IIT CHEMISTRY
SOLUTION & COLLIGATIVE
PROPERTIES
Physical & Inorganic
Organic Chemistry
NV Sir
VKP Sir
B.Tech. IIT Delhi
M.Sc. IT-BHU
NUCLEON CHEMISTRY
Be in Equilibrium with JEE
A-479 Indra vihar, kota Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
NUCLEON CHEMISTRY
CLASSES FOR IIT-JEE
SOLUTION & COLLIGATIVE
PROPERTIES
Contents
Topic
Page No.
Theory
01 - 07
Exercise - 1
08 - 10
Exercise - 2
10 - 12
Exercise - 3
12 - 18
Exercise - 4
19 - 21
Exercise - 5
21 - 22
ANSWER KEY
22 - 23
Name : ____________________________ Contact No. __________________
A - 479 Indra Vihar , Kota
SOLUTION AND COLLIGATIVE PROPERTIES
Concentration Terms :
% Concentration
 % w/w =
weight of solute (g)
× 100
weight of solution (g)
gram of solutes
% w/v = volume of solution in mL  100
Molarity = No. of moles of solute per litre of solution.
1000
n
W
M = V (in L) =  M  x
V (in mL)
 
Molality = No. of moles of solute per kg(1000 g) of solvent.
1000
w
molality =   x
W (g)
M
Normality
No. of equivalents per litre of solution =
no. of equivalents of solute
volume of solution (inL )
Mole Fraction :
For binary mixture
moles of solute
n
Xsolute = total moles in solutions =
nN
moles of solvent
N
XSolvent = Total moles in solutions =
nN
Xsolute + XSolvent = 1
ppm (Parts Per Million).
wt. of solute (in g)
ppm (w/w) = wt. of solution (in g) × 106 = 1 million
Colligative Properties :
 Colligative properties :
The properties of the solution which are dependent only on the total no. of particles or total concentration
of particles in the solution & are not dependent on the nature of particle i.e. shape, size, neutral /
charge etc. of the particles.
There are 4 colligative properties of solution.

Osmotic pressure
 P 
lowering in vapour pressure  P 


 Elevation in boiling point (T b)
 Relative

Depression in freezing point (T f)
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 1
Osmosis :
The spontaneous flow of solvent particles from solvent side to solution side or from solution of low
concentration side to solution of high concentration side through a semipermeable membrane (SPM) is
known as osmosis.
SPM : A membrane which allows only solvent particles to move across it.
(a)
Natural : Semi permeable membrane
Animal/plant cell membrane formed just below the outer skins.
(b)
Artificial membranes also : A copper ferrocyanide.
Cu2[Fe(CN)6] & Silicate of Ni, Fe, Co can act as SPM.
Osmotic Pressure :
Definition : The external pressure which must be applied on solution side to stop the process of
osmosis is called osmotic pressure of the solution.
C1 > C2 particle movement.
Pext. = (1 – 2)
Pext. must be applied on the higher concentration side.

  concentration (molarity)
T
 = CST
S = ideal solution constant
= 8.314 J mol –1 K–1 (exp value)
= R (ideal gas) constant
 = CRT =
 = atm.
C – mol/lit.
R – 0.082 lit.atm. mol–1 K–1
T – kelvin
n
RT (just like ideal gas equation)
V
Type of solutions :
(a) Isotonic solution : Two solutions having same osmotic pressure are consider as isotonic solution.
1 = 2 (at same temperature)
(b) Hypertonic : If 1 > 2 , Ist solution is hypertonic solution w.r.t. 2nd solution
(c) Hypotonic : 2nd solution is hypotonic w.r.t. I st solution.
  Abnormal Colligative Properties :
Vant – Hoff correction
 For electrolytic solutes the No. of particles would be different then the No. of particles actually
added, due to dissociation or association of solute.
The actual extent of dissociation/association can be expressed as a correction factor known as
vant Haff factor (i).
Vant – Hoff factor : i =
moles of particles in solution after dissociation / association
moles of solute dissolved
If solute gets associated or dissociated in solution then experimentable / observed / actual value of
colligative property will be different from theoretically predicted value so it is also known as abnormal
colligative property.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 2
This abnornality can be calculated in terms of Vant-Hoff factor.
i=
exp/ observed / actual / abnormal value of colligative property
Theoretical value of colligative property
exp . / observed no. of particles or concentration
=
Theoritical no. of particles or concentration
Theoretica l mass of subs tan ce
= exp erimental molar mass of the subs tan ce
i > 1 dissociation
i < 1 association
exp .
i= 
theor .
Case - I
: Electrolyte dissociates
Relation between i &  (degree of dissociation) :
Let the electrolyte be A xBy
AxBy (aq.)  xAy+ + yBx–
n=x+y
= no. of particles in which 1 molecules of electrolyte dissociates
C [1  (n  1)]
C
i = 1 + ( n – 1) 
i=
Case - II : Electrolyte associates
Relation between degree of association  & i.
1 
i = 1 +   1 
n 
Relative lowering in vapour pressure ( RLVP) :
Vapour Pressure :
The conversion of a liquid to a vapour takes place in a visible way when the liquid boils, it takes place under
all conditions.
At eq. :
the rate of evaporation = rate of condense
H2 O ()
H2 O (g)
Kp = PH2O(g) eq.

finally
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 3


Since vapour pressure is an equilibrium constant so its value is dependent only on temperature for
a particular liquid
It does not depends on the amount of liquid taken or surface area of the liquid or on volume or shape
of the container. It is a characteristic constant for a given liquid.
Vapour Pressure of a solution
Vapour Pressure of a solutions of a non volatile solute ( solid solute ) is always found to be less than
the vapour pressure of pure solvent .
 Reason :
Some of the solute molecules will occupy some surface area of the solutions so tendency of the
solvent particles to go into the vapour phase is slightly decreased hence
Ps
P
solvent
solution
Ps < P
PS < P
Figure
Lowering in Vapoure pressure = P – PS
= P
P
P
Raoult's law : – ( For non – volatile solutes )
Relative lowering in Vapoure pressure =
Experimentally relative lowering in Vapoure pressure = mole fraction of the non volatile solute in
solutions.
RLVP
=
P - Ps
n
= XSolute =
P
nN
P - Ps
n
=
Ps
N
P - Ps
M
= ( molality ) ×
Ps
1000
(M = molar mass of solvent)
Elevation in Boiling point (T b)
Boiling point of a Liquid :
The temperature at which vapour pressure of a liquid becomes equal to the external pressure present
at the surface of the liquid is called boiling point of liquid at that pressure.
Using Raoult’s law :
Tb = Kb × molality

Kb is dependent on property of solvent and known as molal elevation constant of solvent.

It is also known as ebullioscopic constant.

Kb = elevation in boiling point of 1 molal solution.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 4
Units :
Tb
K

= K kg mol–1
molality mol / kg
Using thermodynamics
2
RTb
1000  L vap

Kb =

LVap – is latent heat of vapourisation in cal/g or J/g
R = 2 cal mol–1 K–1 or 8.314 J mol–1 K–1.
Tb = boiling point of liquid (in kelvin)
Kb = K kg mol–1
2
 Also Kb =
RTb M
1000  Hvap
Hvap – molar enthalpy of vapourisation in cal/mole or J/mole
Here
M  mole wt. of solvent
 Hvap 


 M 
Lvap = 
If solute gets associated/dissociated.
Tb = i × Kb × molality
Depression in freezing point (T f)
Freezing point : Temperature at which vapour pressure of solid becomes equal to vapour pressure of liquid
is called freezing point of liquid or melting point of solid.


