UNIT- 2: SOLUTION

Points to Remember

1. Solution, Solute and Solvent. A homogeneous mixture of two or more chemically non-reacting

substances is known as solution. The substance present in largest amount is called solvent and the

other(s), called solute.

2. Miscible and Immiscible Liquids: Miscible liquids are those, which are completely soluble in each

other in all proportions while immiscible liquids are not soluble in each other.

3. Concentration in per cent.

4. Mole fraction.

Sum of mole-fractions of all components is unity i.e.

5. Solubility. Maximum amount of a solute in a saturated solution per 100 g of a solvent. Like

dissolves like. Ionic and polar substances are generally soluble in polar solvents like water and are

insoluble in non-polar solvents like benzene, chloroform, carbon disulphide etc. Similarly, non-polar

solutes like iodine, sulphur, phosphorus and organic substances are soluble in non-polar solvents like

benzene, chloroform etc. and are insoluble in polar solvents like water.

6. Molarity (M). Number of moles of solute per litre of solution.

7. Molality (m). Number of moles of solute per kg of solvent.

8. Equivalent weight

Mass of solute% by mass = ×100

Mass of solution

Volume of solute% by volume 100

Volume of solution

% by mass-volume = Mass of solute g

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑚𝑙 × 100

M = w ×1000

m× V 𝑚𝑙 g-mol L-1

9. Normality. Number of gram-equivalents of solute per litre of solution.

10. Parts per million (ppm). This unit is convenient unit to express the concentration of solute when

it is present in very small amounts.

11. Relation between strength, molarity and normality of a solution.

12. Molarity of a mixture solution (M) is given by:

13. Henry’s law: It states that the solubility of a gas in a liquid is directly proportional to the

pressure of the gas at a given temperature. If X is the mole fraction of gas in solution, then:

where p is partial pressure of gas and is Henry‟s constant.

14. Vapour pressure. In case of pure liquid, it is the pressure exerted by vapour of liquid over surface

of liquid when vapour and liquid are in equilibrium at a given temperature. It increases with

temperature. In case of a solution of two volatile liquids, each liquid vapour exerts a partial pressure,

and total pressure is equal to sum of partial pressures. Partial pressure of each component is given by

Raoult‟s law.

15. Raoult’s law. For non-volatile solute, relative lowering of vapour pressure,

16. Colligative properties: Those properties of solution, which depend upon the concentration of

solute particles but are independent of nature of solute. These are: relative lowering of vapour

pressure, elevation in boiling point, depression in freezing point and osmotic pressure of solution.

17. Boiling point: The temperature at which vapour pressure of liquid or solution becomes equal to

atmospheric pressure.

18. Molal elevation constant (Kb). The elevation in boiling point of the solution when its molality is

unity. It is also called molal ebullioscopic constant and is represented by Kb. It‟s unit are km–1

or kg

mol–1

.

19. Elevation in Boiling point Boiling point of a solution is always higher than that of a pure

solvent.

20. Freezing point . The temperature at which solid, liquid and vapours are in equilibrium i.e. they

have same vapour pressure.

21. Molal Depression Constant (Kf): It is equal to depression in freezing point of a solution whose

molality is unity i.e. 1 mole of solute is dissolved in one kg of solvent. It is represented by Kf.

.

22. Semi permeable membrane. A membrane which allows a selective passage of solvent molecules

is called a semi-permeable membrane. Cellophane, parchment paper and the walls of living cells are

examples of semi-permeable membrane.

. Osmosis. The passage of solvent molecules from pure solvent (or solution of low concentration) to

the solution of higher concentration through the semi-permeable membrane.

24. Osmotic pressure ( ). The excess pressure that must be applied to the solution side to just

prevent the passage of solvent into the solution through semi-permeable membrane. It is a colligative

property of solution and is given by relation:

where M is molarity of solution.

25. Isotonic solutions. The solutions which have same osmotic pressure.

26. van‟t Hoff factor.

In case of abnormal solutions (solutions showing association or dissociation of solute particles), the

colligative properties are expressed by following relations:

The value of i is equal to 1 in case of normal solution, i > 1 in case of dissociation, and

i < 1 in case of association.

27. The units of are degrees/molality i.e. K/m or K kg mol–1

.

28. Molality, mole-fraction, mass fraction etc. are preferred over molarity, normality, etc. because the

former units involve only masses which do not change with temperature while the latter involve

volumes which change with temperature.

29. A 0.91% NaCl solution is isotonic with red blood corpuscles (RBC). NaCl solution with

concentration greater than 0.91% is called ‘hypertonic’ and less than this is called ‘hypotonic’.

