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2
Solutions
Topic 1 Types of Solutions, Expressing
Concentration of Solution and
Solubility
1. Solutions are homogeneous mixtures of two or more than two components.
2. In solution, a solute is a subs.tance dissolved in another substance, known as solvent
which is present in larger quantity. .
3. The solutions containing two components are binary solutions, e.g., salt solution.
4. Molarity It is defined as the number of moles of solute dissolved in one litre or one
cubic decimeter of the solution.
Number of moles of solute x 1000
MoIarlity (M) = ------------
Volume of solution (ml.)
W
(Moles of solute = _2, where W2 = mass of solute (g) and M 2 = molar mass of solute)
M2
5. Molality It is defined as the number of moles of solute per kilogram of the solvent.
Molality (zn) = Number of moles of solute x 1000
Mass of solvent (g)

6. Normality It is the number of gram equivalents of the solute dissolved in one litre of
the solution.
N orm alit
1y
(N) Number of gram equivalents of solute x 1000
= -----=------'=-----------
Volume of solution (ml.)

(where gram equivalents of solute = .' W2 . )

. Equivalent weight

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20 Chapterwise CBSE Solved Papers Chemistry

7. Mole Fraction It is the number of moles of one component to the total number of
moles of all the components present in the solution. Mole fraction for a binary
solution, if solvent = 1and solute = 2 .
Mole fraction of solute, X 2 = n2
nl + n2
n
Similarly, mole fraction of solvent, Xl = I
nl + n2

.. Xl + X2 = 1
8. Molality and mole fraction do not change with change in temperature.
9. Parts Per Million When a solute is present in trace quantities, the concentration is
expressed in parts per million.
illi Number of parts of the component x 106
P ar t s per ID1 10n = --------=----------=----------
Total number of parts of all the components of the solution
10. Mass Per Cent The mass percentage of a component in a given solution is the mass of
the component per 100 g of the solution.
Mass of solute x 100
Mass per cent = --------
Mass of solution
11. Volume Per Cent The volume percentage is the volume of the component per
100 parts by volume of the solution.
Volume of the component x 100
VoIume per cent = -------~-----
Total volume of the solution
12. Mass by Volume Percentage (wi V) is the mass of solute dissolved in 100 mL of the
solution.
13. Solubility of a substance is its maximum amount that can be dissolved in a specified
amount of solvent. It depends on nature of solvent, temperature and pressure.
14. On dissolving the solid solute in a solvent, its concentration increases, this is
dissolution.
15. Dynamic Equilibrium is the condition when number of solute particles going into the
solutions is equal to the solute particles separating out, i.e., dissolution and
crystallisation occur at the same rate.
Solute + Solvent ~ Solution
16. Saturated Solution is the solution in which no more solute can be dissolved at the
same temperature and pressure.
17. Solubility of gases increase with increase of pressure.
18. (i) Henry's Law It states that, at a constant temperature, the solubility of a gas in a liquid is
directly proportional to the pressure of the gas.
(ii) The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of
the gas (x) in the solution.
p oc X or p = K H . X (where K H is Henry's law constant)

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Solutions 21

19. Higher the value of KH at a given pressure, the lower is the solubility of the gas in the
liquid.
Applications of Henry's Law
(i) To increase the solubility of CO2 is soft drinks and soda water, the bottle is sealed under
high pressure.
(ii) To avoid bends and the toxic effects of high concentrations ofN 2 in the blood, the tanks
used by scuba divers are filled with air diluted with He.
20. Relation between Molarity and Molality
' M x 1000
Molal ity, m = ---------
(1000 x d) - (M x M2)
where M is the molarity and M 2 is the molar mass of component 2 (generally solute)
and d is the density of solution in g em -3.
21. For dilution, MI VI = M 2 V2, similarly N I v;. = N 2 V2 (for all cases like acids-bases,
dilution, etc).

Previous Years' Examinations Questions

L' 1 MarR Questions 7. Differentiate between molarity and
molality for a solution. How does a thange in
1. Define mole fraction.
[All India 2012, 2010C, 2009; Delhi 2012]
temperature influence their values?
[Delhi 2011, 2009; Foreign 2011, 2009]
2. Explain the Henry's law about dissolution of a
gas in a liquid. 8. State Henry's law and mention its two
[All India 2012, 2011; Delhi 2011;Foreign 2011] important applications.
[All India 201OC; Delhi 2008C]
3. 'State the main advantage of molarity over
Or
molality as the unit of concentration.
[Delhi 2010, 2009C] State the law correlating the pressure of a gas
and its solubility in a liquid. State an application
4. What is meant by molality of the solution? of this law. [Foreign 2009; All India 2008]
[All India 2009]
Or
State Henry's law correlating the pressure of a
L' 2 MarRS Questions . gas and its solubility in a solvent and mention
two applications for the law.
5. A solution of glucose (C6H1206) in water is [Delhi 2008; Foreign 2008]
labelled as 10% by weight. What would be the
9. Distinguish between the terms molality and
molarity of the solution? (Molar mass of
molarity. Under what conditions are the
glucose = 180 g rnol"), [All India 2013]
molarity and molality of a solution nearly the
6. If the density of water of a lake is 1.25 g mL-I same? [All India 2008C]

and one kg of lake water contains 92 g of Na+ 10. The solubility of pure nitrogen gas at 25°C and
. ions, calculate the molarity of Na+ ions in this 1 atm is 6.8 x 10-4 mol L-I. What is the
lake water. concentration of nitrogen dissolved in water
(At. mass of Na = 23 g rncl"). under atmospheric conditions? The partial
[HOTS; Foreign 2012, 2008] pressure of nitrogen gas in the atmosphere is
0.78 atm. [HOTS; Delhi 2008]

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22 Chapterwise CBSE Solved Papers Chemistry

••' 3 Marhs Questions

I ~•
• ~. ~ j I _ ~

11. What concentration of nitrogen shouId be 12. Henry's law constant for CO2, dissolvinq in
present in a glass of water at room water is 1.67 x 108 Pa at 298 K. Calculate the
, temperature? Assume a temperature of 25°C,
quantity of CO2 in· 1. L of soda water when
a total pressure of 1 atmosphere and mole
packed under 2.5 atm CO2 pressure at 298 K.
fraction of nitrogen in air of 0.78. (KH for [HOTS; Delhi 2,OO8C)
nitrogen = 8.42xl0-7M/mm Hg). [All India 2009) "

"
Step-by-Step Solutions
1. The mole fraction of a component is the ratio of 5. 109 glucose is present is 100 g solution.
the number of moles of that component to the Number of moles of glucose in 10 g glucose
total number of moles of all the components 10 ' ,
present in the solution. (1/2) =-=0.0555 mol (1/2)
180
Mathematically,
' 100 . .~
Mole fraction of a component, 100 g soIutron = - mL =83.33 mL
1.2 ' ..
Number of moles of the component
Total number of moles of all the components = 0.08333 L (1/2)

For a binary solution, mole fraction of · Numberofmolesof glucose

M o Ianty =-----~-~--
component A, Volume of solution (L)
n
XA =---A = 0.0555 =0,67 M (1)
nA +nB 0.08333
Similarly for 8, nB
XB=--- 6 Given, d = 1.25 g mL-1, W2 = 92 g,
e ,

nA +nB
W1 = 1 kg or 1000 g .\~
and XA + XB =1 (112)
. Mass (m)
; We know that, Density (d) = .
2. Henry's law states that, the partial pressure of Volume (V)
the gas in vapour phase (p) is proportional to
or v=m
the mole fraction of the gas (X) in the solution. d
p =KH·X (1000 g + 92 g)
Here, KH = Henry's law constant. Different 1.25 g mL-1
gases have different KH values at the same
[.: Massof solution = massof solvent + mass
temperature. (1)
of solute]
3. Molality does not change with change in =1092 mL (1)
temperature while molarity decreaseswith rise 1.25
in temperature. (1)
M o Ianity = ----,------'''---:---:----
W2 x 1000
4. Molality is defined as the number of moles of M2 x Volume of solution (ml)
the solute per kilogram of the solvent. It is
represented by m. (1/2) M o larit
an y 92x 1000x 1.25
23x 1092 ,) ,H
or' Molality (m)
Number of moles of glucose = 4.579 M (1)
(1/2)
Volumeof solution(L)