Tf = Kf . Molality
Kf = molal depression constant = cryoscopic constant
2
Kf =
2
RTf
RTf M
=
1000  L fusion
1000  Hfusion
for water Tf = 273 K & LFusion = 80 cal / gm
Kf =
2  273  273
= 1.86 K kg mol–1
1000  80
VOLATILE SOLVENT  Liq. solution

VOLATILE SOLUTE  Liq. solution
SOLUTIONS CONTAINING 
According to Raoult's law (experimentally )
PA  XA
PA = XAPAº
PAº = vapour pressure of pure liquid A = constant (at a particular temperarure)
Similarly.
PB  XB.
º
B

PB = XBP (vapoure pressure of pure liquid B)

According to Dalton's law
PT = PA + PB = XAPA0 + XBPB0
xA' = mole fraction of A in vapour above the liquid / solution.
xB' = mole fraction of B
1
xA '
xB '
=
+
PT
PA º
PB º
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 5
Graphical Representation :
PT
PAº
PA
PB,º
PB,
A is more volatile than B PAº > PBº
XA= 0
XB= 1
XA= 1
XB= 0
Ideal solutions (mixtures) :
The mixtures which follow Raoult's law at all temperature & at all compositions will be known as
ideal mixtures / ideal solution.
Characteristic of an ideal solution

Ideal solution will be obtained only when the forces of attraction between the liquid
molecules are exactly of same nature & almost of same magnitude
A ------ A
 A -------- B,
B ----- B

Hmix = 0

Vmix = 0

Smix = + ve as for process to proceed

Gmix = – ve
eg.
(1 ) Benzene + Toluene.
(2) Hexane + heptane.
(3) C2H5Br + C2H5.

Non - deal solutions :
The mixtures which do not follow Raoult's law will be known as non ideal mixtures/solution.
Non ideal solution can be of two types :
 Non ideal solutions showing positive deviation
 Non ideal solutions showing negative deviation
+ ve deviation


–ve deviation
PT,exp > ( XAPºA + XBPBº )
A    A
B    B > A ---- B


PT exp < xApº + xBpºB

A    A
B    B < A ------ B.

strong force of altraction.
Weaker force of attraction
eg.

Hmix = +ve
energy absorbed

Hmix = –ve

Vmix = +ve
( 1L + 1L > 2L )

Vmix = –ve

Smix = +ve

Smix = +ve

Gmix = –ve

Gmix = –ve
eg.
H2O + HCOOH
H2O + CH3COOH
H2O + HNO3
CHCl3 + CH3OCH3
H2O + CH3OH.
H2O + C2H5OH
C2H5OH + hexane
C2H5OH + cyclohexane.
( 1L + 1L < 2L )
CH3
CHCl3 + CCl4 
C=O
CH3
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
H
C
Cl
Cl
Cl
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 6
Immiscible Liquids :

It is used to purify an organic liquid from impurity.

