Solutions and Colligative Properties

Solution : It is homogeneous mixture of solute and solvent

Solute : Present in less quantity (in moles)

Various ways of expressing conc. of a solutions

mass %, vol %, ppm, ppb, M, N, F, Strength, mole fraction

A. mass percent and ppm

1. A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per
cent of the solute.
2. 50g. of a solute is dissolved in 0.95 kg. of the solvent. The weight percent of the solution is
a) 50 b) 0.95 c) 5 d) 0.095
3. A solution is 0.25% by weight. The weight of solvent containing 1.25 g. of solute would be
a) 500g b) 498.75g c) 500.25g d) 501.25g
4. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g
of a 25% and 400 g of a 40% solution by mass. (NCERT XII 2.7)
5. 400cm3 of water is added to 600 g. of a 6% (weight percent) aqueous solution of a solute. The
final weight percent of the solute is
a) 10 b) 1.5 c) 9 d) 3.6
6. Equal volumes of a 10% aq. solution (by weight) of the solute A and 15% aq. solution (by
weight) of the solute B are mixed. The weight percent of A and B in the mixture would be
respectively (assume densities of both solutions are 1 g ml-1)
a) 5 and 7.5 b) 10 and 15 c) 15 and 10 d) 20 and 30
7. 200 g. of a 5% solution (by weight) of the solute A is mixed with 300g. of a 10% solution
(by weight) of solute B. The weight percent of A and B in the mixture are respectively
a) 3.33 and 15 b) 5 and 10 c) 2 and 6 d) 6 and 2
8. 3 x 10 gms of chlorine dissolved in 1 litre of tap water find out its concentration in terms of
-3

mass % and ppm?

9. Calculate the mass of FeSO4.7H2O. which must be added in 100 kg of wheat to get 10 ppm of Fe.
(IIT-JEE Jan 2020)
10. 10.30 mg of O2 is dissolved into a litre of sea water of density 1.03 g / ml. The concentration of
O2 in ppm is ? ( IIT-JEE Jan 2020)
11. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3,
supposed to be carcinogenic in nature. The level of contamination was 15ppm (by mass).
i) Express this in percent by mass. (NCERT XI 1.17) (NCERT XII 2.9)

AnswerKey
1. 10 2. C 3. B 4. 33.57% 5. D 6. A 7. C 8. 3x10-4 % and 3ppm
9. 4.96 g 10. 10 ppm 11. 1.5 x 10-3 %

B. Molarity (M)
M = moles of solute per litre of solution
1. Calculate the molarity of NaOH in the solution prepared by dissolving its 4g in enough water to
form 250ml of the solution. (NCERT XI)
2. What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water
to make a final volume up to 2L? (NCERT XI 1.11)
3. Concentrated nitric acid used in laboratory work is 68% HNO3 by mass in aqueous solution.
What should be the molarity of such acid if the density of the solution is 1.504g ml-1?
(NCERT XII 2.4)
4. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41
g ml–1 and the mass per cent of nitric acid in it being 69%. (NCERT XI 1.6)

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Solutions and Colligative Properties