30. „Reverse Osmosis‟ is the process of flow of solvent from solution to solvent side when a pressure

higher than osmotic pressure is applied to solution separated from solvent by a semi permeable

membrane.

31. For comparing the values of colligative properties of different solutions, express all given

concentrations into mol L–1

of ions (in case of electrolytes). Greater the molar concentration of

particles, greater is the value of colligative property.

Very Short Answer Type Questions (1mark)

1. What are solutions?

Ans. Homogeneous mixture containing two or more than two components.

2.What are binary solns?

Ans. Binary solution : Solutions containing only two components.

3. Define mole fraction

Ans.. Ratio of number of moles of a component to the total number of moles of solutions.

4 . Define molarity

Ans. Number of moles of solute dissolved in one litre of solution.

5. Define molality

Ans.Number of moles of the solute dissolved in 1 kg of the solvent.

6. State Henry’s law

Ans.The partial pressure of the gas in vapour phase is directly proportional to the molefraction of the

gas in the solution.

P = KH x

7. Mention one application of Henry’s law

Ans.To increase the solubility of CO2 in soft drinks , the bottle is sealed under high pressure.

8. State Raoutlt’s Law

Ans.For a solution of volatile liquids, the partial vapour pressure of each component in the solution is

directly proportional to its mole fraction.

P1 = p01 x1

9. What are Ideal solutions?

Ans.Solutions which obey Raoult‟s law . solute –solvent interaction is nearly equal to solute –solute or

solvent – solvent interaction. Δmix H = 0 and Δmix V = 0

Eg: n- hexane and n-heptane

10. What are non-ideal solutions?

Ans. Solutions which do not obey Raoult‟s law . Inter molecular force of attraction between solute –

solvent is Weaker or stronger than those between solute –solute or solvent – solvent interaction.

Δmix H ≠ 0 Δmix V ≠ 0

11. Define colligative property

Ans.Properties which depend on the number of solute particle irrespective of their nature relative to

the total number of particle present in the solution.

12. Define Ebullioscopic constant

Ans. Elevation of boiling point of one molal solution.

13. Define Cryoscopic constant

Ans. Depression of freezing point of one molal solution.

14. Define Osmosis

Ans. Flow of solvent molecule through semi permeable membrane from pure solvent to the solution.

This process of flow of the solvent is called osmosis.

15. Define Osmotic Pressure

Ans. Excess pressure that must be applied to a solution to prevent osmosis.

16. What is reverse osmosis

Ans. Pure solvent flows out of the solution through the semi permeable membrane when we apply

pressure larger than the osmotic pressure. Application of reverse osmosis is desalination of seawater.

17. What are isotonic solution

Ans.Two solutions having same osmotic pressure at a given temperature.

18. Define van’t Hoff factor

Ans. i = Normal molar mass / abnormal molar mass

i= observed colligative property / calculated colligative property.

Short Answer Type Questions (2mark)

1.18 g of glucose, C6H12O6(Molar Mass = 180 g mol-1

) is dissolved in 1 kg of water in a sauce pan.

Atwhat temperature will this solution boil? (Kbfor water = 0.52 K kg mol–1

, boiling point of pure

water = 373.15 K)

Ans.

2. Explain the following:

(i)Henry’s law about dissolution of a gas in a liquid.

(ii)Boiling point elevation constant for a solvent

Ans. (i)Henry‟s lawstates that the solubility of a gas in a liquid is directly proportional to the pressure

of the gas.

(ii) The boiling point elevation constant may be defined as the elevation in boiling point when

one mole of a non-volatile solute is dissolved in one thousand grams of solvent

3. A solution of glucose (C6H12O6) in water is labeled as 10% by weight. What would be the

Molality of the solution? (Molar mass of glucose = 180 g mol–1)

Ans.

4. Differentiate between molarity and molality for a solution. How does a change in temperature

influence their values?

Ans.

5. Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a

Colligative property? Explain.

Ans.

6. Explain why-

a) Aquatic animals are more comfortable in cold water than in warm water.

b) Deep sea divers are advised not to come to surface immediately from deep waters.

(Ans) (a) This is because “Kh” (Henry constant) values for both N2 and O2 increase with increase in

temperature indicating that the solubility of gases increases with decrease in temperature.

(b) Deep sea divers are advised not to come to surface immediately from deep waters because

sudden change in outside pressure can be fatal for divers because N2 will bubble out of blood vessels

causing severe pain and can be dangerous.

7. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous

solution. What should be the molarity of such a sample of the acid if the density of solution is

1.504 g mL–1

?