When two liquids are mixed in such a way that they do not mix at all then
Ptotal = PA + PB
PA = PA0 XA
XA = 1
PA =
PA0
PB = PB0 XB
XB = 1
PB = PB0
Ptotal = PA0 + PB0
PA0
PB0
nA
= n
B
 Henry Law :
Henry law deal with effect of pressure on the solubility of gas.
Statement :
The solubility of a gas in a liquid at a given temperature is directly proportional to the pressure at which it is
dissolved.
Let
X = Mole fraction of gas at a given temperature as a measure of its solubility.
p = Partial pressure gas in equilibrium with the solution.
Then according to Henry's law.
Xp
or
pX
or
p = KHX
where KH = Henry law constant.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 7
Raoult’s law
1.
At 25°C, the vapour pressure of methyl alcohol is 96.0 torr. What is the mole fraction of CH3OH in a solution
in which the (partial) vapor pressure of CH3OH is 23.0 torr at 25°C?
2.
The vapour pressure of pure liquid solvent A is 0.80 atm. When a nonvolatile substance B is added to the
solvent its vapour pressure drops to 0.60 atm. What is the mole fraction of component B in the solution?
3.
The vapour pressure of pure water at 26°C is 25.21 torr. What is the vapour pressure of a solution which
contains 20.0 glucose, C6H12O6, in 70 g water?
4.
The vapour pressure of pure water at 25°C is 23.76 torr. The vapour pressure of a solution containing 5.40
g of a nonvolatile substance in 90.0 g water is 23.32 torr. Compute the molecular weight of the solute.
Raoult’s law in combinaton with Dalton’s law of P.P. and V.P. lowering
5.
The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution
is prepared at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate total vapour
pressure of the solution.
6.
Calculate the mole fraction of toluene in the vapour phase which is in equilibrium with a solution of benzene
and toluene having a mole fraction of toluene 0.50. The vapour pressure of pure benzene is
119 torr; that of toluene is 37 torr at the same temperature.
7.
At 90°C, the vapour pressure of toluene is 400 torr and that of -xylene is 150 torr. What is the composition
of the liquid mixture that boils at 90°C, when the pressure is 0.50 atm? What is the composition of vapour
produced?
8.
Two liquids A and B form an ideal solution at temperature T. When the total vapour pressure above the
solution is 400 torr, the mole fraction of A in the vapour phase is 0.40 and in the liquid phase 0.75. What are
the vapour pressure of pure A and pure B at temperature T?
9.
Calculate the relative lowering in vapour pressure if 100 g of a nonvolatile solute (mol.wt.100) are dissolved
in 432 g water.
10.
What weight of the nonvolatile solute, urea needs to be dissolved in 100 g of water, in order to decrease
the vapour pressure of water by 25%? What will be the molality of the solution?
11.
The vapour pressure of pure benzene at 25° C is 639.7 mm of Hg and the vapour pressure of a solution of
a solute in C6H6 at the same temperature is 631.7 mm of Hg. Calculate molality of solution.
12.
The vapour pressure of pure benzene at a certain temperature is 640 mm of Hg. A nonvolatile nonelectrolyte
solid weighing 2.175 g is added to 39.0 of benzene. The vapour pressure of the solution is 600 mm of Hg.
What is molecular weight of solid substance?
13.
Benzene and toluene form two ideal solution A and B at 313 K. Solution A (total pressure PA) contains equal
mole of toluene and benzene. Solution B contains equal masses of both (total pressure PB). The vapour
pressure of benzene and toluene are 160 and 60 mm Hg respectively at 313 K. Calculate the value of
PA/PB.
Boiling point elevation and freezing point depression
14.
When 10.6 g of a nonvolatile substance is dissolved in 740 g of ether, its boiling point is raised 0.284°C.
What is the molecular weight of the substance? Molal boiling point constant for ether is 2.11°C·kg/mol.
15.
A solution containing 3.24 of a nonvolatile nonelectrolyte and 200 g of water boils at 100.130°C at
1atm. What is the molecular weight of the solute? (Kb for water 0.513°C/m)
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 8
16.
The molecular weight of an organic compound is 58.0 g/mol. Compute the boiling point of a solution
containing 24.0 g of the solute and 600 g of water, when the barometric pressure is such that pure water
boils at 99.725°C.
17.
An aqueous solution of a nonvolatile solute boils at 100.17°C. At what temperature will this solution freeze?
[Kf for water 1.86°C/m ]
18.
Pure benzene freeze at 5.45°C. A solution containing 7.24 g of C2H2Cl4 in 115.3 g of benzene was observed
to freeze at 3.55°C. What is the molal freezing point constant of benzene?
19.
The freezing point of a solution containing 2.40 g of a compound in 60.0 g of benzene is 0.10°C lower than
that of pure benzene. What is the molecular weight of the compound? (Kf is 5.12°C/m for benzene)
20.
Calculate the molal elevation constant, Kb for water and the boiling of 0.1 molal urea solution. Latent heat
of vaporisation of water is 9.72 kcal mol–1 at 373.15 K.
21.
Calculate the amount of ice that will separate out of cooling a solution containing 50g of ethylene glycol in
200 g water to –9.3°C. (Kf for water = 1.86 K mol1 kg)
22.
A solution of 0.643 g of an organic compound in 50ml of benzene (density ; 0.879 g/ml) lowers its freezing
point from 5.51°C to 5.03°C. If Kf for benzene is 5.12 K, calculate the molecular weight of the compound.
Osmotic pressure
23.
Find the freezing point of a glucose solution whose osmotic pressure at 25oC is found to be 30 atm.
Kf(water) = 1.86kg.mol1.K.
24.
At 300 K, two solutions of glucose in water of concentration 0.01 M and 0.001 M are separated by
semipermeable membrane. Pressure needs to be applied on which solution, to prevent osmosis? Calculate
the magnitude of this applied pressure.
25.
At 10oC, the osmotic pressure of urea solution is 500 mm. The solution is diluted and the temperature is
raised to 25°C, when the osmotic pressure is found to be105.3 mm. Determine extent of dilution.
26.
The osmotic pressure of blood is 7.65 atm at 37°C. How much glucose should be used per L for an
intravenous injection that is to have the same osmotic pressure as blood?
27.
A 250 mL water solution containing 48.0 g of sucrose, C12H22O11, at 300 K is separated from pure water by
means of a semipermeable membrane. What pressure must be applied above the solution in order to just
prevent osmosis?
28.
A 5% solution (w/v) of cane-sugar (Mol. weight = 342) is isotonic with 0.877%(w/v) of urea solution. Find
molecular weight of urea.
29.
10 gm of solute A and 20 gm of solute B are both dissolved in 500 ml water. The solution has the same
osmotic pressure as 6.67 gm of A and 30 gm of B dissolved in the same amount of water at the same
temperature. What is the ratio of molar masses of A and B?
Van’t Hoff factor & colligative properties
30.
A storage battery contains a solution of H2SO4 38% by weight. What will be the Van't Hoff factor if the
Tf(experiment) in 29.08. [Given Kf = 1.86 mol–1 Kg]
31.
A certain mass of a substance, when dissolved in 100 g C6H6, lowers the freezing point by 1.28°C. The
same mass of solute dissolved in 100g water lowers the freezing point by 1.40°C. If the substance has
normal molecular weight in benzene and is completely ionized in water, into how many ions does it dissociate
in water? Kf for H2O and C6H6 are 1.86 and 5.12K kg mol1.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 9
32.
2.0 g of benzoic acid dissolved in 25.0g of benzene shows a depression in freezing point equal to 1.62K.
Molal depression constant (Kf) of benzene is 4.9 K.kg.mol1. What is the percentage association of the
acid?
33.
A decimolar solution of potassium ferrocyanide is 50% dissociated at 300K. Calculate the osmotic pressure
of the solution. (R=8.314 JK1 mol1)
34.
The freezing point of a solution containing 0.2 g of acetic acid in 20.0g of benzene is lowered by 0.45°C.
Calculate the degree of association of acetic acid in benzene. (Kf for benzene = 5.12 K mol1 kg)
35.
0.85 % aqueous solution of NaNO3 is apparently 90% dissociated at 27°C. Calculate its osmotic pressure.