5. Dissolving 120g of a compound of (mol. wt. 60) in 1000g of water gave a solution of density
1.12g/ml. The molarity of the solution is: [IIT-JEE Main 9/04/2014]
a) 1.00M b) 4.00M c) 2.00M d) 2.50M
6. The density of a solution prepared by dissolving 120g of urea (mol. Mass = 60u) in 1000g of
water is 1.15g/ml. The molarity of this solution is: [AIEEE-2012/ IIT-2011]
a) 0.50M b) 1.78M c) 1.02M d) 2.05M
7. The molarity of a solution obtained by mixing 750ml of 0.5(M) HCl with 250ml of 2(M)HCl will
be: [IIT-JEE Main-2013]
a) 1.00M b) 1.75M c) 0.975M d) 0.875M
8. Two solutions of a substance (non electrolyte) are mixed in the following manner. 480ml of 1.5M
first solution + 520mL of 1.2M second solution. What is the molarity of the final mixture?
a) 1.20M b) 1.50M c) 1.344M d) 2.70M [AIEEE-2005]
9. 100 ml of a solution containing 5 g of NaOH are mixed with 200 ml of M/5 NaOH solution.
Calculate the molarity of the resulting solution.
10. How many litres of water must be added to 1 litre of an aqueous solution of HCl with
concentration 0.1 M to create an aqueous solution with concentration 0.01 M?
a) 100 L b) 10 L c) 9 L d) 1 L
11. 29.2%(W/W) HCl Stock solution has a density of 1.25g ml-1. The molecular weight of HCl is
36.5g mol-1. The volume (mL) of stock solution required to prepare a 200mL solution of 0.4M
HCl is: [IIT-JEE 2012]
12. What volume of 95 mass % sulphuric acid (density = 1.85 g/cm3) and what mass of water must
be taken to prepare 100 cm3 of 15 mass % solution sulphuric acid (density = 1.10 g cm-3)?
12’. To 50 ml of 0.5M H2SO4, 150 ml of 0.25 M H2SO4 is added and the mixture is made to volume
250 ml. What is the final Concentration of solution?
a) 0.31M b) 0.37M c) 0.75 M d) 0.25M
13. 6.023 x 1022 molecules are present in 10 g of a substance X. The molarity of solution containing
5 g of substance X in 2 L of solution is …………. X 10-3 (IIT-JEE Sept 2020)
14. Calculate the molarity of a solution of CaCl2 if on chemical analysis it is found that 500 ml of
CaCl2 solution contains 1.505 x 1023Cl- ions.
15. How many mL of 0.1M HCl are required to react completely with 1g mixture of Na2CO3 and
NaHCO3 containing equimolar amounts of both?
16. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its
0.25M solution? (NCERT XI 1.12)

Answer Key
1. 0.4M 2. 0.03M 3. 16.23M 4. 15.44M 5. c 6. d 7. d 8. c
9. 0.55M 10. c 11. 8 12. 9.4ml & 92.6gms 12’ d 13. 25 14. 0.25 M 15. 157.8
ml
16. 25.22ml.

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Solutions and Colligative Properties

C. Normality (N)
N = equivalent moles of solute per litre of solution

𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑚𝑜𝑙𝑒𝑠
N= × 1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.𝑖𝑛 𝑚𝑙

𝑔𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡
𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
N= × 1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.𝑖𝑛 𝑚𝑙

𝑔𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
𝑛−𝑓𝑎𝑐𝑡𝑜𝑟
N= × 1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.𝑖𝑛 𝑚𝑙

𝑔𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡
𝑥 𝑛−𝑓𝑎𝑐𝑜𝑟
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
N= × 1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.𝑖𝑛 𝑚𝑙

N = M x n-factor

About n-factor

About Acids Bases Salts Ions

n- How much moles of H+ How much moles of Total +tive chage or Overall charge
-
factor it gives per mole of OH it gives per mole of total -tive charge on ion
acid base NaCl (n-factor = 1) Na+ 1
H2SO4 (n-factor = 2) NaOH (n-factor = 1) MgCl2 (n-factor = 2) Al3+ 3
2-
HCl (n-factor = 1) Ba(OH)2 (n-factor = 2) Al2(SO4)3(n-factor = 6) SO4 2
H3PO4 (n-factor = 3) Al(OH)3 (n-factor = 3) K2SO4.Al2(SO4)3.24H2O PO43- 3
H3PO3 (n-factor = 2) (n-factor = 8)
Q. Find out normality of 49 gms of H2SO4 in 500 ml of solution?
𝑔𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡
M= × 1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.𝑖𝑛 𝑚𝑙

49
98
M= × 1000 = 1M
500

N = M x n-factor = 1 x 2 = 2 N

D. Formality (F)
F = Formality = number of formula moles per litre of solution

Q = Find out molarity and formality of 120 gm of acetic acid in 500 ml of benzene?
𝑔𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 𝑔𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡
M= × 1000 F= × 1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.𝑖𝑛 𝑚𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙.𝑖𝑛 𝑚𝑙
120 120
60 120
M= × 1000 = 4 M M= × 1000 = 2 F (as acetic acid forms dimer in benzene)
500 500

