CS – 1
SOLUTIONS
C1A Solution : Solution is a mixture of two or more non-reacting pure substances in which composition may be
altered within certain limits.
There are two types of solution. They are :
(a)
Homogeneous solution
(b)
Heterogeneous solution
HOMOGENEOUS SOLUTIONS : In homogeneous solutions, the components are completely miscible
with each other i.e, they are single phase solution. In this case, the components of the solution will lose their
individuality and, therefore, they cannot be separated by filtration, setting or centrifugal action.
HETEROGENEOUS SOLUTION : In heterogeneous solutions, the components are not miscible i.e.,
they are multiple phase solutions. The components of these mixture can be easily separated through filtration, settling or centrifugal action.
C1B
Types of solution : Depending upon the physical state of solvent and solute, there are nine type of
solutions. They
Number
Solute
(S2)
Solvent
(S1)
Examples
1
Gas
Gas
(H2 + N2 + O2 : Air)
2
Liquid
Gas
Water vapours in air, Mist
3
Solid
Gas
Smoke
4
Gas
Liquid
HCl in water, CO2 in water
5
Liquid
Liquid
C2H5OH in water, n-hexane in benzene
6
Solid
Liquid
Sugar in water, urea in water
7
Gas
Solid
Pd/H2 ; Pt/O2 ; Pt/Cl2 ; Ni/H2
8
Liquid
Solid
Hg in Zn, Hg in Au
9
Solid
Solid
Alloys
Out of these nine solutions, there are the most important. They are :
(i)
Gas in Liquids
(ii)
Liquid in Liquid, and
(iii)
Solid in liquid
C1C Methods of Expressing Concentration of a Solution
Name
Symbol
Definition
Formula
Volume Percent
C
amount of solute present in 100 mL of sol.
C
100w 1
V
Weight %
%
amount of solute present in 100 g of sol.
%
100w 1
w1  w 2
Molarity
M
number of mol of solution present in one
M
n 1000
V
m
n 1000
Wsolvent
L (1 dm3, 1000 cm3) of solution
Molality
m
number of mol of solute present
in one kg solvent
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CS – 2
*
Relation between Molarity and Molality : m 
M  1000
, [d is density of solution in g/L and M is
d  MM 
molar mass of solute].
C2A Vapour Pressures : The pressure exerted by the vapours of a liquid which are in equilibrium with it at a
given temperature. log
C2B
H vap  T2  T1 
P2



P1 2.033R  T1T2 
Vapour pressure of solution containing volatile or non-volatile solute : The vapour pressure of a
solution depends upon the nature of the solute i.e., whether it is volatile or not.
Case I : When the solute is non-volatile (Lowering in vapour pressure)
Raoult’s Law : It is stated as, “if the solute is non-volatile, then the vapour pressure of the solution is
directly proportional to the mole fraction of the solvent”, at a given temperature.
Assume nA and nB are the number of solvent and solute in a given solution.
For non-volatile solutes :
PA = P0AXA
PA is the partial pressure of solvent and P0A is the vapour pressure of pure solvent having mole fraction XA
As the solute is non-volatile thus total pressure of solution and partial pressure of solvent are same.
0
A
P = P XA ........
P is the total pressure of solution,
P0  P
P0
nB

,
n A  nB
p0  p
1000

m
p
M solvent
Case II : When the solution is volatile
Raoult’s Law : It is stated as, “At a given temperature, for a solution of volatile liquids, the partial vapour
pressure of each component is directly proportional to their mole fraction”.
For Volatile solutions :
PA = P0AXA
and
PB = P0BXB
(Total vapour pressure of solution) P = PA + PB , P = (P0A – P0B)XA + P0B
Mole fraction of components A and B in vapour phase is determined as
p 0A X A
p 0B X B

YA
YB
*
Ideal solution : Which obey Raoult’s law at all concentrations and follow the condition. Hmix = 0,
Vmix = 0 (for liquid - liquid solution too), Smix > 0.