(R= 0.082 l atm K1 mol1 )
36.
A 1.2% solution (w/v) of NaCl is isotonic with 7.2% solution (w/v) of glucose. Calculate degree of ionization
and Van’t Hoff factor of NaCl.
37.
If relative decrease in V.P. is 0.4 for a solution containing 1 mol NaCl in 3 mole of H2O. Calculate %
ionization of NaCl.
38.
2.56 gm of sulphur in 100 gm of CS2 has depression in F.P. of 0.010°C, Kf = 0.1°C m–1. Calculate atomicity
of sulphur in CS2.
Henry’s Law
39.
1 kg of water under a nitrogen pressure of 1atomsphere dissolves 0.02 gm of nitrogen at 293 K. Calculate
Henry’s law constant.
40.
Calculate the amount of oxygen at 0.20 atm dissolved in 1 kg of water at 293 K. The Henry’s law constant
for oxygen is 4.58 × 104 atomsphere at 293 K.
1.
An aqueous solution containing 288 gm of a non-volatile compound having the stoichiometric composition
CxH2xOx in 90 gm water boils at 101.24°C at 1.00 atmospheric pressure. What is the molecular formula?
Kb(H2O) = 0.512 K mol–1 kg ; Tb(H2O) = 100°C
2.
The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7 gm of the salt per
100 gm of water at 100°C is 70%. If the vapour pressure of water at 100°C is 760 mm. Calculate the vapour
pressure of the solution.
3.
The addition of 3 gm of substance to 100 gm CCl4(M = 154 gm mol–1) raises the boiling point of CCl4 by
0.60°C of Kb (CCl4) is 5.03 kg mol–1 K. Calculate
(a)
the freezing point depression
(b)
the relative lowering of vapour pressure
(c)
the osmotic pressure at 298 K
(d)
the molar mass of the substance
Given Kf(CCl4) = 31.8 kg mol–1K and  (density) of solution = 1.64 gm/cm3
4.
A 10% solution of cane sugar has undergone partial inversion according to the reaction:
Sucrose + Water  Glucose + Fructose. If the boiling point of solution is 100.27°C.
(a)
What is the average mass of the dissolved materials?
(b)
What fraction of the sugar has inverted?
Kb(H2O) = 0.512 K mol–1 kg
5.
1.5 g of a monobasic acid when dissolved in 150g of water lowers the freezing point by 0.165°C. 0.5 g of
the same acid when titrated, after dissolution in water, requires 37.5 ml of N/10 alkali. Calculate the degree
of dissociation of the acid (Kf for water = 1.86°C mol–1).
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 10
6.
Sea water is found to contain 5.85% NaCl and 9.50% MgCl2 by weight of solution. Calculate its normal
boiling point assuming 80% ionisation for NaCl and 50% ionisation of MgCl2 [Kb(H2O) = 0.51kgmol1K].
7.
The vapour pressure of an aqueous solution of glucose is 750 mm Hg at 373 K. Calculate molality and
mole fraction of solute.
8.
A complex is represented as CoCl3 · x NH3. It's 0.1 molal solution in aq. solution shows Tf = 0.558°C. Kf
for H2O is 1.86 K mol–1 kg . Assuming 100% ionisation of complex and coordination no. of Co is six,
calculate formula of complex.
9.
The molar volume of liquid benzene (density = 0.877 g ml1) increases by a factor of 2750 as it vaporizes
at 20°C and that of liquid toluene (density = 0.867gml1) increases by a factor of 7720 at 20°C.
Solution of benzene & toluene has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the
vapour above the solution.
10.
At 100oC, benzene & toluene have vapour pressure of 1375 & 558 Torr respectively. Assuming these two
form an ideal binary solution, calculate the composition of the solution that boils at 1 atm & 100oC. What is
the composition of vapour issuing at these conditions?
11.
Calculate the boiling point of a solution containing 0.61g of benzoic acid in 50g of carbon disulphide assuming
84% dimerization of the acid. The boiling point and Kb of CS2 are 46.2°C and 2.3 K kg mol–1, respectively.
12.
The cryoscopic constant for acetic acid is 3.6 K kg/mol. A solution of 1 g of a hydrocarbon in 100 g of acetic
acid freezes at 16.14°C instead of the usual 16.60°C. The hydrocarbon contains 92.3% carbon. What is
the molecular formula?
13.
Phenol associates in benzene to a certain extent to form a dimer. A solution containing 20 × 10–3 kg phenol
in 1 kg of benzene has its freezing point depressed by 0.69 K. Calculate the fraction of phenol that has
dimerised. Kf for benzene = 5.12 kg mol–1K.
14.
30 ml of CH3OH (d = 0.7980 gm cm–3) and 70 ml of H2O (d=0.9984 gm cm–3) are mixed at 25°C to form a
solution of density 0.9575 gm cm –3. Calculate the freezing point of the solution. K f (H 2O) is
1.86 kg mol–1 K. Also calculate its molarity.
15.
Dry air was drawn thorugh bulbs containing a solution of 40 grams of urea in 300 grams of water, then
through bulbs containing pure water at the same temperature and finally through a tube in which pumice
moistened with strong H2SO4 was kept. The water bulbs lost 0.0870 grams and the sulphuric acid tube
gained 2.036 grams. Calculate the molecular weight of urea.
16.
Vapour pressure of C6H6 and C7H8 mixture at 500C is given by P (mm Hg) = 180 XB + 90, where XB is the
mole fraction of C6H6. A solution is prepared by mixing 936g benzene and 736g toluene and if the vapours
over this solution are removed and condensed into liquid and again brought to the temperature of 500C,
what would be mole fraction of C6H6 in the vapour state?
17.
When the mixture of two immicible liquids (water and nitrobenzene) boils at 372 K and the vapour pressure
at this temperature are 97.7 kPa (H2O) and 3.6 kPa (C6H5NO2). Calculate the weight % of nitrobenzene in
the vapour.
18.
The osmotic pressure of a solution of a synthetic polyisobutylene in benzene was determined at 25°C. A
sample containing 0.20 g of solute/100 cm3 of solution developed a rise of 2.4 mm at osmotic equilibrium.
The density of the solution was 0.88 g/cm3. What is the molecular weight of the polyisobutylene?
19.
A very dilute saturated solution of a sparingly soluble salt A3B4 has a vapour pressure of 20 mm of Hg at
temperature T, while pure water exerts a pressure of 20.0126 mm Hg at the same temperature. Calculate
the solubility product constant of A3B4 at the same temperature.
20.
If the apparent degree of ionization of KCl (KCl =74.5 gm mol–1) in water at 290 K is 0.86. Calculate the
mass of KCl which must be made up to 1 dm3 of aqueous solution to the same osmotic pressure as the
4.0% solution of glucose at that temperature.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 11
21.
An ideal solution was prepared by dissolving some amount of cane sugar (nonvolatile) in 0.9 moles of
water. The solution was then cooled just below its freezing temperature (271 K), where some ice get
separated out. The remaining aqueous solution registered a vapour pressure of 700 torr at 373 K. Calculate
the mass of ice separated out, if the molar heat of fusion of water is 6 kJ.
22..
The specific conductivity of a 0.5 M aq. solution of monobasic acid HA at 27°C is 0.006 Scm–1. It's molar
conductivity at infinite dilution is 200 S cm2 mol–1.
Calculate osmotic pressure (in atm) of 0.5 M HA (aq) solution at 27°C.
Given R = 0.08
atm L
.
mol K
23
What would be the osmotic pressure at 17°C of an aqueous solution containing 1.75 g of sucrose (C12H22O11)
per 150 cm3 of solution?
24.
The vapour pressure of two pure liquids, A and B that form an ideal solution are 300 and 800 torr respectively,
at temperature T. A mixture of the vapour of A and B for which the mole fraction of A is 0.25 is slowly
compressed at temperature T. Calculate
(a) the composition of the first drop of the condensate,
(b) the total pressure when this drop is formed,
(c) the composition of the solution whose normal boiling point is T,
(d) the pressure when only the last bubble of vapour remains, and
(e) the composition of the last bubble.
25.
Tritium, T (an isotope of H) combines with fluorine to form weak acid TF, which ionizes to give T+. Tritium is
radioactive and is a –emitter. A freshly prepared aqueous solution of TF has pT (equivalent of pH) of 1.5
and freezes at –0.3720C. If 600ml of freshly prepared solution were allowed to stand for 24.8 years. Calculate
(i) ionization constant of TF. (ii) Number of –particles emitted.
(Given Kf for water = 1.86 kg mol K–1, t1/2 for tritium = 12.4 years)
1.
For an ideal binary liquid solution with PA > PB , which relation between XA (mole fraction of A in liquid