Strength = gms of solute L- of solution

Strength = M x molecular weight

Strength = N x equivalent weight

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Solutions and Colligative Properties
E. Molality (m)
m = molality = moles of solute kg-1 of solvent
𝑀 1 ƍ 𝑀𝑏
m= & = −
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑚 𝑀 1000

(m = molality, M = Molarity, ƍ = density of solution, Mb is molar mass of solute)

1. 91% H2SO4 w/v (density = 1.84 g ml-1) what is molality of solution.
2. Density of a 2.05M solution of acetic acid in water is 1.02g/ml. The molality of the solution is
a) 1.14mol kg-1 b) 3.28mol kg-1 c) 2.28mol kg-1 d) 0.44molkg-1
3. The density of 3M solution of sodium chloride is 1.252g ml . The molality of the solution will be:
-1

(molar mass, NaCl = 58.5g mol-1) JEE(Main)-2013

a) 2.60m b) 2.18m c) 2.79m d) 3.00m
4. The molality of a solution of ethyl alcohol (C2H5OH) in water is 1.55 m. How many grams of
ethyl alcohol are dissolved in 2 kg of water?
5. The molality of 1 molar solution will be (Given density of solvent = 1.5 kg L-1)
a) 1.5 b) 1 c) 0.66 d) 0.75
6. A solution of sodium sulfate contains 92 g of Na+ ions per kilogram of water. The molality of Na+
ions in that solution in mol kg-1 is
a) 12 b) 4 c) 8 d) 16 jee (main) -2019

NEET-22

Answer Key:
1. 9.98m 2. c 3. c 4. 142.6gm 5c 6b

F. Mole Fraction (ᵡ )
1. Calculate the molality and mole fraction of the solute in aqueous solution containing 3g of urea
per 250 g of water.
2. Calculate the mole fraction of ethanol and water in a sample of rectified spirit which contains
92% ethanol by mass
3. Calculate the mole fraction of water in a mixture of 12 g water, 108 g acetic acid and 92g
ethanol.
4. Calculate mole fraction of 1 molal aq. solution
Answers Key: 1. 0.2m & 0.0036 2. Xethanol = 0.82 3. 0.15 4. 0.0177

C. Raoults law, Ideal and Non Ideal solutions

Vapour Presure : It is pressure exerted by vapors above the surface of a liquid at equilibrium at
constant temperature.

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑉𝑎𝑝𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝐴 𝑖𝑛 𝑣𝑎𝑝𝑜𝑟 𝑝ℎ𝑎𝑠𝑒

Mole Fraction of component A in vapour phase =
𝑇𝑜𝑡𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑝𝑟𝑒𝑠𝑢𝑟𝑒

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑉𝑎𝑝𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝐵 𝑖𝑛 𝑣𝑎𝑝𝑜𝑟 𝑝ℎ𝑎𝑠𝑒

Mole Fraction of component B in vapour phase =
𝑇𝑜𝑡𝑎𝑙 𝑣𝑎𝑝𝑜𝑟 𝑝𝑟𝑒𝑠𝑢𝑟𝑒

1. At 200C, the vapor pressure of pure liquid A is 22 mm Hg and that of pure liquid B is
75mm Hg. What is the composition of the solution of these two components that has a vapor
pressure of 48.5 mm Hg at this temperature (assume ideal behaviour)?

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Solutions and Colligative Properties