*
Non-ideal solution : Which show positive or negative derivations from Raoult’s law.
H mix = +ve; V mix = +ve; b.Pt is smaller than expected
Examples of solutions with positive deviations :
*
(i)
C2H5OH + C6H12
(ii)
(iv)
Acetone (CH3COCH3) + C2H5OH
CH3COCH3 + CS2
(v)
(iii)
CCl4 + CHCl3
C2H5OH + H2O
Non-ideal solution with negative deviation from Raoult’s Law
H mix = –ve; V mix = –ve; b.Pt is higher than expected.
Examplex of solutions with negative deviations :
(i)
Acetone and chloroform
(ii)
Chloroform and ether
(iii)
Chloroform and nitric acid
(iv)
Acetone and aniline
(v)
Water and nitric acid
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CS – 3
*
Azeotrope : A mixture of liquids boils at a constant temperature like pure liquid and has same composition
of component in liquid as well as vapours.
Practice Problems :
1.
Two liquids A and B have vapour pressure in the ratio P0A : P0B = 2 : 3 at a certain temperature.
Assume A and B form an ideal solution and the ratio of mole fraction of A to B in the vapour phase
is 1 : 3, then the mole fraction of A in the solution at the same temperature is
(a)
2.
(c)
1
4
(d)
3
4
180
(b)
90
(c)
270
(d)
200
The vapour pressure of a pure liquid A is 40 mm Hg at 310 K. The vapour pressure of this liquid in
a solution with liquid B is 32 mm Hg. Mol fraction of A in the solution, if it obeys Raoult’s law, is :
(a)
4.
2
3
(b)
The vapour pressure of solution of 5 g of a non-electrolyte in 100 g of water at a particular
temperature is 2985 Nm–2. The vapour pressure of pure water at that temperature is 3000 Nm–2. The
molecular mass (g mol–1) of the solute is
(a)
3.
1
3
0.8
(b)
0.5
(c)
0.2
(d)
0.4
The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to
the solvent. The more fraction of solute in solution is 0.2, what would be the mole fraction of solvent
if decreases in vapour pressure is 20 mm of Hg
(a)
0.8
(b)
0.6
(c)
0.4
(d)
0.2
[Answers : (1) a (2) a (3) a (4) b]
C3A Colligative properties : Properties, whose values depends only on the concentration of solute particles in
solution and not on the identity of the solute are called Colligative Properties.
C3B
Van’t Hoff Factor ‘i’ : There may be change in number of mol of solute due to ionisation or association
hence these properties are also affected. Number of mol of the product is related to degree of ionisation or
association by van’t Hoff Factor ‘i’ is given by
*
In case of dissociation : i = [1 + (n – 1 )]
where n is the number of products (ions or molecules) obtained per mol of the reactant.
*
In case of association : i = [1 + (
1
– 1)]
n
where ‘n’ is the no. of moles of solute undergoes association and  is degree of association
if
i
Theoretical molar mass
Experimental molar mass
i
Observed value of colligative property due to diss. or asso.
Theoretical value of colligative property without diss. or asso.
i
no. of moles of solute at time ' t'
initial no. of moles of solute added
i = 1 (no change),
Einstein Classes,
i > 1 (dissociation),
i < 1 (association)
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CS – 4
Practice Problems :
1.
Which one of the following salt would have the same value of the Van’t Hoff factor (i) as that of
K3[Fe(CN)6]
(a)
2.
NaCl
(b)
Na2SO4
(c)
Al2(SO4)3
(d)
Al(NO3)3
The van’t Hoff factor of a 0.005 M aqueous solution of KCl is 1.96. The degree of dissociation of KCl
at the given concentration is
(a)
92%
(b)
94%
(c)
96%
(d)
98%
[Answers : (1) d (2) c]
C3C There are four Colligative properties :
(i)
Relative lowering in vapour pressure
(ii)
Elevation in boiling point
(iii)
Depression in freezing point
(iv)
Osmotic pressure
*
Colligative properties are defined for only non-volatile solute.