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Solutions 23

7. Molality is defined as the number of moles of KH=8.42x10-7 M/mm Hg

the solute per kilogram of the solvent. It is XN2 = ?
"
represent by m.
XN2=KHXPN2 (1)
Molality (m)
Number of moles of solute x 1000 [Since KH is given in M/mm Hg therefore, this
formula is being used].
Mass of solvent (g)
XN2=8.42 X 10-7 M/mm Hg x 592.8 mm Hg
It does not change with change in temperature.
(1) xN2=4991.376x 1O-7M=4.99x10-4 M
Molarity is defined as the number of moles of
~ (1)
solute dissolved in one litre or one cubic 1000
decimetre of the solution.
8
Molarity (M)
Number of moles of solute xl 000 XN2= 1000 x 4. 99 xl 0-4 =0.0277 M
18
Volume of solution (mL)
=2.77xlO-2 M (1)
It decreaseswith increase in temperature
(asV T). (1)
12. -::I
DC
(i) First find mole fraction of the CO2 gas by
8. Henry's law Refer to ans. 2. (1) using the formula P,= KH X.
Applications of Henry's law (Ii) Find the nCo2 (moles of CO2),
(i) To increase the solubility of CO2 in soft (iii) Find the mass of CO2 from moles of CO2,
drinks and soda water, the bottle is sealed
under high pressure. Given ,KH = 1.67x 108 Pa
(ii) To minimise the painful . effects PC02 = 2.5atm = 2.5x 101325 Pa
accompanying the decompression of
Pco, = KH• XC02
deep sea divers, oxygen diluted with less
soluble helium gas is used as breathing X _ PC02 _ 2.5 x 101325 Pa
(1)
gas. (1/2 + 1/2 = 1) CO2- KH - 1.67x 108 Pa

9. Refer to ans. 7. (1) = 1.517x 10-3

The molarity and molality of a solution will be
nearly same if the mass of solvent is nearly
equal to the volume of solution. (1)

1o·l~~~~~~~~~t~~~~-~-;------------] = 1.517x [for dilute solution]

10-3
1 L soda water = 1L water = 1000 g H20
From Henry's law, 5 DC p.
1000
52 = P2 Moles ofH20=--= 55.55 mol
(1) 18
51 P1
-3 nc~
52--5 1X-P2 Xco2=1.517x10 =--
nH20
P1
_ 6.8x 10-4 x 0.78 nC02 = 55.55 x 1.517 x 10-3
5 2-
1 nC02 = 84.26 x 10-3 mol (1)
= 5.304x 10-4 mol L-1 (1)
Mass of CO2,
11. Given, PN2= 0.78 atm = 0.78 x 760 mm Hg mC02= 84.26x 10-3 x 44= 3.707 g (1)
= 592.8 mm Hg

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Topic 2 Vapour Pressure of Liquid Solutions,

Ideal and Non-ideal Solutions
1. Raoult's law for volatile solute This law states that, for a solution of volatile liquids,
the partial vapour pressure of each component in the solution is directly proportional
to its mole fraction. For component 1, P1 oc xl or P1 = Pl Xl
Similarly for component 2, P2 = P2X2 ; Ptotal = P1 + P2 = r; Xl + P;X2
If Y1 and Y2 are the mole fractions of the components 1 and 2 respectively in vapour
phase, then P1 = Y1 X Ptotal' Similarly, P2 = Y2 X Ptotal'
2. In the solution of a gas in a liquid, one of the components is so volatile that it exists as
a gas, then we can say that Raoult's law becomes a special case of Henry's law in which
K H becomes equal to P~ .
3. Ideal solutions obey Raoult's law over entire range of concentration. For these
solutions, ~mix H = 0 and ~mix V = O.In ideal solutions A-B interactions are nearly
equal to A-A or B-B interactions.
Solution of n-hexane and n-heptane, bromo ethane and chloroethane, benzene and
toluene, etc. are nearly ideal in behaviour.
4. Non-idealsolutions do not obey Raoult's law over entire range of concentration. In case
of positive deviation from Raoult's law (e.g., mixture of ethanol and acetone),A-B
interactions are weaker than those of A-A or B-B interactions, while in case of
negative deviation from the Raoult's law (e.g., mixture of phenol and aniline),A - B
interactions are stronger than those of A-A or B-B interactions.
S. The solutions showing positive deviation and negative deviation from Rault's law are
shown in Fig.(a) and (b) respectively.
Vapour pressure
t of solution
L---..L----_ t

X1 =0 Mole fraction X1 =0 Mole fraction

x =1 X1---+ X1---+
2 +-X2 x2=1
(a) +-X2 (b)

6. For positive deviation mmix = Positive, ~Vmix = Positive

7. For negative deviation m mix = Negative, ~Vmix = Negative
8. The solutions which boil at constant temperature and can distil unchanged in
composition are called azeotropes or azeotropic mixtures.
The solutions which show large negative deviation from Raoult's law, form maximum
boiling azeotropes and the solutions which show large positive deviation from
Raoult's law, form minimum boiling azeotropes.

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Previous Years' Examinations Questions

L'
{~I, ..HooJJ
1 MarR Questions
I.
6. What is meant by positive and negative
deviations from Raoult's law and how is the
,}·,l.;Define ideal solution. sign of Llmix H related to positive and negative
[All India 2013, 2012; Delhi 2010C] deviations from Raoult's law? [All India 2009C]
2. State Raoult's law. [Foreign 2012] 7. State Raoult's law for solutions of volatile
Or liquids. Taking suitable examples, explain the
! .'

Define Raoult's law in its general form in meaning of positive and negative deviations
reference to solutions. from Raoult's law.
[Deihl 2011; All India 2011; Foreign 2011] [Delhi 2008; Foreign 2008; All India 2008]

Or 8. State how the vapour pressure of a solvent is

State Raoult's law for a solution of volatile affected when a non-volatile solute is dissolved
liquids. [Delhi 201OC; Foreign 2009C]
in it? [Foreign 2008]

9. What is meant by negative deviation from

L' 2 MarRs Questions Raoult's law? Draw a diagram to illustrate the

relationship between vapour pressure and
mole fractions of components in a solution to
3. State Raoult's law for a solution containing
volatile components. How does Raoult's law represent negative deviation. [All India 2008C]

become a special case of Henry's law?

. [All India 2013]
,'. 4. Explain why a solution of chloroform and L' 3 Marks Question
acetone shows negative deviation from 10. The partial pressure of ethane over a
Raoult's law? [HOTS; Delhi 2011C] saturated solution containing 656x 10-3g of
;'l
ethane is 1 bar. If the solution were to contains
5. Non-ideal solutions exhibit either positive or
5.0 x 10-2 g of ethane, then what will be the
negative deviations from Raoult's law. What
are these deviations and why are they caused? partial pressure of the gas? [Delhi 2013C]

Explain with one example for each type.

