution of volatile liquids , Raoult's law states that the partial vapour pressure of each
component of the solution is directly proportional to its mole fraction present in solution, i.e., pA∝xA
OR
o
pA = p A x A
According to Henry's Law , the partial pressure of a gas in vapour phase (p)is
Directly proportional to mole fraction (x) of the gas in the solution.
i.e., p = KHx on comparing it with Raoult's Law it can be seen that partial pressure of the volatile component or
gas is directly proportional to its mole fraction in solution
i.e; p ∝x
only the proportionality constant K differs from p0 . Thus, it becomes a special case of
Henry's law in which K = po . H A

b) Substituting the values of various terms involved in equation MB = Kf x WB x 1000

∆Tf x WA
–1
MB = 5.12 x 1.00 x 1000 = 256g mol
0.40 x 50
Q2.a) Calculate the molarity of a sulphuric acid solution in which the mole fraction of water is 0.85.
b) The graphical representation of vapour pressure of two component system as a function of
composition is given alongside.
i) Are the A – B interactions weaker, stronger or of the same magnitude as A – A and B – B
ii) Name the type of deviation shown by this system from Raoult's law.
iii) Predict the sign of ∆mixH for this system.
iv) Predict the sign of ∆mixV for this system.
v) Give an example of such a system.
vi) What type of azeotrope will this system form, if possible ?
Ans. a) nA = 0.85 .......................... i)
nA + nB = 1‐0.85 = 0.15 ........................................ ii
nB
nB + nA
Dividing (ii) by (i), we get
nB 0.15 or 0.15 or nB = 0.15 x 1000
1000/18 =
0.85
nA = 0.85 0.85 x 18
nB
hence molality = 9.8 m