(i)
Relative lowering in vapour pressure
p
p0
 Xsolute .........
[Xsolute is mole fraction of solute
p = (p0 – p) is lowering in vapour pressure]
p
p
(ii)
0

in B
.....
in B  n A
[nB and nA are no. of moles of solutes and solvent respectively]
Elevation in boiling point
Tb = Tb – Tb0 = iKb.m.....
[Kb is called as molal elevation constant, m is the molality of
solution]
In elevation of b.Pt : b.Pt of solution (Tb) > (Tb0) b.Pt of solvent
Kb 
R(Tb0 )2 Msolvent
1000Hvap. .....
[Msolvent is molar mass of solvent and Hv is the enthalpy of
vaporisation per mole of solvent]
(iii)
Depression in freezing point :
Tf = Tf0 – Tf = iKf m....
[Kf is molal depression constant]
F.Pt of solution (Tf) < (Tf0) F.Pt of solvent
Kf 
(iv)
R(Tf 0 ) 2 M solvent
.....
1000 H fus.
[Hf is the enthalpy of fusion per mole of solvent]
Osmotic pressure ()
Spontaneous flow of solvent from dilute solution (it may be pure solvent) into the concentrated solution
through a semi permeable membrane is called osmosis.
Pressure required to prevent osmosis is called as osmotic pressure.
Copper (II) ferrocyanide, Cu2[Fe(CN)6], warm mixture of gelatine and glycerine are common semi-permeable membranes.
 = iCRT .....
*
[C is the molarity or the solution at temperature ‘T’]
For isotonic solution 1 = 2
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CS – 5
*
For a solution 1 > 2 (hypotonic)
*
1 < 2 (hypertonic)
Reverse osmosis : Flow of solvent of high concentrate solution into low by the application of pressure,
greater than osmotic pressure. (For desalination of sea water).
Practice Problems :
1.
The vapour pressure of water at room temperature is lowered by 5% by dissolving a solute in it, then
the approximate molality of solution is
(a)
2.
6.
(c)
4
(d)
3
4 g/l
(b)
8 g/l
(c)
12 g/l
(d)
14 g/l
375.5 K
(b)
374.04 K
(c)
377.12 K
(d)
373.25 K
0
Molal depression of freezing point of water is 1.86 per 1000 g of water. 0.02 mole of urea dissolved in
100 g of water will produce a lowering of temperature of
(a)
5.
1
1.0 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of the solution is
(Kb for H2O = 0.52 K kg/mol)
(a)
4.
(b)
A solution of glucose (C6H12O6) is isotonic with 4 g of urea (NH2 – CO – NH2) per litre of solution. The
concentration of glucose is
(a)
3.
2
0.1860
(b)
0.3720
(c)
1.860
(d)
3.720
Arrange the following aqueous solutions in the order of their increasing boiling points
(i)
10–4 M NaCl
(iii)
10–3 M MgCl2
(a)
(c)
(i) < (ii) < (iv) < (iii)
(b)
(ii) < (i) = (iii) < (iv)
(i) < (ii) < (iii) < (iv)
(d)
(iv) < (iii) < (i) = (ii)
(ii)
10–3 M Urea
(iv)
10–2 M NaCl
1 mol each of following solutes are taken in 5 mol water,
(A)
NaCl
(B)
K2SO4
(C)
Na3PO4
(D)
glucose
Assuming 100% ionisation of the electrolyte relative decrease in vapour pressure will be in the
order :
(a)
7.
8.
10.
(b)
D < C< B <A
(c)
D <A< B < C
(d)
equal
PtCl4.6H2O can exist as a hydrated complex; 1 molal aq. solution has depression in freezing point of
3.720. Assume 100% ionisation and Kf(H2O) = 1.860 mol–1 kg, then complex is :
(a)
[Pt(H2O)6]Cl4
(b)
[Pt(H2O)4Cl2]Cl2.2H2O
(c)
[Pt(H2O)3Cl3]Cl.3H2O
(d)
[Pt(H2O)2Cl4].4H2O
The osmotic pressure of a 5% (wt./vol.) solution of cane sugar at 1500C is
(a)
9.