phase) and YA(mole fraction of A in vapour phase) is correct?
(A) YA < YB
2.
(B) XA > XB
(C)
YA X A

YB X B
(D)
YA X A

YB X B
Mole fraction of A vapours above the solution in mixture of A and B (XA = 0.4) will be
[Given : PA = 100 mm Hg and PB = 200 mm Hg]
(A) 0.4
3.
(C) 0.25
(D) none of these
The exact mathematical expression of Raoult’s law is
(A)
4.
(B) 0.8
P 0  Ps
P0
n

N
(B)
P 0  Ps
P0
N

n
P 0  Ps n

(C)
Ps
N
(D)
P 0  Ps
P0
=n×N
A mixture contains 1 mole of volatile liquid A ( PA =100 mm Hg) and 3 moles of volatille liquid
B ( PB = 80 mm Hg). If solution behaves ideally, the total vapour pressure of the distillate is
(A) 85 mm Hg
(B) 85.88 mm Hg
(C) 90 mm Hg
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
(D) 92 mm Hg
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 12
5.
The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the
solvent. The mole fraction of solute in solution is 0.2, what would be mole fraction of the solvent if decrease
in vapour pressure is 20 mm of Hg
(A) 0.2
(B) 0.4
(C) 0.6
(D) 0.8
6.
Two liquids A & B form an ideal solution. What is the vapour pressure of solution containing 2 moles of A
and 3 moles of B at 300 K? [Given : At 300 K, Vapour pr. of pure liquid A ( PAo ) = 100 torr,, Vapour pr. of
pure liquid B ( PBo ) = 300 torr ]
(A) 200 torr
(B) 140 torr
(C) 180 torr
(D) None of these
7.
The solubility of common salt is 36.0 gm in 100 gm of water at 20°C. If system I, II and III contains 20.0,
18.0 and 15.0 g of the salt added to 50.0 gm of water in each case, the vapour pressure would be in the
order.
(A) I < II < III
(B) I > II > III
(C) I = II > III
(D) I = II < III
8.
18 g of glucose (C6H12O6) is added to 178.2 of water. The vapour pressure of water for this aqueous
solution at 100°C is
(A) 7.60 torr
(B) 76.00 torr
(C) 752.40 torr
(D) 759.00 torrp
9.
Benzene and toluene form nearlyideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and
that of toluene is 22 torr. The partial vapour pressure of benzene at 20°C for a solution containing 78 g of
benzene and 46 g of toluene in torr is:
(A) 50
(B) 25
(C) 37.5
(D) 53.5
10.
At 300 K, the vapour pressure of an ideal containing 3 mole of A and 2 mole of B is 600 torr. At the same
temperature, if 1.5 mole of A & 0.2 mole of C (non-volatile) are added to this solution the vapour pressure
of solution increases by 30 torr. What is the value of P°B ?
(A) 940
(B) 405
(C) 90
(D) None of these
11.
Which of the following plots represents an ideal binary mixture?
(A) Plot of Ptotal v/s 1/XB is linear (XB = mole fraction of 'B' in liquid phase).
(B) Plot of Ptotal v/s YA is linear (YB = mole fraction of 'A' in vapour phase)
1
(C) Plot of P
v/s YA is linear
total
1
(D) Plot of
Ptotal v/s YB is non linear
12.
The vapour pressure of a solution of a non-volatile electrolyte B in a solvent A is 95% of the vapour
pressure of the solvent at the same temperature. If the molecular weight of the solvent is 0.3 times the
molecular weight of solute, the weight ratio of the solvent and solute are
(A) 0.15
(B) 5.7
(C) 0.2
(D) 4.0
13.
At a given temperature, total vapour pressure in Torr of a mixture of volatile components A and B is given
by PTotal = 120 – 75 XB hence, vapour pressure of pure A and B respectively (in Torr) are :
(A) 120, 75
(B) 120, 195
(C) 120, 45
(D) 75, 45
14.
Which of the following aqueous solution will show maximum vapour pressure at 300 K?
(A) 1 M NaCl
(B) 1 M CaCl2
(C) 1 M AlCl3
(D) 1 M C12H22O11
15.
The Van’t Hoff factor for a dilute aqueous solution of glucose is
(A) zero
(B) 1.0
(C) 1.5
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
(D) 2.0
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 13
16.
The correct relationship between the boiling points of very dilute solution oif AlCl 3 (T 1K) and
CaCl2 (T2K) having the same molar concentration is
(A) T1 = T2
(B) T1 > T2
(C) T2 > T1
(D) T2  T1
17.
A 0.001 molal solution of a complex [MA8] in water has the freezing point of –0.0054°C. Assuming 100%
ionization of the complex salt and Kf for H2O = 1.86 km–1, write the correct representation for the complex
(A) [MA8]
(B) [MA7]A
(C) [MA6]A2
(D) [MA5]A3
18.
Assuming each salt to be 90 % dissociated, which of the following will have highest boiling point ?
(A) Decimolar Al2(SO4)3
(B) Decimolar BaCl2
(C) Decimolar Na2SO4
(D) A solution obtained by mixing equal volumes of (B) and (C)
19.
Elevation of boiling point of 1 molar aqueous glucose solution (density = 1.2 g/ml) is
(A) Kb
(B) 1.20 Kb
(C) 1.02 Kb
(D) 0.98 Kb
20.
What will be the molecular weight of CaCl2 determined in its aq. solution experimentally from depression of
freezing point?
(A) 111
(B) < 111
(C) > 111
(D) data insufficient
21.
1.0 molal aqueous solution of an electrolyte A2B3 is 60% ionised. The boiling point of the solution at 1 atm
is ( K
b ( H 2O )
(A) 274.76 K
 0.52 K kg mol 1 )
(B) 377 K
(C) 376.4 K
(D) 374.76 K
22.
The freezing point depression of a 0.1 M aq. solution of weak acid (HX) is –0.20°C.
What is the value of equilibrium constant for the reaction?
HX (aq)
H+(aq) + X¯ (aq)
[Given : Kf for water = 1.8 kg mol–1 K. & Molality = Molarity ]
(A) 1.46×10–4
(B) 1.35 × 10–3
(C) 1.21 × 10–2
(D) 1.35 × 10–4
23.
The van’t Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is
(A) 91.3%
(B) 87%
(C) 100%
(D) 74%
24.
The vapour pressure of an aqueous solution is found to be 750 torr at certain temperature 'T'. If 'T' is the
temperature at which pure water boils under atmospheric pressure and same solution show elevation in
boiling point Tb = 1.04 K, find the atmospheric pressure (Kb = 0.52 K kg mol–1 )
(A) 777
(B) 779
(C) 782
(D) 746
25.
A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution
is (Given Kf = 1.86° C kg mol–1 for water) :
(A) –0.45°C
(B) –0.90°C
(C) –0.31°C
(D) –0.53°C
26.
In a 0.2 m aqueous solution of a weak acid HX, the degree of ionization is 0.3. Taking Kf for water as 1.85,
the freezing point of the solution will be nearest to :
(A) –0.480°C
(B) 0.360°C
(C) –0.260°C
(D) +0.480°C
27.
Freezing point of an aqueous solution is –0.186°C. Elevation of boiling point of the same solution is Kb =
0.512°C m–1, Kf = 1.86°C m–1. The increases in boiling point is :
(A) 0.186°C
(B) 0.0512°C
(C) 0.0092°C
(D) 0.2372°C
28.
Which one of the following aqueous solutions will exhibit highest boiling point ?
(A) 0.01 M Na2SO4
(B) 0.01 M KNO3
(C) 0.015 M urea
(D) 0.015 M glucose
29.
If  is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for calculating the molecular
mass is :
(A) 1 + 
(B) 1 – 
(C) 1 + 2
(D) 1 – 2
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 14
30.
Which has the highest boiling point ?
(A) 0.1 M Na2SO4
(C) 0.1 M MgCl2
(B) 0.1 M C6H12O6 (glucose)
(D) 0.1 M Al (NO3)3
31.
Which combination of (I) vapour pressure, (II) intermolecular forces and (III) and Hvap(latent heat of
vaporisation) is matched correctly
I
II
III
I
II
III
(A)
low
strong small
(B)
high
strong
large
(C)
low
weak large
(D)
high
weak
small
32.
The molecular weight of benzoic acid in benzene as determined by depression in freezing point of the
solution is :
(A) ionization of benzoic acid
(B) dimerization of benzoic acid
(C) trimerization of benzoic acid
(D) solvation of benzoic acid
33.
If liquids A and B form an ideal solution, the
(A) enthalpy of mixing is zero
(B) entropy of mixing is zero
(C) free energy of mixing is zero
(D) free energy as well as the entropy of mixing are each zero
34.
Equimolar solutions in the same solvent have :
(A) same boiling point but different freezing point
(B) same freezing point but different boiling point
(C) same boiling and same freezing points
(D) different boiling and freezing points
35.
1 mol each of the following solutes are taken in 9 mol water,
(I) NaCl
(II) K2SO4
(III) Na3PO4
relative decrease in vapour pressure will be in order :
(A) I < II < III < IV
(B) IV < III < II < I
(C) IV < I < II < III
(IV) glucose
(D) equal
36.
In the depression of freezing point experiment, it is found that
(I) The vapour pressure of the solution is less than that of the pure solvent.
(II) The vapour pressure of the solution is more that of pure solvent.
(III) Only solute molecules solidify at the freezing point.
(IV) Only solvent molecules solidify at the freezing point.
(A) I, II
(B) II, III
(C) I, IV
(D) I, II, III
37.
There are some of the characteristics of the supersaturated solution
I. Equilibrium exists between solutions and solid state
II. If a crystal of solute is added to supersaturated solution, crystallisation occurs
III. Supersaturated solutions contain more solute than they should at a particular temperature
Correct characteristics of supersaturated solution are :
(A) I, II, III
(B) II, III
(C) I, III
(D) I, II
38.
Select correct statements :
(A) The fundamental cause of all colligative properties is the higher entropy of the solution relative to that of
the pure solvent.
(B) The freezing point of hydrofluoride solution is larger than that of equimolal hydrogen chloride solution
(C) 1M glucose solution and 0.5 M NaCl solution are isotonic at a given temperature