[All India 2011, 2010; Delhi 2010; Foreign 2010]

Step-by-Step Solutions
1. The solutions which obey Raoult's law over the 2. Raoult's law For a solution of volatile liquids,
entire range of concentration are known as the partial vapour pressure of each component
ideal solutions. For ideal solutions, in the solution is directly proportional to its
LlH (mixing)= °
and LlV (mixing)= 0.
e.g., solution of n-hexane and n-hepta'ne; r
mole fraction. Thus, for any component, partial
vapour pressure, p oc x
bromoethane and chloroethane, etc. (1/2) ',' p = p" x
In these solutions (binary solutions) A-8 type" <: where p" is the vapour pressure of pure
,interactions are nearly equal to the A-A and , 'I component and x is the mole fraction of that
8-8 interactions. (1/2) -,' component. (1)

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26 Chapterwise CBSE Solved Papers Chemistry

3. Refer to ans. 2. (1) In case of negative deviation, A-8

interactions are stronger than A-A and 8-8
Henry's law If gasis the solute and liquid is the
interactions. Due to this, vapour pressure
solvent, then according to Henry's law, decreases which results in negative deviation.
p =KH x Example, phenol and aniline and
i.e., partial pressure of the volatile component chloroform + acetone show negative
(gas) is directly proportional to the mole deviation. (1/2)
fraction of that component (gas)in the solution.
(1/2)
Hence, Raoult's law and Henry's law has been
identical except that their proportionality - pO,
constants are different. It is equal to pOfor
Raoult's law and KH for Henry's law.
Therefore, Raoult's law becomes a special case
of Henry's law in which KH·becomes equal to
vapour pressure of pure component p", (1/2)
x, = 0 Mole fraction
x2 = 1 x, ---+
4. ~~INegative deviation means A-8 interactions are +-x2
,. stronger than A-A and 8-8 interactions.
Plot for non-ideal solution showing
positive deviation
A mixture of chloroform and acetone shows (1/2)
negative deviation from Raoult's law because
chloroform molecule forms H-bonding with
acetone molecule. As a result of this A-8
interaction becomes stronger than A-A and
8-8 interactions. This leads to the decrease in
vapour pressure and resulting in negative
deviation. (1)
H3C", /CI
C=O····H-C-CI
H3C / '<CI (1) x, = 0 Mole fraction x, = 1
x2 = 1 x, ---+ x2 = 0
+-x2
5. Positive deviation means A-8 interactions are
Plot for non-ideal solution showing
weaRer than A-A and 8-8 interaction while
negative deviation
opposite is true in case of negative deviation. (1/2)

6. Refer to ans. 5. (1)

For non-ideal solutions, vapour pressure is
For positive deviation, flH (mixing) = Positive
either higher or lower than that predicted by
For negative deviation, flH (mixing) = Negative
Raoult's law. If it is higher, the solution exhibits
(1/2 +1/2 =1)
positive deviation and if it is lower, it exhibits
negative deviation from Raoult's law. 7. Refer to ans. 2 and ans. 5. (2)
In case of positive deviation, A-8 8. On dissolving, non-volatile solute into solvent,
interactions are weaker than A-A and 8-8 vapour pressure decreases. When a
interaction. Due to this, vapour pressure non-volatile solute is added to a solvent, its
increaseswhich results in positive deviation. vapour pressuredecreasesbecausesome of the
Example, ethanol + acetone and CS2 + acetone surface sites are occupied by solute molecules.
Thus, less space is available for the solvent
show positive deviation. (1/2)
molecule to vaporise. (2)

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Solutions 27

9. Refer to ans.5. (Negative deviation including 6.56x10-3g=kx1bar

diagram). . (2)
k = 6.56
10-3g bar "
X (1)

10.~? (i) First, we will find out the value of constant, Again, when, m' = 5.00 x 10-2 g, p' = ?
k using the formula, m = kp
m'=kxp'
(ii) Then, we will find out the value of p' using 5.00 x 10-2g = 6.56 x 10-3g bar"!« p' (1)
formula, m' = k x p' by taking the value of k
from (i). , 5.00 X 10-2
p=-------::-
6.56 x 10-3
According to Henry's law.
= 7.62 bar (1)
mocp m=kp

Topic 3 Colligative Properties, Determination of

Molecular Mass and Abnormal Molar Mass
1. The properties of solutions which depend only on the number of solute particles, not
on the nature of the solute particles are known as colligative properties.
Colligative properties oc number of particles in the solution oc 1
. Molar mass of solute
Colligative properties are
(i) relative lowering of vapour pressure of solvent.
(ii) depression of freezing point of the solvent.
(iii) elevation of boiling point of the solvent and.
(iv) osmotic pressure of the solution.
2. Relative lowering in vapour pressure of an ideal solution containing the non-volatile
solute is equal to the mole fraction of the solute at a given temperature. Relative
lowering of vapour pressure (Raoult's lawfor non-volatilesolute)
o
PI - PI _
o - X2
PI
o
PI - PI _ n2
P; ni + n2
o
PI - PI = n2
(For dilute solutions n2 < < nl)

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28 Chapterwise CBSE Solved Papers Chemistry

3. The boiling point of a liquid is that temperature at which its vapour pressure becomes
equal to the atmospheric pressure.
Elevationof boilingpoint

Ebulliscope

(where Kb = boiling point elevation constant or molal elevation constant or

ebullioscopic constant).
Cryoscope M2 = Kb xW2 X 1000
!!.Tb xWl (g)

4. When a non-volatile solute is added to a solvent, the freezing point of the solution is
always lower than that of pure solvent as the vapour pressure of the solvent decreases
in the presence of non-volatile solute. This difference in freezing point is known as
depression of freezing point.
Depression of freezing point !!.T r oc m or !!.T r = K rm
Kr xW2 x 1000
!!.T r = ---'------
M2 X Wl (g)

(where, K r = freezing point depression constant or molal depression constant or

cryoscopic constant).
Kr xW2 x 1000
M2 = ---'-----
-r,
xWl(g)

RxMlXT;
Kr=---------'--
1000 x !!.fus H

where Rand Ml are gas constant and molar mass of the solvent, respectively. Tr and
Tb are freezing point and boiling point of the pure solvent, respectively (in K). !!.fusH
and !!.vapHare enthalpies for the fusion and vapourisation of the solvent, respectively.
5. Osmosis and osmotic pressure The process of flow of solvent molecules from lower
concentration solution to solution of higher concentration is known as osmosis and
osmotic pressure of the solution is the pressure which just prevents the flow of solvent
molecules.
6. Osmotic pressure, 1t = CRT or 1t = n2 RT (where, C = ~ )
V
1tV= W2RT or M2 = W2RT
M2 1tV

7. Two solutions having same osmotic pressure at a given temperature are called isotonic
solutions.
8. A solution having lower osmotic pressure than the other is called hypotonic while the
one with higher osmotic pressure is called hypertonic.
9. People taking salty food experience water retention in tissue cells and intercellular
spaces due to osmosis. The resulting puffiness is called edema.

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Solutions 29

10. If a pressure larger than the osmotic pressure is applied to the solution side, then the
pure solvent flows out of the solution through the semi-permeable membrane. This
phenomenon is called reverse osmosis.
11. For the substances undergoing association, dissociation, etc. in the solution, molecular
mass determined from colligative properties is different from expected value.
This is known as abnormal molecularmass.
12. van't Hofffactor is the ratio of the experimental value of colligative property to the
calculated value of the colligative property. It is used to find out the extent of
dissociation or association.
. Normal molar mass
van't Hoff factor, t = ---------
Abnormal molar mass
Total number of moles of particles after association / dissociation
Number of moles of particles before association / dissociation
Observed colligative property
Calculated colligative property
If i > 1 solute undergoes dissociation, and if i < 1, solute undergoes association.
13. For association, when molecules of solute forms dimer, n = 2

Degree of association, ex = (~ - .1)

- -1
n
For dissociation of the solute AB type, m = 2, and for the solute AB 2 or A2B type
(CaCI2, Na2SO 4)' m = 3.
.. i-I
Degree 0fdi .ssociation, ex = --
m-l

14. Inclusion of van't Hoff factor modifies the equations for colligative properties as
follows
o

(i) Relative lowering of vapour pressure of solvent, PI - PI =i n2

p; nI

(ii) Elevation of boiling point, t::.Tb= iKbm

(iii) Depression of freezing point, t::.Tf=iKfm

(iv) Osmotic pressure of solution, 1t = £n2 RT

V

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Previous Years' Examinations Questions

L' 1 MarR Questions

1. Define the term azeotrope. [All India2013) 11. Define the term osmosis and osmotic
pressure. Is the osmotic pressure of a solution
2. Define the term osmotic pressure.
a colligative property? Explain. [DeIhl2011)
[All India2013; Delhi201OC,2OO9C)