A<B<C<D
4 atm
(b)
3.4 atm
(c)
5.078 atm
(d)
2.45 atm
Following solutions at the same temperature will be isotonic
(a)
3.42 g of cane sugar in one litre water and 0.18 g of glucose in one litre water
(b)
3.42 g of cane sugar in one litre water and 0.18 g of glucose in 0.1 one litre water
(c)
3.42 g of cane sugar in one litre water and 0.585 g of NaCl in one litre water
(d)
3.42 g of cane sugar in one litre water and 1.17 g of NaCl in one litre water
Relative decrease in V.P. of an aqueous glucose dilute solution is found to be 0.018. Hence
elevation in boiling point is : (it is given 1 molal aq. urea solution boils at 100.540C at 1 atm.
pressure).
(a)
0.0180
Einstein Classes,
(b)
0.180
(c)
0.540
(d)
0.030
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CS – 6
11.
Two solutions (A) containing FeCl 3(aq) and (B) containing K 4Fe(CN) 6 are separated by
semipermeable membrane as shown below. If FeCl3 on reaction with K4Fe(CN)6 produces blue
colour of Fe4Fe(CN)6, the blue colour will be noticed in
12.
(a)
(A)
(b)
(B)
(c)
In both (A) and (B)
(d)
Neither in (A) nor in (B)
10 g. of solute A and 20 g. of solute B are both dissolved in 500 ml. of water. The solution has the same
osmotic pressure as 6.67 g of A and 30 g of B are dissolved in the same volume of water at the same
temperature. Thus the ratio of molar masses of A and B is
(a)
13.
3.33
(b)
3
(c)
10
(d)
0.33
0
At 25 C, the osmotic pressure of human blood due to the pressure of various solutes in the blood is
7.65 atm. Assuming that molarity and molality are almost same. The freezing point of blood is
[Kf = 1.86 K kg/mol].
(a)
0.5820C
(b)
–0.5820C
1.5820C
(c)
(d)
–1.5820C
[Answers : (1) d (2) c (3) b (4) b (5) c (6) c (7) c (8) c (9) b (10) c (11) d (12) b]
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
The mathematical expression of Raoult’s law is
(a)
P0  P n

P0
N
(b)
P0  P N

P0
n
(c)
P0  P n

P
N
(d)
P0  P
 nN
P0
The freezing point of equimolal aqueous solutions
will be highest for
(a)
C6H5NH3Cl (aniline hydrochloride)
(b)
Ca(NO3)2
(c)
La(NO3)3
(d)
C6H12O6 (glucose)
4.
– 0.450C
(b)
0.900C
(c)
–0.310C
(d)
–0.530C
Einstein Classes,
(a)
x2 
mM 1
mM 1
(b) x 2 
1000  mM 1
1000  mM 1
(c)
x2 
1000  mM 1
1000  mM 1
(d) x 2 
mM 1
mM 1
where M1 is the molar mass of solvent.
5.
A 0.2 molar aqueous solution of a weak acid (HX)
is 20 percent ionised. The freezing point of the
solution is (Kf for H2O = 1.86 K mol–1 kg).
(a)
The expression relating molality (m) and mole fraction of solute (x2) in a solution is
6.
For an ideal binary liquid solution with p0A > p0B,
which of the following relations between xA (mole
fraction of A in liquid phase) and yA (mole fraction
A in vapour phase) is correctly represented ?
(a)
xA = yA
(b)
xA > yA
(c)
xA < yA
(d)
xA and yA cannot be correlated with each
other
2.56 gm of sulphur in 100 gm of CS2 has depression
in F.P. of 0.010 0 , K f = 0.10(molal) –1 . Hence
atomicity of sulphur in CS2 is :
(a)
2
(b)
4
(c)
6
(d)
8
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 7
7.