Sol. Percentage mass (%) and molarity (M) of the solution are related as

Molarity (M) = )(HNO solute of mass Molar

10density mass %

3

= )mol (g 68

10)mL (504.1681-

1 g = 16.23 M

8. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3,

supposed to be carcinogen. The level of contamination was 15 ppm (by mass)

(i) Express this in percent by mass.

(ii) Determine the molality of chloroform in the water sample.

Sol. Mass of solution (wA + wB) = 106 g

Mass of solute (wB) = 15 g

% mass = 6

BA

B

10

10015

ww

100w

= 1.5 × 10

–3%

Mass of solvent (wA) = 106 – 15 = 999985 g

Molality (m) = g 999985)mol (g 5.119

(g) 10)g(15

wMM

10w

1-

3

AB

3B

= 1.25 × 10–4

m

9. Non-ideal solutions exhibit either positive or negative deviations from Raoult's law. What

are these deviations and why are they caused? Explain with one example for each type.

Ans.

10.State Raoult’s law for a solution containing volatile components. How does Raoult’s

law become a special case of Henry’s law?

Ans.

Short Answer Type Questions (3 Mark)

1. 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution

freezes at – 0.34°C. What is the molar mass of the material?

(Kf for water = 1.86 K kg mol-1

)

Ans.

WB= 15.0g;WA= 450g; ΔKf= 1.86 K kg molL-1

ΔTf = 0°C– (– 0.34°C) = 0.34°C or 0.34 K

2.Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in 2

L of water at 25 oC, assuming that it is completely dissociated.

(R= 0.082 L atm K–1 mol–1, Molar mass of K2SO4 = 174 g mol–1).

Ans.

3) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of

benzene by 0.40 K. Find the molar mass of the solute. (Kf for benzene = 5.12 kg mol–1

)

ans.

4. Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.00 g of

water. (Kb for water = 0.512 kg mol–1),(Molar mass of NaCl = 58.44 g)

ans.

5.A solution prepared by dissolving 1.25 g of oil of wintergreen (methyl salicylate) in 99.0 g of

benzene has a boiling point of 80.31°C . Determine the molar mass of this compound.

(B.P.of pure Benzene =80.10°C and Kb for benzene = 2.53°C kg mol–1)

Ans.

6. What mass of NaCl (molar mass= 58.5 g mol–1

) must be dissolved in 65 g of water to lower the

freezing point by 7.5°C ? The freezing point depression constant, Kf , for water is 1.86 K kg mol–

1.Assume van’t Hoff factor for NaCl is 1.87.

Ans.

7. What concentration of nitrogen should be present in a glass of water at room

Temperature ? Assume a temperature of 25° C, total pressure of 1 atmosphere and mole

fraction of nitrogen in air of 0.78. [KH for nitrogen = 8.42 × 10–7

M/mm Hg]

Ans.

8. Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2 in 200 g of

water. (Molar mass of MgBr2= 184 g/mol)(Kf for water = 1.86 K kg /mol )

Ans.

9. Vapour pressure of water at 298 K is 17.535 mm Hg. Calculate vapour pressure of water

at 293 K when 25 g of glucose is dissolved in 450 g of water.

Ans. p 0A

= 17.535 mm Hg. Let vapour pressure of solution be ps

B0A

s0A xp

pp

or 1 -

0A

s

p

p= xB or ps = )x1(p B

0A

or ps = 0A

p × xA = 17.535 × 18/450180/25

18/450

= 17.535 × 0.9945 = 17.438 mm Hg

17.44 mm Hg.

10.(a) What happens to blood cells when they are placed in water containing less than 0.99%

(mass/volume) salt?

(b) What care should be taken while preparing intravenous injection?

(c) Why osmotic pressure is better technique for determination of molar mass of

biomolecules?

Ans.

(a) The blood cells collapse due to loss of water by osmosis when placed in water containing less

than 0.9% (mass/volume) salt.

(b) They should be prepared in aquatic medium and salt concentrations should be same as blood

plasma levels.

(c) Osmotic pressure is better technique for the determination of molar mass of biomolecules because

they are generally not stable at higher temperatures and have poor solubility. Therefore, other

techniques like elevation in boiling point and depression in freezing point cannot be used to determine

molecular masses of biomecules.

HOTS QUESTIONS

Q.1)Two elements A and B, form compounds having molecular formula AB2 and AB4. When

dissolved in 20 g of C6 H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB4

lowers it by 1.3 K. The molal depression constant for benzenes is 5.1 K kg mol–1

. Calculate

atomic mass of A and B.

Sol. We know, MB = fA

Bf

Tw

wK1000

3.220

11.51000M

2AB

= 110.87 g mol

–1

3.120

11.51000M

4AB

= 196.15 g mol

–1

If a and b are respectively atomic masses of A and B then,

2ABM = a + 2b or a + 2b = 110.87

4ABM = a + 4b or a + 4b = 196.15

Solving the simultaneous equations we have

a = 25.58 u; b = 42.64 u.