1’. At 880C VP of pure benzene and toluene are 900 mm Hg and 360 mm Hg respectively. If the
solution bolis at 880C. Find the mole fraction of benzenein the solution
2. The vapor pressures of ethanol and methanol are 44.5 and 88.7 mm Hg respectively. An ideal
solution is formed at the same temperature by mixing 60 g of ethanol and 40 g of methanol.
Calculate the total vapor pressure of the solution and the mole fraction of methanol in the vapor
phase. [I.I.T.1986]
3. Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing
1mol of X and 3mol of Y is 550mm Hg. At the same temperature, if 1mol of Y is further added to
this solution, vapour pressure of the solution increases by 10mm Hg. Vapour pressure (in
mmHg) of X and Y in their pure states will be, respectively: [AIEEE-2009]
a) 200 and 300 b) 300 and 400 c) 400 and 600 d) 500 and 600
4. At 800C, the vapour pressure of pure liquid ‘A’ is 520mm Hg and that of pure liquid ‘B’ is
1000mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 800C and 1atm pressure, the amount of
‘A’ in the mixture is (1atm = 760mm Hg) [AIEEE-2008]
a) 52mol percent b) 34mol percent c) 48mol percent d) 50mol percent
5. The vapour pressure of water is 12.3kPa at 300K. Calculate vapour pressure of 1 molal solution
of a non-volatile solute in it.
6. Benzene and toluene form nearly ideal solutons. At 200C, the vapour pressure of benzene is 75
torr and that of toluene is 22torr. The partial vapour pressure of benzene at 200C for a solution
containing 78g of benzene and 46g of toluene in torr is
a) 50 b) 25 c) 37.5 d) 53.5
7. For an ideal solution of two compounds A and B, which of the following is true?
[IIT-JEE 19/04/2014]
a) A – A, B – B and A – B interactions are identical
b) ∆Hmixing > 0 (zero)
c) ∆Hmixing < 0 (zero)
d) A – B interaction is stronger than A – A and B – B interactions

Answers Key:
1. 0.5 2. 66.15 mm and 0.657 3. C 4. D 5. 12.08 kPa 6. A 7. A

D. Henry’s Law
Saturated Solution : A solution in which no more solute can be dissolved at the same
temperature and pressure is called a saturated solution
Solubility: It is concentration of substance in mol L-1 in its saturated solution at constant
temperature.
Effect of temperature on solubility: the dissolution process is endothermic (ΔHsol > 0), the
solubility should increase with rise in temperature and if it is exothermic (ΔH sol < 0) the
solubility should decrease with rise in temperature, These trends are also observed
experimentally.
Effect of pressure on solubility: Effect of pressure Pressure does not have any significant
effect on solubility of solids in liquids. It is so because solids and liquids are highly
incompressible and practically remain unaffected by changes in pressure.
Solubility of a Gas in a Liquid: Solubility of gases in liquids is greatly affected by pressure
and temperature. The solubility of gases increase with increase of pressure.

Henry’s Law = The pressure of a gas over a solution in which a gas is dissolved is
directly proportional to the mole fraction of the gas dissolved in the solution
P = KHx
p is partial pressure of the gas in vapour phase (equilibrium with the solution) and x is mole
fraction of the gas in the solution
Different gases have different KH values at the same temperature (Table 2.2). This suggests
that KH is a function of the nature of the gas.