b) i) Stronger
ii) Negative deviation
iii) Negative
iv) Negative
v) 20% acetone and 80%
chloroform by mass
vi) maximum boiling
azeotrope
 ASSIGNMENT
Q1. Define mole fraction
Q2. What type of intermolecular attractive interaction exists in the pair of methanol and acetone?
Q3. What do you understand by “colligative properties” ?
Q4. Why is the vapour pressure of a solution of glucose in water lower than that of water?
Q5. State any two characteristics of ideal solutions.
Q6. Some liquids on mixing form “azeotrpoes”. What are azeotropes ?
Q7. Define molal elevation constant or ebullioscopic constant .
Q8. What is “reverse osmosis “ ?
Q9. Derive an equation to express that relative lowering of vapour pressure for a solution is equal to the
mole fraction of the solute in it when the solvent alone is volatile.
Q10. State Raoult's law for the solution containing volatile components. What is the similarity
between Raoult's law and Henry's law ?
Q11. Boiling point of water at 750 mm Hg is 99.63oC. How much sucrose is to be added to 500g of
water such that it boils at 100oC ?
Q12. 18 g of glucose , C6 H12 O6 (Molar Mass = 180 g mol 1) is dissolved in 1 kg of water in a sauce pan.
At what temperature will this solution boil ? (Kb for water = 0.52 K kg mol 1, boiling point of pure
water = 373.15 K )
Q13. After removing the outer shell of the two eggs in dil. HCl, one is placed in distilled water and the other
in a saturated solution of NaCl . What will you observe and why ?
Q14. Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid
components are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure of a
mixture of 26.0g of heptane and 35.0 g of octane ?
(MULTIPLE CHOICE QUESTION)MCQ
1. Which of the following units is useful in relating concentration of solution with its vapour
pressure?
(i) mole fraction
(ii) parts per million
(iii) mass percentage
(iv) molality
2. On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the
following cases dissolution of sugar will be most rapid?
(i) Sugar crystals in cold water.
(ii) Sugar crystals in hot water.
(iii) Powdered sugar in cold water.
(iv) Powdered sugar in hot water.
3. At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.
(i) less than the rate of crystallisation
(ii) greater than the rate of crystallisation
(iii) equal to the rate of crystallisation
(iv) zero
4. A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when
small amount of ‘A’ is added to the solution. The solution is _________.
(i) saturated
(ii) supersaturated
(iii) unsaturated
(iv) concentrated
5. Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid
solvent does not depend upon ____________.
(i) Temperature
(ii) Nature of solute
(iii) Pressure
(iv) Nature of solvent
6. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to
____________.
(i) low temperature
(ii) low atmospheric pressure
(iii) high atmospheric pressure
(iv) both low temperature and high atmospheric pressure
7. Considering the formation, breaking and strength of hydrogen bond, predict which of the
following mixtures will show a positive deviation from Raoult’s law?
(i) Methanol and acetone.
(ii) Chloroform and acetone.
(iii) Nitric acid and water.
(iv) Phenol and aniline.
8. Colligative properties depend on ____________.
(i) the nature of the solute particles dissolved in solution.
(ii) the number of solute particles in solution.
(iii) the physical properties of the solute particles dissolved in solution.
(iv) the nature of solvent particles.
9. Which of the following aqueous solutions should have the highest boiling point?
(i) 1.0 M NaOH
(ii) 1.0 M Na2SO4
(iii) 1.0 M NH4NO3
(iv) 1.0 M KNO3
10. The unit of ebulioscopic constant is _______________.
(i) K kg /mol or K (molality)–1
(ii) mol kg/ K or K–1(molality)
(iii) kg mol–1 K–1 or K–1(molality)–1
(iv) K mol kg–1 or K (molality)
11. In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M
MgCl2 solution is _____________.
(i) the same
(ii) about twice
(iii) about three times
(iv) about six times
12. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because
_____________.
(i) it gains water due to osmosis.
(ii) it loses water due to reverse osmosis.
(iii) it gains water due to reverse osmosis.
(iv) it loses water due to osmosis.
13. At a given temperature, osmotic pressure of a concentrated solution of a substance
_____________.
(i) is higher than that at a dilute solution.
(ii) is lower than that of a dilute solution.
(iii) is same as that of a dilute solution.
(iv) cannot be compared with osmotic pressure of dilute solution.
14. Which of the following statements is false?
(i) Two different solutions of sucrose of same molality prepared in different solvents will have the
same depression in freezing point.
(ii) The osmotic pressure of a solution is given by the equation Π = CRT ( where C is the molarity
of the solution).
(iii) Decreasing order of osmotic pressure for 0.01 M aqueous solutions of barium chloride,
potassium chloride, acetic acid and sucrose is BaCl2 > KCl > CH3COOH > sucrose.
(iv) According to Raoult’s law, the vapour pressure exerted by a volatile component of a solution is
directly proportional to its mole fraction in the solution.
15. The values of Van’t Hoff factors for KCl, NaCl and K2SO4, respectively, are _____________.
(i) 2, 2 and 2
(ii) 2, 2 and 3
(iii) 1, 1 and 2
(iv) 1, 1 and 1
16. Which of the following statements is false?
(i) Units of atmospheric pressure and osmotic pressure are the same.
(ii) In reverse osmosis, solvent molecules move through a semipermeable membrane from a region
of lower concentration of solute to a region of higher concentration.
(iii) The value of molal depression constant depends on nature of solvent.
(iv) Relative lowering of vapour pressure, is a dimensionless quantity.
17. Value of Henry’s constant KH ____________.
(i) increases with increase in temperature.
 (ii) decreases with increase in temperature.
(iii) remains constant.
(iv) first increases then decreases.
18. The value of Henry’s constant KH is _____________.
(i) greater for gases with higher solubility.
(ii) greater for gases with lower solubility.
(iii) constant for all gases.
(iv) not related to the solubility of gases.
19. If two liquids A and B form minimum boiling azeotrope at some specific composition then
_______________.
(i) A–B interactions are stronger than those between A–A or B–B.
(ii) vapour pressure of solution increases because more number of molecules of liquids A and B can
escape from the solution.
(iii) vapour pressure of solution decreases because less number of molecules of only one of the
liquids escape from the solution.
(iv) A–B interactions are weaker than those between A–A or B–B.
20. We have three aqueous solutions of NaCl labeled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1M,
0.01M and 0.001M, respectively. The value of van’t Hoff factor for these solutions will be in the
order______.
(i) iA < iB < iC
(ii) iA > iB > iC
(iii) iA = iB = iC
(iv) iA < iB > Ic
Answers:
(i) 2. (iv) 3. (iii) 4. (ii), [Hint : If added substance dissolves, the solution is unsaturated. If it
does not dissolve solution is saturated. If precipitation occurs solution is supersaturated.]
5. (iii) 6. (ii), [Hint : Body temperature of human beings remains constant.] 7. (i) 8. (ii) 9.
(ii) 10. (i) 11. (iii) 12. (iv) 13. (i) 14. (i) 15. (ii) 16. (ii) 17. (i) 18. (ii) 19. (i) 20. (iii)
 FLOW CHART