(D) All are correct statements
39.
In cold countries, ethylene glycol is added to water in the radiators of cars during winters. It results in
(A) lowering of boiling point
(B) reduced viscosity
(C) reduced specific heat
(D) lowering of freezing point
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 15
40.
A liquid mixture having composition corresponding to point z in the figure
shown is subjected to distillation at constant pressure. Which of the
following statement is correct about the process
(A) The composition of distillate differs from the mixture
(B) The boiling point goes on changing
(C) The mixture has highest vapour pressure than for any other composition.
(D) Composition of an azeotrope alters on changing the external pressure.
41.
Boiling point-composition diagram of the liquid-vapour equilibrium for propanol-2
(a) and-2-methyl-propanol-1
(b) is shown a side. If binary liquid mixture of a and b is distilled fractionally,
which of the following would be correct observation ?
(A) composition of the still (residue) will approach pure liquid a
(B) composition of the distillate will approach pure b
(C) composition of distillate and residue will approach pure a and b respectively
(D) Neither of the components can be obtained in pure state.
42.
At 27°C, 41 ml of ozone dissolves in 100 of water at a pressure of 1.00 atm. What mass of ozone dissolves
in 200 ml of water at a pressure of 3.00 atm and at the same temperature ?
(A) 0.240 g
(B) 0.480 g
(C) 0.600 g
(D) 0.720 g
43.
The solubility of gases in liquids :
(A) increase with increase in pressure and temperature
(B) decrease with increase in pressure and temperature
(C) increase with increase in pressure and decrease in temperature
(D) decreases with increase in pressure and increase in temperature
44.
Select incorrect statements :
(A) NH3 is solube in water due to hydrogen bonding as well as due to formation of ions
(B) Gases which can be liquefied easily are more soluble in water than the gases which cannot be liquefied
(C) The solute follows Henry law at all pressure of gas
(D) The solute follows Henry law at low pressure of gas
45.
At 20°C and a total pressure of 760 torr, 1 L of water dissolves 0.043 gm of pure oxygen 0.019 gm of pure
nitrogen. Assuming that dry air is composed of 20% oxygen and 80% nitrogen (by volume). The masses
(in g/L) of oxygen and nitrogen dissolved by 1 L of water of 20°C exposed to air at a total pressure of 760
torr are respectively.
(A) 0.0086, 0.015
(B) 0.015, 0.0086
(C) 0.15, 0.086
(D) 0.03, 0.8
46.
2 litre solution of 0.1 M NaCl and 4 litre of 0.1 M AgNO3 are mixed. The depression of freezing point of the
resultant solution will be [Kf(H2O) = 1.86 K kg mol–1]
(A) 0.372 K
(B) 1.488 K
(C) 0.124 K
(D) 0.248 K
47.
How many grams of glucose should be dissolved in 100 gm water in order to produce a solution with
101.19°C difference between the freezing point & boiling point temperatures ?
(Given Kf = 1.86 K kg mol–1 ; Kb = 0.52 K. kg mol–1)
(A) 9 gm
(B) 18 gm
(C) 180 gm
(D) None of these
48.
From separate solutions of four sodium salts NaW, NaX, NaY and NaZ had pH 7.0, 9.0, 10.0 and 11.0
respectively. When each solution was 0.1 M. Van’t Hoff factor (i) is maximum for acid
(A) HW
(B) HX
(C) HY
(D) HZ
49.
The solubility of N2(g) in water exposed to the atomsphere, when the partial pressure is 593 mm is
5.3 × 10–4 M. Its solubility at 760 mm and at the same temperature is :
(A) 4.1 × 10–4 M
(B) 6.8 × 10–4 M
(C) 1500 M
(D) 2400 M
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 16
50.
The following graph represents variation of boiling point with composition of liquid and vapours of binary
liquid mixture. The graph is plotted at constant pressure.
Which of the following statement(s) is incorrect. Here X & Y stands for mole fraction in liquid and vapour
phase respectively
(A) Xbenzene = 0.5 and Ytoluene = 0.2
(C) Xbenzene = 0.3 and Ytoluene = 0.4
51.
(B) Xtoluene = 0.3 and Ybenzene = 0.6
(D) if Xbenzene = 0.7 than Ytoluene < 0.3
Which of the following represents correctly the changes in thermodynamic properties during the formation
of 1 mol of an ideal binary solution.
(A)
(B)
(C)
(D)
52.
FeCl3 on reaction with K4[Fe(CN)6] in aqueous solution gives blue colour. These are separated by a
semipermeable membrane AB as shown. Due to osmosis there is
(A) blue colour formation in side X.
(B) blue colour formation in side Y.
(C) blue colour formation in both of the sides X and Y.
(D) no blue colour formation.
53.
Statement-1 : The freezing point of water of depressed by the addition of glucose.
Statement-2 : Entropy of solution is less than entropy of pure solvent.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
54.
Statement-1 : The difference in the boiling points of equimolar solution of HCl and HF decreases are their
molarity is decreased.
Statement-2 : The extent of dissociation decreases with increases dilution.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 17
56.
In a mixture A and B components show negative deviation as :
(A) Vmix > 0
(B) Vmix < 0
(C) A – B interaction is stronger than A–A and B–B interaction
(D) None of the above reasons is correct
57.
Acetone and carbon disulphide form binary liquid solution showing positive deviation from Raoult law. The
normal boiling point (Tb) of pure acetone is less than that of pure CS2. Pick out the incorrect statements
among the following.
(A) Boiling temperature of mixture is always less than boiling temperature of acetone.
(B) Boiling temperature of Azeotropic mixture is always less than boiling temperature of pure CS2.
(C) When a small amount of CS2 (less volatile component) is added to excess of acetone boiling point of
resulting mixture increases.
(D) A mixture of CS2 and CH3COCH3 can be completely separated by simple fractional distillation.
58.
Which one of the following statements are true ?
(A) Raoult’s law states that the vapour pressure of a component over a solution is proportional to its mole
fraction
(B) The osmotic pressure () of a solution is given by the equation  = MRT, where M is the molarity of the
solution.
(C) The correct of osomotic pressure for 0.01 M aqueous solution of each compound is
BaCl2 > KCl > CH3COOH > sucrose
(D) Two success solution of same molality prepared in different solvents will have the same freezing point
depression.
59.
The vapour pressure of a dilute solution of a solute is influenced by :
(A) Temperature of solution
(B) Mole fraction of solute
(C) Mole fraction of solvent
(D) Degree of dissociation of solute
60.
According to Henry’s law, partial pressure of gas (Pg) is directly proportional to mole fraction of gas in
dissolved state, i.e., Pgas = KH.Xgas where KH is Henry’s constant. Which are correct ?
(A) KH is characteristics constant for a given gas-solvent system
(B) Higher is the value of KH, lower is solubility of gas for a given partial pressure of gas
(C) KH has temperature dependence
(D) KH increase with temperature
61.
Assuming 50% dissociation. Van’s Hoff’s factor of Na2SO4 ?
62.
A 0.2 percent aqueous solution of a non-volatile solute exerts a vapour pressure of 1.004 bar at 100°C.
What is the molar mass of the solute ? (Give your answer in nearest integer)
(Give : Vapour pressure of pure water at 100°C is 1.013 bar and molar mass of water is 18 g mol–1)
63.
Calculate the mass of ascrobic acid (C6H8O6) to be dissolved in 75 g pf acetic acid to lower its melting point
by 1.5°C. [For acetic acid Kf = 3.9 K kg mol–1] (Give answer in nearest integer)
64.
The molar freezing point depression constant of benzene (C6H6) is 4.90 K kg mol–1. Selenium exists as a
polymer of the type Sex. When 3.26 g of selenium in dissolved in 226 g of benzene, the observed freezing
point is 0.112°C lower than for pure benzene. Deduce the molecular formula of selenium.
(At. Mass of Se = 78.8 g mol–1)
65.
Moles of K2SO4 to be dissolved in 12 mol water to lower its vapour pressure by 10 mm Hg at a temperature
at which vapour pressure of pure water is 50 mm Hg is :
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 18
IIT-JEE PROBLEMS (PREVIOUS YEARS)
* Marked Questions are having more than one correct option.
1.
To 500 cm3 of water, 3.0 × 10–3 kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be
the depression in freezing point? kf and density of water are 1.86 K kg–1 and 0.997 g cm–3 respectively:
[JEE 2000]
(A) 0.186 K
(B) 0.228 K
(C) 0.372 K
(D) 0.556 K
2.
The vapour pressure of two miscible liquids A and B are 300 and 500 mm of Hg respectively. In a flask 10
moles of A are mixed with 12 moles of B. However as soon as B is added, A starts polymerizing into a
completely insoluble solid. The polymerization follows first-order kinetics. After 100 minutes, 0.525 mole of
a solute are dissolved which arrests the polymerization completely. The final vapour pressure of the solution
is 400 mm of Hg. Estimate the rate constant of the polymerization reaction. Assume negligible volume
change on mixing and polymerization and ideal behaviour for the final solution.
 100 
(Given : ln  99  = 10–2)