3. Define the following terms. 12. What is van't Hoff factor? What possible values
can it have if the solute molecules undergo
(I) Isotonic solutions
dissociation? [Delhi 2011C)
(i0 van't Hoff factor [Deihl 2012; All India2012)
13. The molecular masses of polymers are
4. Explain boiling point elevation constant for a
determined by osmotic pressure method and
solvent or Ebullioscopic constant.
not by measuring other colligative properties.
[All India2012; Foreign2012)
Give two reasons. [All India2011C)
5. What is meant by reverse osmosis?
[All India2013, 2011; Foreign2009) 14. Define the term osmosis and osmotic
pressure. What is the advantage of using
6. Define the term van't Hoff factor. [All India2009) osmotic pressure as compared to other
7. What is meant by colligative properties? colligative properties for the determination of
[All India2009) molar masses of solutes in solutions?
[All India 2010)

15. Find the boiling point of a solution containing

L' 2 MarRs Questions 0.520 g of glucose (C6H1206) dissolved in
8. An aqueous solution of sodium chloride 80.2 g of water. [Given, Kb for water
1
freezes below 273 K. Explain the lowering in = 0.52 Km- ]. [All India2010C)
freezing point of water with the help of a
suitable diagram. [DelhI2013C) 16. Find the freezing point of a solution containing
0.520 g glucose (C6H1206) dissolved in 80.2 g
9. 18 g glucose, C6H1206 (molar mass = 180 g
of water [Given, Kf for water = 1.86 Km-1].
rnol ") is dissolved in 1kg of water in a sauce [AllIndia20lOC)
pan. At what temperature will this solution
boil? 17. Define the term osmotic pressure. Describe
how the molecular mass of a substance can be
(Kb for water =0.52 K kg mol-l, boiling point of
determined by a method based on
pure water =373.15 K). [Delhi 2013) measu rement of osmotic pressu re?
. 10. A 1.00 molal aqueous solution of [Deihl 2008;All India 2008; Foreign2008)
trichloroacetic and (CCI3COOH) is heated to its
18. The depression in freezing point of water
boiling point. The solution has the boiling observed for the same molar. Concentrations
point 100.18°e. Determine the van't Hoff
of acetic acid, trichloroacetic acid and trifluoro
factor for trichloroacetic acid. (Kb for water acetic acid increases in the order as stated
. = 0.512 K kg rnol'"), [Delhi 2012) above. Explain. [Delhl2oo8C)

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Solutions 31

Or
I. 3 Marks Questions Calculate the boiling point of a solution

19. Determine the osmotic pressure of a solution

prepared by adding 15.00 9
of NaCI to 250.0 g
of water.
prepared by dissolving 2.5xl0-2 g of K2S04 (Kb for water = 0.512 K kg mol-l, molar mass
. 1
in 2 L of water at 25°C, assuming that it is of NaCI = 58.44 g mol- ). [Delhi 2011]
completely dissociated. (R = 0.0821 L atm K-l
mol-l, molar mass of K2S04 =174 g mol "). 25. What would be the molar mass of a compound
if 6.21 g of it dissolved in 24.0 g of chloroform
[Delhi 2013]
form a solution that has a boiling point of
20. 1.00 g of a non-electrolyte solute when 68.04°(, The boiling point of pure chloroform
dissolved in 50 g of benzene lowered the is 61.7°C and the boiling point elevation
freezing point of benzene by 0.40 K. Find the constant, Kb for chloroform is 3.63°C/m.
molar mass of the solute. [Delhi 2011]

(Kf for benzene = 5.12 Kkg rnol ") [All India 2013]
Or
A solution prepared by dissolving 8.95 mg of a
21. Calculate the amount of KCI which must be gene fragment in 35.0 mL of water has an
added to 1 kg of water so that the freezing osmotic pressure of 0.335 torr at 25°('
point is depressed by 2 K.
Assuming the gene fragment is
(Kf for water= 1.86 K kg me!"). [Delhi 2012] non-electrolyte, determine its molar mass.
[Delhi 2011; All India 2011]
22. At 25° C, the saturated vapour pressure of
water is 3.165 k Pa (23.75 mm Hg). Hnd the 26. What mass of NaCI must be dissolved in 65.0 g
saturated vapour pressure of a 5% aqueous of water to lower the freezing point of water by
solution of urea (carbamide) at the same 7.50°(7 The freezing point depression
temperature. (Molar mass of urea constant (Kf) for water is 1.86°C/m. Assume
= 60.05 g marl). [Foreign 2012] van't Hoff factor for NaCI is 1.87.
(Molar mass of NaCI = 58.5 g rnol'"),
23. 15.0 g of unknown molecular material is
[All India 2011, 2010; Foreign 2010]
dissolved in 450 g of water. The resulting
solution freezes at - 0.34°(' What is the molar 27. A 0.561 m solution of unknown electrolyte
mass of the material? depresses the freezing point of water by
(Kf for water = 1.86 K kg marl). 2.93°(, What is van't Hoff factor for this
electrolyte? The freezing point depression
[All India 2012, 2010]
constant (Kf) for water is 1.86°C kg rno!".
Or
[Foreign 2011]
A solution of glycerol (C3Hs03) in water was
28. Phenol associates in benzene to a certain
prepared by dissolving some g!ycerol in 500 g
extent to form a dimer. A solution containing
of water. This solution has a boiling point of
20 g of phenol in 1.0 kg of benzene has its
100.42°C, what mass of glycerol was dissolved
freezing point lowered by 0.69 K. Calculate the
to make this solution?
fraction of phenol that has dimerised (Given Kf
(Kb for water = 0.512 K kg mor'").
for benzene = 5.1 Km-l). [HOTS]
[All India 2012; Delhi 2012, 2010]
Or
24. Calculate the freezing point of an aqueous
An aqueous solution containing 12.48 g of
solution containing 10.50 g of MgBr2 in 200 g
barium chloride in 1.0 kg of water boils at
of water (molar mass of MgBr2 = 184 g, Kf for 373.0832 K. Calculate the degree of
water 1.86 K kg mor '). [Delhi 2011] dissociation of barium chloride.
(Given, Kb for Hp = 0.52 Km-l, molar mass of
BaCI2 = 208.34 g mo!"). [Delhi 2011C]

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32 Chapterwise CBSE Solved Papers Chemistry

29. At 300 K, 36 g of glucose, C6H1206 present per 36. 100 mg of a protein is dissolved in just
litre in its solution has an osmotic pressure of enough water to make 10.0 mL of solution. If
4.98 bar. If the osmotic pressure of another this solution has an osmotic pressure of
glucose solution is 1.52 bar at the same 13.3 mm Hg at 25°C, what is the molar mass
temperature, calculate the concentration of the of the protein? (R = 0.0821 L atm mol'" K-1
other solution. [HOTS; All India 2011C] and 760 mm Hg = 1 atm).
Or [Delhi 2009; All India 2009)