Q.2) If density of some lake water is 1.25 g mL–1

and it contains 92.0 g of Na+ ions per kg of

water. Calculate the molality and molarity of Na+ ions in lake.

Sol. Molality = (kg) w aterof Mass

ions Na of .Mol

= (kg) 1

)mol g(23/)g(92 -1

= 4 m

Molarity = (L) w aterof .Vol

ionsNa of .Mol

= )(gL 10 1.25 / (g) 1000

)mol (g 23 / )g(92

1-3

-1

= (g) 1000)mol (g 23

)L (g 101.25 )g(92

1-

-13

= 5M

Q.3) 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point

observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of the acid.

Kf for H2O = 1.86 K kg mol–1

and density of solution is 1.124 g mL–1

.

Sol. Mass of solution = 500 + 19.5 = 519.5 g

Volume of solution = )mL (g 124.1

(g) 5.519

1- = 462.18 mL

Molarity of solution = 18.46278

10005.19

= 0.541 mo L

–1

Tf = i × Kf × m

or i =

10005.1986.1

500781

= 1.0753

diss. = 12

10753.1

1m

1i

= 0.0753

Also diss. = c/Ka or Ka = 2

diss. × c

= (0.0753)2 0.541 = 3.07 × 10

–3

.4) The air is a mixture of a number of gases. The major components are oxygen and nitrogen

with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium

with air at a pressure of 10 atm. At 298 K, if the Henry’s law constant for oxygen and nitrogen

are 3.30 x 107 mm and 6.51 x 10

7 mm respectively, calculate the composition of these gases in

water.

Sol. Consider 1 mol of air and its volume at 10 atm be V

Volume of O2 = 100

20V = 0.2 V; Volume of N2 =

100

79V = 0.79 V

Partial pressure of O2 )p(2O =

V

V2.010= 2 atm.

Partial pressure of N2 )p(2N =

V

V 79.0.10 = 7.9 atm .

Solubility of O2 2Ox =

(mm) 1030.3

(mm) 7602p

7HK/2O

= 4.606 × 10

–6 .

Solubility of N2 2Nx =

(mm) 1051.6

(mm) 7602p

7HK/2N

= 9.22 × 10

–5

Q.5) How many mL of a 0.1 M HCl are required to react completely with 1 g mixture of

Na2CO3 and NaHCO3 containing equimolar amounts of two?

Sol. Let Na2CO3 be x g

Mol of Na2CO3 (n1) = 106

xmol

NaHCO3 = (1- x) g

Mol of NaHCO3 (n2) = 84

)x1( mol

OHCO NaCl 2HCl CONa 22mol 2mole 1

32

OHCO NaCl HCl NaHCO 22mol 1mole 1

3

HCl required for 106

xmol of Na2CO3 =

106

x2 mol

HCl required for 84

)x1( mol of NaHCO3 =

84

)x1( mol

Since given mixture contains equimolar amount

84

)x1(

106

x or x = 0.56

Total HCl required = 84

56.01

106

56.02

84

x1

106

x2

= 0.01578 mol

If V is the volume of HCl required, then

V(L) × 0.1 = 0.01578

or VL = 0.1578 L or 157.8 mL

VALUE BASED QUESTIONS

1. Bharath went to his grandfather’s house in winter this year. As usual he went for fishing.

His grandmother told him there will be no fishes in the lake. He noticed that it was more

difficult to find fishes in winter. The fishes were deep inside the river. Whereas in summer

they were on the surface and hence he was able to catch fishes.

a) Why are fishes on the surface in water than in the depth in summer?

b) What value can be derived from this?

Ans: a) According to Henry‟s law at low temperature gases are more soluble and hence as

more oxygen gets dissolved in water fishes survive better even in depth of the river.

In summer as the oxygen is less in water the fishes come to the surface.

b) The value that I derive from this is wisdom is superior to knowledge

2. Scuba divers when come towards the surface, the pressure gradually decreases resulting

in the release of dissolved gases leading to formation of bubbles of nitrogen gas in the

blood which blocks the capillaries and thus harmful effects are created. Toavoid bends

and toxic effects of high concentration of nitrogen gas, the air is diluted with helium.

After reading the above passage , answer the following questions.

(i) Which law is associated with this?

(ii) Why is the condition of bends overcome by the use of Helium

(iii) Mention the value associated with providing divers air diluted with helium.

Ans: (i) Henry‟s Law

(ii) Solubility inversely proportional to KH.

(iii )Critical thinking and self awareness