[JEE 2001, 10/100]
3.
During depression of freezing point in a solution, the following are in equilibrium :
(A) Liquid solvent-solid solvent
(B) Liquid solvent-solid solute
(B) Liquid solute-solid solute
(D) Liquid solute-solid solvent
[JEE 2003]
4.
Match the boiling point with Kb for x, y and z, if molecular weight of x, y and z are same.
[JEE 2003]
b.pt.
Kb
x
100
0.68
y
27
0.53
z
253
0.98
5.
A 0.004 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at same temperature. The
apparent degree of dissociation of Na2SO4 is
(A) 25%
(B) 50%
(C) 75%
(D) 85%
[JEE 2004]
6.
1.22 g of benzoic acid is dissolved in (i) 100 g acetone (Kb for acetone = 1.7) and (ii)100 g benzene
(Kb for benzene = 2.6). The elevation in boiling points Tb is 0.17°C and 0.13°C respectively.
(a) What are the molecular weights of benzoic acid in both the solutions?
(b) What do you deduce out of it in terms of structure of benzoic acid?
[JEE 2004]
7.
The elevation in boiling point of a solution of 13.44 g of CuCl2 in 1kg of water using the following information
will be (Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal–1) :
[JEE 2005]
(A) 0.16
(B) 0.05
(C) 0.1
(D) 0.2
8.
72.5 g of phenol is dissolved in 1 kg of a solvent (kf = 14) which leads to dimerization of phenol and freezing
point is lowered by 7 kelvin. What percent of total phenol is present in dimeric form?
[JEE 2006]
9.
When 20 g of naphtholic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a freezing
point depression of 2 K is observed. The van’t Hoff factor (i) is
(A) 0.5
(B) 1
(C) 2
(D) 3
[JEE 2007]
Paragraph for Question No. Q.10 to Q.12
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute
molecules are added to get homogeneous solution. These are called colligative properties. Applications of
colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and
water mixture as anti-freezing liquid in the radiator of automobiles.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 19
A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.
K  = 1.86 K kg mol
Freezing point depression constant of ethanol 
 = 2.0 K kg mol
K
Boiling point elevation constant of water 
 = 0.52 K kg mol
K
Boiling point elevation constant of ethanol 
 = 1.2 K kg mol
K
Given : Freezing point depression constant of water
–1
water
f
ethanol
f
–1
–1
water
b
–1
ethanol
b
Standard freezing point of water = 273 K
Standard freezing point of ethanol = 155.7 K
Standard boiling point of water = 373 K
Standard boiling point of ethanol = 351.5 K
Vapour pressure of pure water = 32.8 mm Hg
Vapour pressure of pure ethanol = 40 mm Hg
Molecular weight of water = 18 g mol–1
Molecular weight of ethanol = 46 g mol–1
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be
non-volatile and non-dissociative.
[JEE 2008, 3/163]
10.
11.
The freezing point of the solution M is
(A) 268.7 K
(B) 268.5 K
(C) 234.2 K
[JEE 2008]
(D) 150.9 K
The vapour pressure of the solution M is
(A) 39.3 mm Hg
(B) 36.0 mm Hg
(C) 29.5 mm Hg
[JEE 2008, 3/163]
(D) 28.8 mm Hg
12.
Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The
boiling point of this solution is
[JEE 2008]
(A) 380.4 K
(B) 376.2 K
(C) 375.5 K
(D) 354.7 K
13.
The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of
N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm
pressure is
[JEE 2009]
–4
–5
–4
(A) 4.0 × 10
(B) 4.0 × 10
(C) 5.0 × 10
(D) 4.0 × 10–5
14.
The freezing point (in ºC) of a solution containing 0.1 g of K3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water
(Kf = 1.86 K kg mol–1) is :
[JEE 2011]
(A) – 2.3 × 10–2
(B) – 5.7 × 10–2
(C) – 5.7 × 10–3
(D) – 1.2 × 10–2
15.
For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation
in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than the
concentration of solvent, the vapour pressure (mm of Hg) of the solution is (taken Kb = 0.76 K kg mol–1)
[JEE 2012]
(A) 724
(B) 740
(C) 736
(D) 718
16.*
Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s)
is(are)
[IIT-JEE- 2013]
(A) G is positive
(B) Ssystem is positive
(C) Ssurroundings = 0
(D) H = 0
17.
Consider spearate solution of 0.500 M C2H5 OH (aq). 0.100 M Mg3 (PO4)2 (aq) 0.250 M KBr (aq) and
0.125 M Na3 PO4 (aq) at 25ºC . Which statement is ture about these solutions, Assuming all salts to be
strong electrolytes ?
[JEE-Mains 2014]
(A) 0.125 M Na3 PO4 (aq) has the highest osmotic presure.
(B) 0.500 M C2H5OH (aq) has the highest osmotic presure.
(C) They all have the same osmotic presure.
(D) 0.100 M Mg3 (PO4)2 (aq) has the highest osmotic presure.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 20
18.
MX2 dissociates into M2+ and X– ions in an aqueous solution, with a degree of dissociation () of 0.5. The
ration of the observed depression of freezing point of the aqueous solution to the value of the depression
of freezing point in the absence of ionic dissociation is :
NCERT QUESTIONS
1.
The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution
contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas ?
2.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of
the solvent. What is the molar mass of the solute ?
3.
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 26.0 g of
heptane and 35g of octane ?
4.
Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g
octane to reduce its vapour pressure to 80%.
5.
A solution containing 30g 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.
6.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of
5% glucose in water if freezing point of pure water is 273.15 K.
7.
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.