Calculate the boiling point of one molar aqueous 37. Calculate the freezing point depression
solution. Density of KBr solution is 1.06 g mL-1. expected for 0.0711 m aqueous solution of
(Kb for H20= 0.52 K kg mol-l, atomic mass; Na2S04' If this solution actually freezes at
K = 39, Br = 80). [HOTS; All India 2011C] - 0.3 20°C, what would be the value of van't
Hoff factor? (Kf for water is 1.86°C rnol ").
30. A solution prepared by dissolving 1.25 g of oil of
winter green (methyl salicylate) in 99.0 g of [Deihl 2009; Foreign 2009)
benzene has a boiling point of 80.31°(, 38. Calculate the freezing point of a solution
Determine the molar mass of this compound. containing 18 g glucose, C6H1206 and 68.4 g
(Boiling point of pure benzene = 80.10°C and Kb sucrose, C12H22011 in 200 g of water. The
for benzene = 2.53°C kg rno!"). freezing point of pure water is 273 K and Kf
[Delhi 2010; Foreign 2010) for water is 1.86 Km-1. [HOTS; All India 2009C)
31. What mass of ethylene glycol (molar mass 62.0 39. Calculate the temperature at which a
g rno!") must be added to 5.50 kg of water to solution containing S4 g of glucose,
lower the freezing point of water from O°C to - (C6H1206) in 250 g of water will freeze.
10.0°C? (Kf for water = 1.86 K kg rnol'").
(Kf for water '= 1.86 K kg rnol " and molar
[All India 2010) mass of glucose = 180 g mot").
32. Calculate the amount of sodium chloride which [Delhi 2008; All India 2008; Foreign 2008)
must be added to one kilogram of water so that
the freezing point of water is depressed by 3 K.
40. A solution containing 8 g of a substance in
(Given, Kf = 1.86 K kg mol-l, atomic mass of 100 g of diethyl ether boils at 36.86°C,
whereas pure ether boils at 35.60°(,
Na = 23, CI = 35.5). [Delhi 2010C; Delhi 2009)
Determine the molecular mass of the solute
33. A solution of urea in water has a boiling point of (For ether, Kb = 2.02 K kg rno!").
373.128 K. Calculate the freezing point of the
[All India 2008; Foreign 2008)
same solution. (Given, for water Kf = 1.86 Km-1
and Kb = 0.52 Km-1). [HOTS; Delhi 201OC, 2009C]
41. A S% solution (by mass) of cane sugar in
water has a freezing point of 2Z1 K. Calculate
34.' 0.1 mole of acetic acid was dissolved in 1 kg of the freezing point of 5% (by mass) solution of
benzene. Depression in freezing point of glucose in water. The freezing point of pure
benzene was determined to be 0.256 K. What water is 273.15 K. (Molar mass of cane sugar
conclusion can you draw about the state of the = 342 g mol'? and molar mass of glucose
solute in solution? (Given, Kf for benzene
= 180 g mer"). [Foreign 2008)
= 5.12 Km- 1). [Delhi 2010C]
42. Calculate the mass of a non-volatile solute'
35. Calculate the mass of ascorbic acid (C6Hs06) to
(molar mass 40 g rnor '), which should be
be dissolved in 75 g of acetic acid to lower its
dissolved in 114 g of octane to reduce its
melting point by 1.5°(' (Kf for acetic acid is
vapour pressure to 80%.
3.9 K kg rnol'). [All India 2010C]
(Molar mass of octane = 114 g mol ").
[Foreign 2008)

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Solutions 33

43. The boiling point elevation of 0.30 g acetic acid (ii) At 300 K, 36 g of glucose, C6H1P6 present
in 100 g benzene is 0.0633 K. Calculate the per litre in its solution has an osmotic
molar mass of acetic acid from this data. What pressure of 4.98 bar. If the osmotic pressure
< c conclusion can you draw about the molecular of another glucose solution is 1.52 bar at
state of the solute in the solution?
the same temperature, calculate the
(Given, Kb for benzene = 2.53 K kg rnol'").
concentration of the other solution.
[HOTS; All India 2008C)
[All India 2011C]
44. Calculate the depression in freezing point of 48. (i) List any four factors on which the
" . water when 20.0 g of CH3CH2CHCICOOH is colligative properties of a solution
added to 500 g of water. depend.
(Given, Ka = 1.4x 10-3, Kf = 1.86 K kg rnol'"), (ii) Calculate the boiling point of one molar
[HOTS; Delhi 2008C) aqueous solution (density 1.06 g mL -1) of
KBr. (Given, Kb for H20 = 0.52 K kg mol-l,
45. The freezing point of a solution containing
0.2 g of acetic acid in 20.0 g of benzene is At. mass: K =39, Br = 80).
lowered by 0.45°C. Calculate [HOTS; All India 2011C)

(i) the molar mass of acetic acid from this data 49. (i) Define the terms osmosis and osmotic
(ii) van't Hoff factor. pressure. What is the advantage of using
osmotic pressure as compared to other
(For benzene, Kf = 5.12 K kg rnol ").
colligative properties for the
What conclusion can you draw from the value determination of molar masses of
of van't Hoff factor obtained? [All lndla 2008C] solutes in solutions?
(ii) A solution prepared from 1.25 g of oil of
wintergreen (methyl salicylate) in 90.0 g
•..' 5 Marks Questions of benzene has a boiling point of 80.31°C.
Determine the molar mass of this
46. (i) Define the terms osmosis and osmotic
compound. (Boiling point of pure
pressure. Is the osmotic pressure of a
benzene = 80.10°C (Kb for benzene =
solution a colligative property? Explain.
2.53°C kg rnol "). [All India 2010]
(ii) Calculate the boiling point of a solution
50. (i) What is van't Hoff factor? What possible
prepared by adding 15.00 g of NaCI to
values can it have if the solute molecules
250.0 g of water.
undergo dissociation?
(Kb for water =0.512 K kg mol ", molar
(ii) An aqueous solution containing 12.48 g
mass of NaCI = 58.44 g). [Delhi 2011]
of barium chloride in 1.0 kg of water boils
47. (i) The molecular masses of polymers are at 373.0832 K. Calculate the degree of
determined by osmotic pressure method dissociation of barium chloride.
and not by measuring other colligative (Given Kb for H20 = 0.52 K m-1,
properties. Give two reasons. Molar mass of BaCI2 208.34 g mo!").
[HOTS; Delhi 2011C)

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Step-by-Step Solutions
1. These are constant boiling mixtures which There are four colligative properties' (1/2)
distill out unchanged in their composition, (i) Relative lowering in vapour pressure
e.g., ethanol (C2HsOH) and water (H20) are (ii) Elevation of boiling point
azeotropes. (1) (iii) Depression of freezing point
(iv) Osmotic pressure (1/2)
2. Osmotic pressure It is the extra pressurewhich
is applied on the solution to just prevent the 8. ':\1 When a non-volatile solute is added to a
flow of solvent into the solution through a •• solvent, the freezing' point of the solution Is
always lower than that of pure solvent as the
semipermeable membrane. (1) vapour pressure of the solvent decreases In
the presence of non-volatile solute.
3. (i) Isotonic solution Solutions having the same
osmotic pressure are called isotonic Plot for the lowering in freezing point of water
solutions. For such solutions, concentration when NaCI is added to it is shown below
},.-!,e<;-.\.
is also same. (1/2)
'0 "',0
(ii) van't Hoff factor It is defined as the ratio 0o.0~
of the experimental value of colligative
property to the calculated value of the
t
Q)
,
,so~\:i
..
\oG

:;
colligative property and is used to find
(/)
(/)
,,
--~
Q)
0. , Tf = Freezing point
out the extent of dissociation or ,
:; " of sodium chloride
association. Mathematically, it is o
a.
L':.Tf~ solution
represented as §! t, :, T;
i = Observed colligative property (1/2) 273 K
Calculated coil igative property Temperature/K -
(2)

4. We know that, ~ Tb =Kbm 9. (i) First, find out L':.Tb by using formula.
L':.T kb XWb x 1000
When, m=l, ~Tb =Kb. Thus, boiling point b
MaxWa
elevation constant is equal to the elevation in
boiling point when 1 mole of a solute is (ii) Then, the value of T b by using formula,
dissolved in 1 kg of solvent. It is also called L':.Tb = Tb - T;
ebullioscopic constant. (1)
Given, Wa = weight of H20 (solvent) = 1 kg
5. The direction of osmosis can be reversed if a Wb = weight of C6H1206 (glucose-solute)
pressure larger than the osmotic pressure is = 18 g
applied to the solution side. That is, now the JOb=373.15K
pure solvent flows out of the solution through Kb =0.52 K kg rnol"
the semipermeable membrane. This Mb = Molar massof solute (glucose)
phenomenon is called reverse osmosis. It is
= 180 g rnol'"
used in desalination of sea water. (1)
~Tb=KbXl000XWb (1)
6. Refer to ans. 3 (ii). (1) Mb xw,
7. Colligative properties The properties of _ (0.52K kg mon xl 000 x18 g
solution which depend upon the number of - (180gmonxl000g
solute particles and not upon the nature of the
solute are known as colligative properties. 9360 =0.052 K
180000

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Solutions 35

As we know that, 13. The osmotic pressure method has the

LlTb = Tb - rob advantage over other methods because
=> 0.052 = Tb - 373.15 (i) pressure measurement is around the room
=> Tb=373.15+0.o52 temperature and the molarity. of the
= 373.202 K solution is used instead of molality. (1)
'" 373.20 K (approx.) (1) (i;) its magnitude is large as compared to
other colligative properties even for very
dilute solutions. (1)