8.
9.
10.
11.
Relative decrease in vapour pressure of an
aqueous solution containing 2 mol [Cu(NH3)3Cl]
Cl in 3 mol H2O in 0.50. On reaction with AgNO3,
this solution will form :
(a)
1 mol AgCl
(b)
0.25 mol AgCl
(c)
2 mol AgCl
(d)
0.40 mol AgCl
When mercuric iodide is added to the aqueous
solution of potassium iodide, the :
0.01 M NaCl
(b)
0.005 M C2H5OH
(c)
0.005 M MgI2
(d)
0.005 M MgSO4
freezing point is raised
(b)
freezing point is lowered
(c)
freezing point does not change
(a)
–1.860C
(b)
1.860C
(d)
boiling point does not change
(c)
–3.920C
(d)
2.420C
(a)
Sugar beet will lose water from its cells
(b)
Sugar beet will absorb water from
solution
(c)
Sugar beet will neither absorb nor lose
water
(d)
Sugar beet will dissolve in solution
Which is not a colligative property
16.
17.
18.
The depression in f.pt. of 0.01 m aqueous solution
of urea, sodium chloride and sodium sulphate is in
the ratio
(a)
1:1:1
(b)
1:2:3
(c)
1:2:4
(d)
2:2:3
Lowering in vapour pressure is the highest for
(a)
0.2 m Urea
(b)
0.1 m Glucose
(c)
0.1 m MgSO4
(d)
0.1 m BaCl2
(a)
Lowering of vapour pressure
(b)
Freezing point
A solution of one mole of benzoic acid in 15 moles
of benzene produces a relative lowering in vapour
pressure of 1/31. The mol. wt. of benzoic acid is
(c)
Osmotic pressure
(a)
122
(b)
244
(d)
Elevation of boiling point
(c)
61
(d)
189
Osmosis is a spontaneous flow through a
semipermeable membrane of
(b)
(c)
(d)
13.
(a)
(a)
If a thin slice of sugar beet is placed in concentrated
solution of NaCl then
15.
Which aqueous solution has minimum freezing
point
The molal freezing point constant of water is 1.86
K molality–1. If 342 g of cane sugar (C12H22O11) are
dissolved in 1000 g of water, the solution will freeze
at
(a)
12.
14.
19.
A less concentrated solution into more
concentrated solution
The plot of
1
1
(where xA and yA aree
versus
yA
xA
the mole fraction of A in liquid and vapour phases,
respectively) is linear with slope and intercept
respectively are given as
The solvent from a solution of lower con
centration to one of higher
concentration

, p
, p
, p

/ p
/ p
/ p
(a)
p 0A / p 0B , p 0A  p 0B / p 0B
Solute particles from a solution of higher
concentration to one of lower
concentration
(b)
p 0A / p 0B
None
(c)
p 0B / p 0A
(d)
p 0B / p 0A
When the vapour pressures of solutions of two
liquids are less than those expected from ideal
solutions, they are said to show
(a)
Positive deviations from ideal behaviour
(b)
Negative deviations from ideal behaviour
(c)
Positive deviations for lower
concentrations and negative deviations
for higher concentration
(d)
None
The ratio of the value of any colligative property
for KCl solution to that of sugar solution is
(a)
1
(b)
0.5
(c)
2
(d)
5
Einstein Classes,
20.
0
B
 p 0A
0
A
 p 0B
0
B
 p 0A
0
B
0
A
0
B
The vapour pressure of a solution of a non-volatile
solube 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 the solute, what is the ratio of
weight of solvent to solute.
(a)
0.15
(b)
5.7
(c)
0.2
(d)
none of these
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CS – 8
21.
22.
23.
24.
25.
26.