8.
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 ?
9.
Henry's law constant for the molality of methane is benzene at 298 K is 4.27 × 105 mm Hg. Calculate the
solubility of methane in benzene at 298 K under 760 mm Hg.
10.
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.
11.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure
benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction
to benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
12.
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.
13.
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.
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 21
14.
Methanol and ethanol form nearly an ideal solution at 300 K. A solution is made by mixing 32 g methanol
and 23 g ethanol. Calculate the partial pressures of its constituents and total vapour pressure of the solution at 300 K.
(P°CH3OH = 90 mm Hg ; P°C2H5OH = 51 mm Hg)
15.
Vapour pressures of pure benzene and toluene at 293 K is 75 mm Hg and 22 mm Hg respectively. 23.4 g
of benzene and 64.4 g of toluene are mixed. If the two form and ideal solution, calculate the mole fraction
of benzene in the vapour phase assuming that the vapours are in equilibrium with the liquid mixture at this
temperature.
16.
The boiling point of water becomes 100.52°C if 1.5 g of a non-volatile solute is dissolved in 100 mL of it.
Calculate the molecular mass of the solute. (Kb for water = 0.6 K m–1)
17.
On dissolving 0.25 g of a non-volatile substance in 30 mL benzene (density = 0.8 mL–1), its freezing point
decreases by 0.40°C. Calculate the molecular mass of the substance (Kf = 5.12 K kg mol–1)
18.
Osmotic pressure of a solution contaning 7 g of a protein per 100 mL of a solution is 25 mm Hg at 37°C.
Calculate molar mass of the protein.
19.
The osmotic pressure of a dilute aqueous solution of compound (X) containing 0.12 g per litre is twice the
osmotic pressure of a dilute aqueous solution of another compound (Y) containing 0.18 g per litre. What is
the ratio of molecular mass of (X) to that of (Y) ? Both (X) and (Y) remain in molecular form in solution.
20.
1 litre aqueous solution of sucrose (molar mass = 342) weighing 1015 g is found to record an osmotic
pressure of 4.82 atm at 293 K. What is the molality of the sucrose solution ?
(R = 0.0821 litre atm K–1 mol–1)
21.
Assuming complete ionisation, calculate the expected freezing point of solution prepared by dissolving 6 g
of Glauber's salt, Na2SO4.10H2O in 0.1 kg of H2O. (Kf for H2O = 1.86 K kg mol–1)
22.
Calculate the van't Hoff factor of CdSO4 (mol. mass = 208.4) if the dissociation of 5.21 g of CdSO4 in half
litre water gives a depression in freezing point of 0.168°C. (Kf for water = 1.86 kg mol–1)
23.
How much urea (molar mass = 60 g mol–1) should be dissolved in 50 g of water so that its vapour pressure
at room temperature is reduced by 25% ? Calculate molality of the solution obtained.
EXERCISE # 1
1.
0.24
2.
0.25
3.
24.5 torr
5.
66.13 mm Hg
6.
0.237
7.
92 mol% toluene; 96.8 mol % toluene
8.
PA = 213.33 torr, PB = 960.0 torr
9.
0.04
10.
111.1g, 18.52 molal
11.
0.162 m
12.
65.25
13.
0.964
14.
106 g/mol
15.
64.0 g/mol
16.
100.079°C
17.
–0.62°C
18.
5.08°C/m
19.
2050 g/mol
20.
Kb= 0.512 kg mol K–1, Tb = 373.20 K
21.
38.71 g
22.
156.06
23.
Tf = –2.28oC
24.
P = 0.2217 atm should be applied
25.
(Vfinal = 5.Voriginal)
26.
54.2 g
27.
13.8 atm
28.
59.99
29.
MA/MB = 0.33
30.
i = 2.5
31.
3 ions
32.
 = 99.2%
33.
7.482 ×105 Nm–2
34.
94.5 %
35.
4.64 atm
36.
0.95; 1.95
37.
100
38.
8
39.
7.7 × 104 atm
40.
7.77 × 10–6 kg
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
4.
57.24 g/mol
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 22
EXERCISE # 2
1.
C44H88O44
2.
746.24 mm/Hg
3.
(a) 3.79°C, (b) 0.018, (c) 4.65 atm, (d) 251.5
4.
(a) 210.65, (b) 62.35%
o
5.
18.34%
6.
Tb = 102.3 C
7.
0.741 m, 0.013 8.
9.
0.73
10.
xb = 0.2472, Yb = 0.4473
12.
C6H6
13.
a = 0.7333
14.
–19.91°C, 7.63 M
15.
M = 53.8
16.
0.93
17.
20.11 %
18.
2.4 × 105 g/mol
19.
5.4 × 10–13
20.
8.9 gm
21.
12.54
22..
12.72
23
0.81 atm
24.
(a) 0.47, (b) 565 torr, (c) xA=0.08, xB= 0.92, (d) 675 torr, (e) x'A= 0.11, x'B= 0.89
25.
(i) Ka = 7.3 × 10–3 (ii) 4.55 × 1022
11.
[Co(NH3)5Cl]Cl2
46.33°C
EXERCISE # 3
1.
(C)
2.
(C)
3.
(C)
4.
(B)
5.
(C)
6.
(D)
7.
(D)
8.
(C)
9.
(A)
10.
(C)
11.
(C)
12.
(B)
13.
(C)
14.
(D)
15.
(B)
16.
(B)
17.
(C)
18.
(A)
19.
(D)
20.
(B)
21.
(D)
22.
(B)
23.
(B)
24.
(A)
25.
(A)
26.
(A)
27.
(B)
28.
(A)
29.
(C)
30.
(D)
31.
(D)
32.
(B)
33.
(A)
34.
(C)
35.
(C)
36.
(C)
37.
(B)
38.
(D)
39.
(D)
40.
(D)
41.
(C)
42.
(B)
43.
(C)
44.
(C)
45.
(A)
46.
(D)
47.
(A)
48.
(A)
49.
(B)
50.
(B)
51.
(C)
52.
(D)
53.
(C)
54.
(C)
56.
(BC)
57.
(ACD)
58.
(ABD) 59.
2
62.
4
63.
5
64.
8
65.
1
1.
(A)
4.
Kb(x) = 0.68, Kb(y) = 0.53, Kb(z) = 0.98
6.
(a) 122,
(b) It means that benzoic acid remains as it is in acetone while it dimerises in benzene as
(A)
15.
(A)
19.
M1 : M2 : : 1 :3
(ABCD) 60.
(ABCD) 61.
EXERCISE # 4
1.0 × 10–4
2.
O
H
O
O
H
O
C
3.
(A)
5.
(C)
C
7.
(A)
8.
35% phenol is present in dimeric form
9.
(A)
10.
(D)
11.
(B)
16.
(BCD) 17.
(C)
18.
2
12.
(B)
13.
(A)
14.
EXERCISE # 5
14.
PCH3OH = 60 mm Hg ; PC2H5OH = 17 mm Hg ; Total vapour pressure = 77 mm Hg
15.
0.59
20.
0.2112 m
23.
Mass of urea = 55.56 g ; molality of solution = 18.52
16.
100.28°C
17.
133.33 18.
21.
22.
1.806
271.95 K
Physical & Inorganic By
NV Sir
B.Tech. IIT Delhi
A-479 Indra vihar, kota
Ph. - 9982433693 (NV Sir) 9462729791(VKP Sir)
54094
Organic Chemistry By
VKP Sir
M.Sc. IT-BHU
Solution & Colligative
Properties # 23