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Solutions and Colligative Properties
KH values for both N2 and O2 increase with increase of temperature indicating that the
solubility of gases increases with decrease of temperature. It is due to this reason that
aquatic species are more comfortable in cold waters rather than in warm waters.
T↑ S↓ x↓ KH↑
Scuba divers must cope with high concentrations of dissolved gases while breathing air at
high pressure underwater. Increased pressure increases the solubility of atmospheric gases
in blood. When the divers come towards surface, the pressure gradually decreases. This
releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood.
This blocks capillaries and creates a medical condition known as bends, which are painful
and dangerous to life To avoid bends, as well as, the toxic effects of high concentrations of
nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium
(11.7% helium, 56.2% nitrogen and 32.1% oxygen).
At high altitudes the partial pressure of oxygen is less than that at the ground level. This
leads to low concentrations of oxygen in the blood and tissues of people living at high
altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to
think clearly, symptoms of a condition known as anoxia.
1. Example 2.4 Page no. 40 (NCERT)
2. The Henry’s law constant for oxygen dissolved in water is 4.34 x 10 4atm at 250C. If
partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate
the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with
air at 250C. (CBSE Sample Paper)
Ans: According to Henry’s Law, PO2=KHxO2
𝑃𝑂2 0.2
xO2 = = = 4.6 x 10-6
KH 4.34 𝑥104
Changing mole fraction into molarity
xO2 = nO2 / nO2 + nH2O = nO2 / nH2O
nO2 = xO2 x nH2O = 4.6 x 10-6 x 55.5 = 2.55 x 10-4 these are the moles present in
1000cm3 of water hence this is molarity
3. The Henry’s law constant for the solubility of N2 gas in water at 298K is 1.0 x 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 298K and 5 atm pressure is [IIT-JEE-2009]
a) 4.0 x 10 -4 b) 4.0 x 10 -5 c) 5.0 x 10 -4 d) 4.0 x 10-6
Sol. (A)
P = KH 𝑋𝑁2
0.8 x 5 = 1 x 105 x 𝑋𝑁2
𝑛𝑁2
𝑋𝑁2 = 4 x 10-5 (in 10 moles of water) ⇒ 4 x 10-5 = ⇒ 𝑛𝑁2 = 4 x 10-4
𝑛𝑁2 +10
Henry’s Law : The mass of a gas dissolved per unit volume of the solvent at a
given temperature is proportional to the pressure of the gas in equilibrium with
the solution.
mαP m=Kp
m is the mass of the gas dissolved in a unit volume and p is the pressure of the gas
at equilibrium
4. 2.13 page no. 60 NCERT
The partial pressure of ethane over a solution containing 6.56 x 10 -3 g of ethane is 1
bar. If the solution contains 5.00 x 10-2 g of ethane, then what shall be the partial
pressure of the gas?
Ans: According to Henry’s law, m = Kp
𝑚 6.56 𝑥10−2 (𝑔)
𝐾 = = = 6.56 𝑋10−2 𝑔𝑏𝑎𝑟 −1
𝑃 1(𝑏𝑎𝑟)
𝑀 5.00 𝑋10−2 (𝑔)
In the second case 𝑝 = = = 0.762 bar
𝐾 6.56 𝑋10−2 𝑔(𝑏𝑎𝑟 −1 )

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Solutions and Colligative Properties
E. Relative lowering of vapour pressure
1. The vapor pressure of pure benzene at a certain temperature is 0.842 bar. A non-volatile solute
weighing 2.175 g is added to 39.0 g of benzene. The vapor pressure of the solution is 0.790 bars.
What is the molar mass of the solute?
2. At 250C, the vapor pressure of pure water is 23.76 mm of Hg and that of an aqueous dilute
solution of urea is 22.98 mm of Hg. Calculate the molality of the solution.

3. The vapour pressure of acetone at 200C is 185 torr. When 1.2g of a non-volatile substance was
dissolved in 100g of acetone at 200C, its vapour pressure was 183torr. The molar mass (g mol-1)
of the substance is: [IIT-JEE Offline 2015]
a) 32 b) 64 c) 128 d) 488
4. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004bar at the normal boiling
point of the solvent. What is the molar mass of the solute ( 1atm = 1.01325 bar)

b
Answers Key:
1) 70.4 3) b)
2) 1.8m 4) 40.24g mol-1

F. elevation in boiling point

1. Calculate the molar mass of a substance 1.3 g of which when dissolved in 169 g of water gave a
solution boiling at 100.0250C at a pressure of one atmosphere (Kb for water = 0.52 K m-1).
2. The boiling point of water (1000C) becomes 100.520C if 3g of a non-volatile solute is dissolved in
20ml of it. Calculate the molar mass of the solute (Kb for water = 0.52 Km-1)

Answers Key:
1) 160gms 2) 150gms
G. depression in freezing point
1. A solution containing 25.6 g of sulphur dissolved in 1000 g of naphthalene whose melting point is
80.10C gave the freezing point lowering of 0.6800C. Calculate the formula of sulphur. (K f for
naphthalene = 6.8 K/m, atomic mass of sulphur = 32).

2. When 30.0 g of a non-volatile solute having the empirical formula CH2O are dissolved in 800g of
water, the solution freezes at -1.160C. What is the molecular formula of the solute? (Kf for water
= 1.86 K m-1).

3. In winter, the normal temperature in a Himalayan’s valley was found to be -100C. Is a 30% by
mass of aqueous solution of ethylene glycol (molar mass = 62) suitable for car radiator?
(Kf for water = 1.86 K/m).