Expressing Conc of Solution

SOLUTIONS

Ideal Solution Non Ideal Solution

i) P =Po X
A A A

P =Po X
B B B

ii) ∆H=0
iii) ∆ V=0 Positive Deviations Negative Deviations
(i) P > Po X i) P < Po X
A A A A A A
o o
P >P X P<P X
B B B B B B

(ii) ∆H > O (+ve) ii) ∆H< O (‐ve)

(iii) ∆V> O (+ve) iii) ∆V < O (‐ve)
 APPENDIX –A
Important formula (from unit 1 ‐ 3)
UNIT- 1 SOLUTIONS
HENRY'S LAW P=KHX KH is Henry's law constant .
RAOULT'S LAW :‐ PA= P0 AX A ; PB= P0 BXB Ptotal= PA + PB
5. RELATIVE LOWERING OF VAPOUR PRESSURE
P0 – P /P0 = X X = n / n +n
A A A B
B A A B

For dilute solution, nB << nA, hence nB is neglected in the denominator.

P0A – PA/ P0A = nB /nA
P0A – PA/ P0A = WB*MA/MB*WA
6. ELEVATION OF BOILING POINT
∆Tb= kb m
Where , ∆Tb = Tb – T0b
M= kb 1000 WB/∆Tb WA
7. DEPRESSION IN FREEZING POINT
∆Tf = Kf m
Where , ∆T = T0 – T
f f f
M= kf 1000 WB/∆Tf WA
8. OSMOTIC PRESSURE
Π= CRT Π= n/VRT
1
R= 0.0821 Latm mol ;
i = normal molecular mass/ observed molecular mass
VAN'T HOFF FACTOR (i)
i = observed colligative properties/ calculated value of colligative properties i<1 (for association)
i>1 (for dissociation)
MODIFIED FORMS OF COLLIGATIVE PROPERTIES
5) P0 – P / P0 = i n /n
A A A B A
6) ∆Tb = i Kb m
7) ∆Tf = i Kf m
8) Π= i CRT
UNIT 2 ELECTROCHEMISTRY
1.) ƿ= R a\l
2.) K = 1\ƿ
3.) K = 1\R l\a
4.) ᴧm = k100\c
5.) E = E0 ‐ 2.303R.T\nF log K
cell cell
6.) At 250 C E =E0 ‐ 0.0591\n log K 7.) At equilibrium
cell cell

E0 =cell
0.0591 \n log K
8.) ∆ G = ‐ nFE0
UNIT 3 CHEMICAL KINETICS
1. 1. Rate of reaction:‐ For a reaction R → P,

Rate of reaction = change of conc. of R or P / T time interval

D
2. Order of reaction: 2. For the reaction aA +bB→ cC+ d

3. Rate = K[A]x [B]y

4. Units of rate constants and graph between rate and conc. of reactat

Order of zero first second third

reaction

Unit of rate molL‐ 1s ‐ 1 s‐ 1 mol‐ 1L+1s‐ 1 mol‐ 2L+2s‐ 1

constt.

Relation b/w R α [A]0 R α [A]1 R α [A]2 R α [A]3

rate & conc of

Reactant

Graph b/w rate & R R R R

conc of
Reactant [A]1 [A]2 [A]3

[A]
5. Integrated rate equation for zero order and first order reaction
for zero order reaction for first order reaction
Integrated rate equation

Half life t1/2 = [R]0 /2K t1/2 = 0.693/K

Graph b/w half life & conc of

Reactant

t1/2 t1/2

[R] [R]
Graph b/w conc.of reactant & time
[R] log[R]

Time Time

1‐n
6. t α [conc]
1/2
where n = order of reaction.
7. Arrhenius equation

log K = log A – Ea/2.303RT