The freezing point of aqueous solution that contains
5% by mass urea, 1.0% by mass KCl and 10% by
mass of glucose is
(a)
290.2 K
(b)
285.5 K
(c)
269.3 K
(d)
250 K
25 mL of an aqueous solution of KCl was found to
require 20 mL of 1 M AgNO3 solution when titrated
using a K2CrO4 as indicator. Depression in freezing point of KCl solution with 100% ionisation will
be : [Kf = 2.00 mol–1 kg and molarity = molality]
(a)
5.00
(c)
0
1.6
29.
1.1 g of CoCl3 . 6NH3 (mol. wt. = 267) was dissolved
in 100 g of H2O. The freezing point of the solution
was –0.290C. Approximately how many moles of
solute particles exist in solution for each mole of
solute introduced is [Kf for H2O = 1.860C. m–1].
(a)
2
(b)
3
(c)
6
(d)
4
A 5 per cent aqueous solution by mass of a
non-volatile solute boils at 100.150C. The molar
mass of the solute is [Kb = 0.52 K kg mol–1].
(b)
3.20
(a)
182.4
(b)
100
(d)
0
(c)
50
(d)
52
0.8
Osmotic pressure of 40% (wt./vol.) urea solution is
1.64 atm and that of 3.42% (wt./vol.) cane sugar is
2.46 atm. When equal volumes of the above two
solutions are mixed, the osmotic pressure of the
resulting solution is
(a)
1.64 atm
(b)
2.46 atm
(c)
4.10 atm
(d)
2.05 atm
Dry air was passed successively through a solution
of 5 g of a solution in 180 g of water and then
through pure water. The loss in weight of solution
was 2.50 g and that of pure solvent 0.04 g. The
molecular weight of the solute is
(a)
31.25
(b)
3.125
(c)
312.5
(d)
None
30.
The vapour pressure of a solution containing 2 mol
of liquid A and 3 mol of liquid B at 600C is 500 mm.
When 2 mol of A is further added to the above
solution, the vapour pressure decreases to 400 mm
Hg at the same temperature.The vapour pressures
of the pure liquids A and B are respectively in mm
Hg
(a)
149.12, 733.92
(b)
109.12, 733.92
(c)
149.12, 703.90
(d)
data is insufficient
The molecular weight of NaCl determined by
studying freezing point depression of its 0.5%
aqueous solution is 30. The apparent degree of
dissociation of NaCl is
(a)
0.95
(b)
0.50
ANSWERS (SINGLE CORRECT
(c)
0.60
(d)
0.30
CHOICE TYPE
One litre of milk weighs 1.035 kg. The butter fat it
contains to the extent of 4% by volume has a
density of 875 kg/m3. The density of the fat-free
‘skimmed’ milk is
(1 m3 = 103 litre).
27.
28.
(a)
1042 kg/m3
(b)
1.042 kg/m3
(c)
0.010 kg/m3
(d)
0.0145 kg/m3
A 0.001 m aqueous solution of K3[Fe(CN)6] freezes
at –0.062 0 C. The apparent percentage of
dissociation is
(Kf for water = 1.86)
(a)
60%
(b)
50%
(c)
78%
(d)
90%
Einstein Classes,
1.
c
11.
b
21.
c
2.
d
12.
b
22.
b
3.
a
13.
c
23.
d
4.
a
14.
a
24.
a
5.
c
15.
a
25.
a
6.
d
16.
b
26.
a
7.
a
17.
d
27.
c
8.
a
18.
b
28.
d
9.
a
19.
c
29.
a
10.
b
20.
b
30.
a
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 9
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
Find out the osmotic pressure exerted by a solution
of 0.1 M of electrolyte AB2 in Pascals at 300K.
2.
0.5 g of a non-volatile organic substance was
dissolved in 100 ml of CCl4 at 300C. The vapour
pressure of the solution was found to be 141.9 torr.
Calculate the molar mass of the substance if the
vapour pressure of CCl4 at 300C is 143 torr and its
density is 1.58 g ml–1.
3.