4. Addition of 0.643 g of a compound to 50 ml of benzene (density 0.879 g/ml) lowers the freezing
point from 5.510C to 5.030C. If Kf for benzene is 5.12, calculate the molecular weight of the
compound. [I.I.T. 1992]

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Solutions and Colligative Properties

5. Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20g of
benzene (C6H6), 1g of AB2 lowers the freezing point by 2.3K whereas 1.0g of AB4 lowers it by
1.3K. The molar depression constant for benzene is 5.1K kg mol-1. Calculate atomic masses of
A and B.

6. A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing
point of 5% glucose in water if freezing point of pure water is 273.15K.

Answers Key:
1) S8 4) 156.06
2) C2H4O2 5) A = 25.58gm B = 42.64gms
3) -12.86 6) 269.07K

H. Osmotic Pressure
1. Osmotic pressure of a solution containing 7g of a protein per 100 cm3 of solution is
3.3 x 10-2 bar at 370C. Calculate the molar mass of protein.

2. A 5% solution of cane sugar (molar mass = 342) is isotonic with 0.877% solution of urea.
Calculate the molar mass of urea. (assume density of both the solutions is 1 gm/cc)
(Haryana S.B. 1988, Pb. S.B. 1999S)

Answers Key:
1) 54644.54gms 2) 59.99g

I. Abnormal molar mass

1. Decimolar solution of NaCl developed an osmotic pressure of 4.6 atm at 300 K. Calculate the
degree of dissociation. (Punjab S.B. 1994)

2. A solution containing 1.5 g of Ba(NO3)2 in 0.1 kg of water freezes at 272.720 K. Calculate the
apparent degree of dissociation of the salt. (Kf for water = 1.86 K/m and molar mass of
Ba(NO3)2 = 261).

3. Phenol associates in benzene to form a dimer (C6H5OH) 2. The freezing point of a solution
containing 5g of phenol in 250 g of benzene is lowered by 0.700. Calculate the degree of
association of phenol in benzene (KF for benzene = 5.12 K m-1).

4. 2.0 g of benzoic acid dissolved in 25.0 g of benzene show a depression in freezing point equal to
1.62 K. Molal depression constant (KF) of benzene is 4.9 K kg mol-1. What is the percentage
association of the acid? (REE– 90)
5. The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by
0.450C. Calculate the degree of association of acetic acid in benzene (Kf for benzene = 5.12 K
mol-1 kg). (REE 1994)

6. The degree of dissociation (α) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the
expression: [AIEEE-2011]
𝑖−1 𝑥+𝑦−1 𝑥+𝑦+1 𝑖−1
a) α = b) α = c) α = d) α =
𝑥+𝑦+1 𝑖−1 𝑖−1 (𝑥+𝑦−1)

7. A 5.2molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of
methyl alcohol in the solution? [AIEEE 2005]
a) 0.190 b) 0.086 c) 0.050 d) 0.100

8. Calculate the freezing point of a one molar aqueous solution (density 1.04 g ml-1) of KCl.
(Kf for water = 1.86 kg mol-1, atomic masses of K=39, Cl=35.5).

Dr. M Raja Tel: 98 69 69 69 20 www.iitking.com

Solutions and Colligative Properties

9. Calculate the normal freezing point of a sample of sea water containing 3.8% NaCl and 0.12%
MgCl2 by mass. (Kf for water = 1.86 K m-1)

10. 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in the freezing point of water
observed is 1.00C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

11. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic
pressure is 0.75 atm at 270C.

12. Determine the osmotic pressure of a solution prepared by dissolving 25mg of K2SO4 in 2 litre of
water at 250C, assuming that it is completely dissociated.

Answers Key:
1) 87% 2) 81% 7) 8) -3.8520C
3) 72% 4) 99.2% 9) -2.59 C
0
10) i = 1.0753 & Ka = 3.07 x 10-3
5) 94.6% 6) d 11) 0.03mol 12) 5.27 x 10-3 atm.

At 880C VP of pure benzene and toluene are 900 mm Hg and 360 mm Hg respectively. If the solution bolis at
880C. Find the mole fraction of benzenein the solution

Dr. M Raja Tel: 98 69 69 69 20 www.iitking.com