" ,I Molality of solution, m = 1.00 m 14. Refer to ans. 11 and ans. 13. (2)
---------_._------
"(Boiling point of solution, 15. ;(;/ (i) First, find out zr, by using formula
Tb, =100.18° C = 373.18 K
,. tlTb= Kb X Wb X 1000
" -'Boiling point of water (solvent) Mb xW Q

"..' .T; = 100.00°

.,.1")"" 1"'
C = 373 K
(II) Then, find out the value of Tb by using
,LlTb = Tb - Tb = 373.18K - 373 K = 0.18 K L- fo_rm_ula tl._T!<..b
_=-'Tb'-.-_T,b'--- ---'
(1/2)
LlTb =; Kbm (1/2) Given, W2 = 0.520 g,
0.18K =; x 0.512 K kg mol' x1 mol kg" W, = 80.2 g, Kb = 0.52 Km-'
· 0.18 K M2 of C6H1206 = 6 x 12 + 12 x 1+ 6 x 16
1 = ------,.--------,-
0.512 K kg rnol " x 1 mol kg" = 180 g mot"
;=0.35 (1)
LlTb = KbW21000 (1)
11. Osmosis The process of flow of solvent M2W,
molecules through semipermeable membrane
from solvent to the solution or from solution of _ 0.52 Km-' x 0.520 gx 1000
ts, lower concentration to the solution of higher - 180 g mol" x 80.2 g
concentration is known as osmosis. (1/2)
= 0.019 K (1/2)
Osmotic pressure Refer ans 2. (1/2)
Boiling point of solution
It is a colligative property because it depends
only upon the number of solute particles but = 373 K + 0.019 K =373.019 K (1/2)
------------- .-----
d"hoton their nature, e.g., 0.1 molarKCland NaCI
have the same osmotic pressure under same 16. '9 (i) First, find out tlTf
KfW2 X 1000
using the formula

. conditions of temperature and pressure. (1) tlTf

M2XWj
1~./The-extent to which a solute is dissociated or
(ii) Then, find out the value of Tf by using
associated can be expressed by van't Hoff formula, tlTf = T," - Tf
factor, ;. It is defined as
· Normal molar mass Given, W2 (glucose) = 0.520 g
1=---------------
. Abnormal molar !!lass W, (H20) = 80.2 g
· Observed colligative property x, (H20) = 1.86 Km-'
I=------~-~-~~
Calculated colligative property
1 ; M2 of C6H,206 (glucose) = 180 g mol'
Total number of moles of particles
, :after association / dissociation LlT = KfW21000
1=------------- (1) f
. Number of moles of particles M2W,
before association / dissociation 1.86 Km-' x 0.520 g x 1000
(1)
In case of dissociation, the value of; is greater 180 g mol" ' x 80.2 g
than 1. (1) = 0.0669 K (1/2)

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36 Chapterwise CBSE Solved Papers Chemistry

Freezing point of solution 2.5 xl 0-2 x 0.0821 x 298

1t = (1)
= 273 - 0.0669 174x2
= 272.933K (1/2) 2
=61.1645x 10- =0.1758x 10-2 atm
17. Osmotic pressure Refer ans. 2. (1) 348
Relation between osmotic pressure and molar = 1.76 x 10-3 atm (approx.) (1)
mass
20. Given, W2 =1.00 g, W, =50 g,
1t V = n2 RT
«, =5.12 K kg mol-'

1tV = W2 RT I1Tf =0.40 K

M2 I1T =KfW2Xl000
f
M2 = W2 RT M2XW, (1)
1tV M =Kf xW2xl000
2
(where 1t is the osmotic pressure and M2 is the W, x I1Tf
molar mass of solute). (1) 5.12 xlxl000
= (1)
50 x 0.40
18. =<; Depression in freezing point of a solvent due to
•• the dissolution of a solute is dependent on the =256 g rnol " (1)
degree of dissociation (a). tJ.T, oc (a).

The depression in freezing points are in the

21. ~~?Apply tJ.T = Kf x W2
f
X 1000 to find out the

M2 xWj
order value ofW2•
Acetic acid < trichloroacetic
< trifluoroacetic acid Given, W, = 1000 g,I1Tf = 2 K
H", CI", Kf = 1.86 K kg mol-'
H-C-COOH< CI-C-COOH
H/ CI/ M2 of KCI = 39.09 + 35.45
= 74.54 rnol' (1)
<~"'C-COOH
F/ (1) I1T = Kf x W2 X 1000 (1)
f
W,XM2
Fluorine has the highest electron withdrawing
2 = 1.86 X W2 x 1000
inductive effect so trifluoroacetic acid is the
strongest acid and acetic acid is the weakest 1000 x 74.54
acid. Therefore, trifluoroacetic acid ionises to W _ 2 x 74.54
the greater extent and acetic acid ionises to the 2 - 1.86
minimum extent. Greater the number of ions
W2 =80.15 g (1)
produced, greater is the depression in freezing
point. (1) o

22. ~ Apply formula, P -:.Ps =:2.

WRT P
Apply 1t = _b - to find out the value of 1t. nj

MbV
pO - ps _ _ n
-X2 -- 2
Given, mass ofK2S04, Wb =25 xl 0-2 g p" n,
Molar mass of K2S04, Mb =174 g mol " [for a dilute solution, n2« n,l
V=2L, T=25°C=298 K pO - ps _ W2 X M,
(1)
R=O .0821L atm K-' mol-' pO M2 X W,
We know, osmotic pressure, Given, pO = 3.165 kpa, W2 = 5 g, W, = 95 g
1t = WbRT (1) M2 = 60.05 g rnol'", M, = 18 g mol "
MbV

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Solutions 37

3.165 - Ps = 5 x 18 = 0.0158 (1) I'..Tf=iKf·m, Ll

AT __ i K, X W2 X 1000
f
3.165 60.05 x 95 M2 x W, (1/2)
3.165 - Ps = 0.049 3 x 1.86 K kg mol-1 x 10.50 gx 1000
Ps=3.115 kpa
/',.Tf = 1
(1)
184 g mol- x 200 s
23. I'..t, = Kf X W2 X 1000 = 1.592 K (1/2)
M2XW,
Freezing point of solution = 273 - 1.592
Given, W, = 450 g, W2 = 15.0 g = 271.408 K (1)
I'..Tf = (273 K- 272.66 K)= 0.34 K Or
K, = 1.86 K kg mol'
~9 Apply van't Hoff equation, ~ Tb = iKbm, where
M2 = Kf X W2 X 1000 (1) ; = number of ions produced by NaCI.
I'..Tf x w,
Given, W2 = 15.00 g, W, = 250.0 g
M2 = 1.86 K kg mol-' x 15 g x 1000 (1)
0.34 K x 450 g M2 = 58.44 g rnol'"

M2 = 182.35 g rnol'" (1) Kb = 0.512 K kg rnol'"

Or NaCI(aq) ~ Na+(aq) + CI-(aq)
Given, W, = 500 g i=2 (1)
Boiling point of solution (Tb) = 100.42 ° C
I'..Tb =iKb·m
Kb for water = 0.512 K kg rnol'
= I. Kb
AT x W2 X 1000
M2(C3H803) = (3 x 12) + (8 x 1) + (3 x 16) Ll b
M2W, (1/2)
= 92 g mol'
I'..Tb = Tb - T; /',.Tb = 2x 0.512 K kg mol-' x 15.0 gx 1000
58.44 g mol" ' x 250 g
= 373.42K- 373K
= 0.42 K (1)
I'..Tb=1.051K (1/2)

I'..Tb = Kb X W2 X 1000 (1) Boiling point of solution

M2XW, =373K+1.051K

W _I'..TbxM2xW, = 374.051 K (1\

2 - Kb x1000 25. Given, W2 = 6.21 g, W, = 24.0 g,

0.42 K x 92 g mol-' x 500 g
Tb = 68.04°C, T; = 61.7°C
0.512 K kg mol" ' x 1000
and Kb = 3.63° C / m
= 37.73 g (1)
I'..Tb = Kb X W2 X 1000 (1)
24. Apply van't Hoff equation, M2XW,
'.
~ Tf = iKr m, where i= number of ions M2 = Kb X W2 x1000
or
produced by MgBr2' I'..Tb xw,