Acetic acid (CH3COOH) associates in benzene to
form double molecules. 1.65 g of acetic acid when
dissolved in 100 g of benzene raised the boiling point
by 0.360C. Calculate the van’t Hoff factor and the
degree of association of acetic acid in benzene.
(Kb = 2.57 K kg mol–1).
5.
The vapour pressure of a solution containing 0.05
mole of sodium sulphate in 450 g of water is 756.2
mm of Hg at 1000C. What is the apparent degree of
ionization of Na 2SO 4m the solution ? (Vapour
pressure of pure water at 1000C is 760 mm of Hg).
12.
1.25 g benzoic acid (molar mass = 121) when
dissolved in 100 cm3 of benzene produces osmotic
pressure of 1.73 atm at 300 K. Benzoic acid is known
to form dimer in benzene. Calculate (i) percentage
of benzoic acid in the associated state (ii) the
equilibrium constant of the dimerization reaction.
13.
Calculate the composition of the vapour in
equilibrium with an ideal solution of ethylbenzene
(P0e = 10.0 torr) and methylbenzene (P0m = 37.0 torr)
in which the mole fraction of ethylbenzene in the
liquid is 0.35. Calculate the total vapor pressure of
the solution.
14.
Phenol associates in benzene to a certain extent to
form a dimer. A solution containing 20 × 10–3 kg of
phenol in 1.0 kg of benzene has its freezing point
depressed by 0.69 K. Calculate the fraction of
phenol dimerized. Kf for benzene = 5.120 mole kg–1.
15.
The vapour pressure of pure benzene at 250C is
639.7 mm of Hg and the vapour pressure of a
solution of a solute in C 6 H 6 at the same
temperature is 631.9 mm of Hg. Calculate molality
of solution.
16.
A 0.025M solution of monobasic acid had a
freezing point of –0.060C. Calculate Ka for the acid.
Kf (H2O) = 1.86.
An aqueous solution of a non-volatile solute boils
at 100.170C. At what temperature would it freeze ?
(Kb = 0.52 K kg mol–1 and Kf = 1.88 K kg mol–1).
4.
11.
At 300 K, the vapour pressure of an ideal solution
containing one mole of A and 3 moles of B is 550
mm of Hg. At the same temperature, if one mole of
B is added to this solution, the vapour pressure of
the solution increases by 10 mm of Hg. Calculate
the V.P. of A and B in their pure state.
6.
The vapour pressure of 2% aqueous solution of an
electrolyte XY at 373 K is 755 mm of Hg. Calculate
the molecular mass of the solute. Vapour pressure
of pure water at 373 K is 760 mm of Hg.
7.
The vapour pressures at a certain temperature of
two liquids A and B are 39 mm Hg and 27.5 mm Hg
respectively. The molecular weights are MA = 72 and
MB = 87. The liquids are completely miscible. What
is the vapour pressure at the same temperature for
a mixture of A and B in the weight ratio 1 : 2 ?
8.
A solution containing 30 gm of a non-volatile
solute in exactly 90 gm water has a vapour of
21.85 mm of Hg at 250C. Further 18 gm of water is
added to solution, the new vapour pressure becomes
22.15 mm of Hg at 250C. Calculate (i) Mol. wt of
solute., (ii) Vapour pressure of water at 250C.
9.
Chloroacetic acid has Ka = 1.36 × 10–3M. Calculate
the freezing point of 0.1 M solution of this acid.
Assume that molarity and molality have identical
values.
Kf(water) = 1.86 K kg mol–1.
10.
Two solution containing equal quantities of water,
one having 0.5 mole of sugar and other have 22.2
gm of calcium chloride. If both the solutions freezes
at the same temperature then calculate the
apparent degree of ionization of calcium chloride
in the solution.
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 10
FINAL STEP EXERCISE
(SUBJECTIVE)
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 mol–1 and 0.997 g
cm–3 respectively.
2.
Solution of two liquids X and Y obey Raoult’s law.