Given, W2 = 10.50 g, W, = 200 g (I'..Tb = Tb - T; = 68.04° C- 61.7° C

M2(MgBr2)= 184 g mol'" = 6.34° C) (1)

Kf = 1.86 K kg rnol" ' _ 3.63° C m-' x 6.21 g x 1000

M 2-
MgBr2(aq) ~ Mg2+(aq) + 2Br-(aq) 6.34° Cx 24.0 g
= 148.15 g rnol' (1)
i= 3 (1)

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38 Chapterwise CBSE Solved Papers Chemistry

Or
28. ~9 (I) First calculate M2 (observed) by using the
Given, W2 = 8.95 mg = 8.95 x 10-3 g, Kf XW2 X 1000
formula, M2 = -'----=--
35 6Tf·WI
V = 35.0 ml=-- l ,
1000 (II) Calculate 1by using the formula,
0.335 ; = M2 (Calculated)
1t= 0.335 torr=-- atm
760 M2 (Observed)
(',: Horr= 1 mm of Hg) (iii) Calculate a by using the formula,
/-1
T=273+ 25= 298 K
1tV = nRT a=G-l)
or 1tV = W2RT (1)
M2 Here, n:;= 2 because phenol forms dimer on
association.
M2 = W2RT
1tV Given, W2 = 20 g, WI = 1 kg= 1090 g,
t.Tf = Q.69 K '
8.95x 10-3 g x 0.082l atm mol-l K-l
x 298 K Kf = 5.1 Km-l
=-~---~~-~~-----
0.335 t 35.0 l t.T = Kf x W2 X 1000
--amx-- f
760 1000 (1) M2Wl
M2= 14193 g rnol'" _ K W2 X 1000 X
M 2- f
=1.4193x 104 g mof" (1)
st, X WI
, M _ 5.1 Km-l'x 20 ~ x 1000
26. Given, WI = 65.0 g, t.Tf = 7.50° C, 2- 0.69 Kx 1000 g
K, = 1.86°C/m, ;= 1.87
M2 (observed) = '147.82 (1)
and M2 = 58.5 g rnol'" (1)
M2 (calculated)
st, =; x Kf X W2 X 1000 (1)
C6H50H= 6x 12+ 6x 1 + 16
M2xWl
",; 94 g mal-l
_ t. Tf x M2 X WI
W 2-
;x «, x 1000
",) ,
; = M2 (Calculated)
_ 7.50° Cx 58.5 g mol-l x 65 g M2 (Observed)
- 1.87x 1.86°C/mx 1000
';,' .' " -' '," 94 ' - '
'=--=0.635 (1)
= 8.199 g (1) 147.82
27. Given, m = 0.561 m, t. T, = 2.93° C and 2C6H50H ~ (C6H50H)2'
Kf = 1.86° C kg marl ;-1 0.635- 1
t. T,
.
= ;Kfm
t.Tf
(1) U=-(l
,~-1
)= 1 '
2'-1
1=-
Kfm
= 0.365 = 0.73
(1) 0.5
.1.86° C kg mal-l x 0.561 m '\ ' .
U= 73% (1)
= 2.807 (1)
1; , '

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Solutions 39

Or For first solution,

(4.98 bar) = (36 g) x R ~ T ... (i) (1/2)
:t:; (/)First calculate M2 observed by using the (180 g mol- ) x V
f ormu Ia, M 2 =- --'Kb'-x_W----"-2
_x_l000_ For second solution,
Arb xWl
(1.52 bar) = W2 x R x T ... (ii) (1/2)
(II) Calculate / by using the formula, M2 xV
/ M2 (Calculated)
On dividing Eq. (ij) by Eq. (i), we get
M2 (Observed)
(iii) Calculate a by using the formula, (1.52 bar) = W2 x R x T x 180 x V (1)
/-1 (4.98 bar) M2xV 36xRxT
a=--
m-l
W2 = 1.52 = 0.0610 mol L-' (1)
M2 4.98x 5
. Here, m = 3 because 1 molecule of BaCI2 on
dissociation gives three ions. Or
Given, W2 = 12.48 g, W, = 1.0 kg-=l 000 g,
Q (i) Calculate the molality of solution by using
Tb (solution) = 373.0832 K the formula,
Kb for H20 = 0.52 Km-' Mxl000
m
l000xd-MxM2
and M2(BaCI2)= 208.34 g rnol'"
(where, M = molarity, d = density and
Il. Tb = Tb - Pb = 373.0832 K - 373 K M2 = molar mass of solute)
= 0.0832 K (iI) Calculate Ll.Tb by using the formula,
Kb x W2 X 1000 Ll.Tb=Kbm
M 2 (0 b serve d) =
sr; xW, (iii) Calculate Tb (boiling point of solution) by
0.52 Km-' x 12.48g x 1000 using the formula, T; + Ll.Tb= Tb
M (o b serve d) = ---------=:..---
2 0.0832 K xl 000 g
Given, concentration of the solution = 1 molar
M2 (observed) = 78 g rnol" ' (1)
Density of the solution = 1.06 g mL-'
.
I
M2 (Calculated)
= ~--'------'- M2' molar mass of KBr = 39 + 80
M2 (Observed) = 119 g mol "
= 208.34 g mol-' = 2.67 (1) Kb for H20 = 0.52 K kg morl
78 g rnol" '
· M xl000
Mo Ia Iitv, m = ------- (1/2)
For BaCl2,. m = 3 as it gives 3 ions on 1000xd-MxM2
dissociation.
1 x 1000
a=~=2.67- 1= 1.67 = 0.835 m=--------
1000x 1.06- 1 x 119
m-l 3-1 2
= 1.0626 mol kg-~ (112)
a = 83.5% (1)
Il.Tb = Kbm
29·1;.0•. Apply n=CRTand c=~=~
U M~
= 0.52 K kg rnol!x 1.0626 mol kg " (1/2)
Il.Tb = 0.5525 K"" 0.553 K (1/2)

1t = CRT= W2 x Rx T Tb = 373 K+ 0.553 K

M2x V = 373.553 K (1)
For both solutions, R, T and V are constant.

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40 Chapterwise CBSE Solved Papers Chemistry

30. = Kfm
Refer to ans. 25. ~Tf
~T _1.86Km- xO.128K
, .

Ans. M2 = 152 g mar' (3)

f - 0.52 Km-'
31. Given, M2 (ethylene glycol) = 62 g rno!",
= 0.458 K (1)
W, = 5.50 kg = 5500 g, ~ T, = 10 K
[O°C- (-10 C)= 0
lO°C= 10 K] 34. ~~?(i) First find M2
K xUJ xlOOO
observed by using the

and K, = 1.86 K kg mar' formula, M2 = -'f_-'2=-- __

X UJj st,
~Tf=KfW21000 (1)
(ii) Compare it with calculated M2 to hnow
M2W, either it Is associated or dissociated.
W = ~TfM2W, = 10 Kx 62 g mol-' x 5500 g
2 Given, W, = 1 kg= 1000 g (benzene)
Kf1000 1.86 K kg mar' x 1000
. (1) ~Tf = 0.256 K, x, = 5.12 Km-'
W2 = 1833.33 g = 1.833 kg (1) 0.1 mole of acetic acid
= 0.1 x 60= 6.0 g(W2)
32. "9 Apply the formula, Mf
iK UJ x 1000
--.Lf-=-2
M2UJj
--
[.,' Molar mass of CH3COOH = 60 g man
and find out the value of UJ2. (1/2)
M2 = Kf X W2 X 1000 (112)
Given, W, =1 kg=1000 g, ~Tf =3K
st, X W,
K, =1.86 K Kg mar'
5.12 Km-' x 6.0 g x 1000
M2(NaCI)=23+35.5= 58.5 g mol-'
0.256 Kx 1000 g
_iKf xW2x1000
~ Tf - ---'-----"--- (1) = 120 g mol ' (1/2)
M2XW,
_~TfXM2XW,
i = M2 (Calculated) = 60 = -'!. = 0.5 (1/2)
W 2- M2 (Observed) 120 2
t-:«, x 1000
The value of i is 0.5. It is less than 1. Hence,
[For NaCl, i = 2 because 1 mole of NaCI on
the solute i.e., CH3COOH gets dimerise in
dissociation gives 2 moles of ions]
benzene. (1)
3 Kx 58.5 g mar' x 1000 g
(1) 35. Refer to ans. 31. Ans. W2 = 5.08 g (3)
2 x 1.86 K kg mar' x 1000
=47.18g (1)
36. Given, W2 = 100 mg= 0.1 g,
V= 10.0 mL= 0.01 L,
(i) First find molality of the solution by using
13.3
the formula, t>Tb =Kbm 1t = 13.3 mm Hg=-- atm
(ii) Find Mf by using the formula, t>Tf = Kfm 760
(iii) Find freezing point of the solution by using T = 25° C= 273+ 25= 298 K
the formula, Tf = Tf t>Tf 0 -
R= 0.0821 L atm mol-' K-'