At a certain temperature, it is found that when the
total pressure above a given solution is 400 mm Hg,
the mole-fraction of X in the vapour is 0.45 and in
the liquid is 0.65. What are the vapour pressures of
the two pure liquids at the given temperature ?
3.
A certain liquid mixture of the liquids A and B
behaving ideally shows a vapour pressure of 70 mm
Hg at 250C for a certain mole fraction X of A. For
the same mole fraction X for B in the mixture the
vapour pressure of the mixture is 90 torr at 250C if
the difference between the vapour pressure of the
pure liquid i.e. PA0 – PB0 = 40 torr, Calculate X, PA0
and PB0.
4.
When 3.24 gm of mercuric nitrate Hg(NO3)2 is
dissolved in 1 kg of water, the freezing point of the
solution is found to be –0.05580C. When 10.84 g of
mercuric chloride HgCl2 is dissolved in 1 kg of
water, the freezing point of the solution is –0.07440C
kf = 1.86
mol –1 K kg. Will either of these
dissociate into ions in an aqueous solution.
5.
1 g of a monobasic acid when dissolved in 100 g of
water lowers the freezing point by 0.1680C. 0.2 g of
the same acid when dissolved and titrated required
15.1 ml of N/10 alkali. Calculate degree of
dissociation of the acid. (Kf for water is 1.86).
6.
Liquids X and Y form an ideal solutions. The vapour
pressures of pure X and Y at 1000C are 300 and 100
mm of Hg respectively. Suppose that the vapour,
above a solution composed of 1 mole of X and
1 mole of Y at 1000C is collected and condensed.
This condensate is then heated to 1000C and vapour
is again condensed to form a liquid A. What is the
mole-fraction of X in A ?
7.
The molar volume of liquid benzene (density = 0.877
g ml–1) increases by a factor of 2750 as it vapourizes
at 200C and that of liquid toluene (density 0.867 g
ml –1) increases by a factor of 7720 at 20 0C. A
solution of benzene and toluene at 20 0C has a
vapour pressure of 46.0 torr. Find the mole
fraction of benzene in the vapour above the
solution.
Einstein Classes,
8.
A very small amount of a non volatile solute (that
does not dissociate) is dissolved in 56.8 cm3 of
benzene (density 0.889 g cm–3). At room temper
ature, vapour pressure of this solution is 98.88 mm
Hg while that of benzene is 100 mm Hg. find the
molality of this solution. If the freezing
temperature of this solution is 0.73 degree lower
than that of benzene, what is the value of molal
freezing point depression constant of benzene ?
9.
A mixture which contains 0.550 g of camphor and
0.045 g of an organic solute freezes at 1570C. The
solute contains 93.46% of C and 6.54% of H by
weight. What is the molecular formula of the
compound ? (Freezing point of camphor = 178.40C
and Kf = 37.70)
10.
1000 g of 1 m aqueous solution of sucrose is cooled
and maintained at –3.5340C. Find how much ice will
separate out. Kf(H2O) = 1.86 K. kg. mol–1.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CS – 11
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1.
7.45 × 105 Pa
2.
M = 62.86 g/mol
3.
– 0.6080C
4.
0.5
5.
P0A = 400 mm, P0B = 600 mm
6.
55.8
7.
31.83 mm Hg
9.
– 0.210C
10.
75%
11.
75%
12.
63.8%, K = 23.56
13.
Xethyl benzene = 0.127, Xmethyl benzene = 0.873, P = 27.55 torr
14.
73.4%
16.
2.96 × 10–3
15.
8.
67.83, 23.78
0.15
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
0.230C
2.
276.9 mm Hg, 628.57 mm Hg
3.
X = 0.25, P0A = 100 torr and P0B = 60 torr
4.
Only Hg(NO3)2 will dissociate
5.
0.19
6.
0.9
7.
0.73
8.
0.1452, 5.027 K molality–1
9.
C12H10
10.
352.98 g
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111