Given, Tb = 373.128 K, 1tV = W2 RT (1)

M2
K, = 1.86 Km-l, K~ = 0.52 Km-'

st; = 373.128 K- 373 K= 0.128 K M2 (Protein) = W2RT

1tV
[.,' Tbo for water = 373 K]
0.1 gx 0.0821 L atm mar' K-' x 298 K (1)
~Tb =Kbm (1)
13.3 1 L
-- atmx 0.0
m = ~ Tb = 0.128 K 760
or (1)
Kb 0.52 Km-' M2 = 13980.45 g rnol' (1)

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Solutions 41

37. ;'9 Find out the value of expected freezing point

depression, L'l.Tf by using formula, L'l.Tf = iKfm,
Freezing point of aqueous solution
= 273- 2.79 = 270.21 K (1)
then find the value of 'i' by using formula,
., L'l.Tf 39. Given, M2 (glucose, C6H1206)= 180 g rnol" '
1=-.
Kfm W2 = 54 g, W, = 250 g, K, = 1.86 K kg rnol "

llT = Kf x W2 X 1000 (1)

Given, m= 0.0711 molal solution, f
M2XW,
llTf = 0 x 320° C= 0.320 K,
K, = 1.86° C mol-l = i.86 Km-l 1.86 K kg mol-' x 54 gx 1000 2.23 K
For Na2S04, ; = 3 (because three ions are 180 g rnol ' x 250 g (1)
produced when it dissociates completely).(1/2) Freezing point of solution = 273 K- 2.23 K
Expected freezing point depression, = 270.77 K (1)
IIT, =; Kf x m = 3 x 1.86 x 0.0711
= 3 x 0.132 "C = 0.397°C (1)
40. Refer to ans. 25.

Further, when the solution freezes at -0.320°C

Ans. M2 = 128.25 g mot ' (3)

IIT, = ;'Kfm (1/2)

41. ~~I Freezing point of pure water is 273.15 K(O°C)
., IIT, 0.320 K and that of given solution is 271 K, i.e., we have
or I - -- - ----,-----
- Kfm - 1.86 Km-l x 0.0711 m depression in freezing point. So,
(i) Apply the formula of depression in freezing
= 2.42 (1)
point for cane sugar and find the value of

38. ~9 (i) First find total number of moles of solute

• (i. e., nglucose + n sucrose) present in the
Kf·
(ii) Use thisvalue of Kf to find out the value of
L'l.Tf for glucose solution. Then, find the
solution. freezing point of solution.
L- _
(ii) Find L'l.Tf by using the formula

(iii) Tr"'T,
L'l.Tf = K

-L'l.Tf
f
x~::'1
1000
.'

~
I Given, 5% sugar solution means, W2 = 5 g,
W, = 95 g, M2 = 342 g rnol'

Similarly, for 5% glucose solution,

W2 (glucose) = 5 s: WI' = 95 g
Molar mass of glucose, C6H1206
M2 (glucose) = 180 g rnol'"
= 6 x 12 + 12 xl + 6 x 16 = 180 g mol-l
Molar mass of sucrose, C12H22011 llTf(cane sugar) = 273.15 K- 271 K = 2.15 K

= 12x 12+ 22x1+11x16 st, = Kf x W2 X 1000

. = 342 g mol-l M2W,
18 2.15 = Kf x 5 xl 000
nglucose = - = 0.1
180 342x 95
68.4
nsucrose=--= 0.2 K =2.15 x342 x95 =13.9707
f
342 5 x 1000 (1)
Total moles of solute, n2 = 0.1 + 0.2 = 0.3 (1) «, xW2'x1000
AT f (Ig ucose ) = -'--~---
'-'
AT
'-'
_
f -
«, x W2 xl 000 [
n2--
_ W2] M2'W"
M2W, M2 AT (' I ) 13.9707x 5 x 1000
'-' f g ucose = -------
IIt, = Kf x n2 xl 000 180x 95
W, = 4.085 K (1)

1.86 Km-' x 0.3 mol x 1000 = 2.79 K Freezing point of the solution
200 g (1) = 273.15 K- 4.085 K = 269.065 (1)

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42 Chapterwise CBSE Solved Papers Chemistry

42. Given, vapour pressure is reduced to 80% CH3CH2CHCICOOH (aq) ~

when non-volatile solute is dissolved in octane. CH3CH2CHClCOO-(aq) + H+(aq)
It means if o" = 100 atm then Ps = 80 atm.
Initial cone. C o 0
M2 = 40 g rnol'", Equili. cone. C - Co. Co. Co.
W,=114g CO-a)
and M, (C6H'6) = 114 g rnol '. (1/2) (a is the degree of dissociation of acid).
2
pO_Ps=~ (112)
K = (Ca) = Ca2
p" n2 +n, a C (1- a) (1/2)

pO - Ps _ W2 / M2
1.4 x 10-
3
= 0.065
p" - W2 W, a=!i=
-+- 0.3264 (1/2)
M2 M, (1)
i -1
a=--
lOO - 80 _ W2/ 40 m-l (1/2)
100 - W2 114
-+-
40 114 0.065 = i -1
2 -1
0.2 = W2
W2 +40 [For the given acid m = 2 because 1 molecule
gives 2 particles on dissociation.]
W2 =10 g (1)
i = 1 + 0.065 = 1.065 (1/2)
43. Refer to ans. 34.
!l.Tf = iKfm
i = 0.5. Here i < 1, therefore the solute, acetic
= 1.065 x 1.86 K kg rnol " x 0.3264 m
acid is associated in benzene. (3)
!l. t, = 0.6465", 0.65 K (1/2)

44. ~I (i) First calculate molality, m and degree of

45. Refer to ans. 34.
dissociation, using the formula 0.= ~
M2 = 113.77 g mol'", i = 0.52
(ii) Calculate van't Hoff factor, I by using i < 1, therefore the non-volatile solute
;-1 (acetic acid) is associated. (3)
formula, a = -- (here, m = 2)
m-l
(Iii) Find d Tf by using, d Tf = ; K( m
46. (i) Refer to ans. 11. (2)
(ii) Refer to ans. 24 (or part). (3)
Molar mass of CH3CH2CHCICOOH 47. (i) Refer to ans. 13. (2)
= 4x 12+ 7x 1 + 35.5+ 2x16 (ii) Refer to ans. 29. (3)
= 122.5 g rnol'"
48. (i) (a) Number of particles of solute
Number of moles of CH3CH2CHCICOOH • (b) Association or dissociation of solute
20 (c) Concentration of solution
--- (d) Temperature (2)
122.5
= 0.1632 mol (ii) Refer to ans. 29 (or part). (3)

· f so
soluti Moles of solutex 1000 49. (i) Refer to ans. 11. (2)
. 0 Ia I Ity
M 0 ution = --------
Mass of solvent (g) (ii) Refer to ans. 30. (3)
0.1632 x 1000 50. (i) Refer to ans. 12. (2)
m=-----
500 (ii) Refer to ans. 28 (or not). (3)
=0.3264m (1/2)

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