Class XII CHEMISTRY

Solutions CHEMISTRY
Chapter – 2: Solutions

Key features of Drill Exercises:

• It is a document that focuses on extensive and exhaustive practice of

Questions as per latest CBSE Board Curriculum.

• Drill Exercises are Class wise/ Subject wise/ Chapter wise format.

• It contains four sections, namely:

(A) Practice Questions: Questions covering all the categories along

with their difficulty level; viz. Very Short Answer (V.S.A.) Type,
Short Answer (S.A.) Type, Long Answer (L.A.) Type.

(B) Previous Year Questions: Questions asked in the previous years

of CBSE Examination on this chapter.

(C) Must Do Questions: Questions from topics/ sub-topics regularly

asked in the Board Examination.

(D) HOTS (Higher Order Thinking Skills): Questions to assess the

students’ understanding, analytical ability and interpretation.

NOTE: Detailed answers to all the questions are also provided for the
student’s self assessment.

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DRILL EXERCISES

A. Practice Questions

The questions in this section indicate weightage and their difficulty level.

Answers will be provided at the end of each section given below:

1 Mark: Very Short Answer Type Difficulty level: Easy and Average

Q1. Give two examples of a solid solution in which solute is solid.

Q2. Why does the molarity of a solution changed with temperature?

Q3. Why does the molality of a solution remain unaffected with temperature?

Q4. Which concentration terms remain unaffected by temperature?

Q5. 12% by mass of a solution means 12 g of the solute in …….g of solvent.

Q6. Define supersaturated solution.

Q7. What is sum of the mole fraction of all components in a two component system?

Q8. What is the difference between molarity and molality?

Q9. What is the molarity of pure water?

Q10. Why are soft drinks and soda water bottles sealed under pressure?

Q11. Name the solid that separates first when saline water is slowly frozen.

Q12. Why is oxygen diluted with helium as a breathing air for deep sea diving?

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Q13. State Henry’s law and mentions one if its application.

Q14. What temperature change is expected during the mixing of two liquids whose
solution shows a positive deviation from Raoult’s law?

Q15. Why liquid ammonia bottle is first cooled on the ice before opening it?

Q16. What is the enthalpy and volume change, when an ideal solution is formed?

Q17. What type of azeotropes is formed on the mixing of ethanol and water?

Q18. What type of behavior is expected on adding water to nitric acid?

Q19. What is the condition applied in reverse osmosis?

Q20. What is the Van’t Hoff factor of the compound K 4 Fe(CN) 6 ?

Q21. Examine the following diagram and give the name of process involved.

Q22. Why is an increase in temperature observed on mixing methanol with acetic

acid?

Q23. Calculate the molarity of solution containing 0.2 moles of NaCl in 250 ml solution.

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Q24. Two liquids A and B boil at 120o C and 150oC respectively at atmospheric
pressure. Which of them is expected to have higher pressure at 50o C?

Q25. What is the Van’t Hoff factor for a compound which undergoes dimerization in
organic solvents?

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Answers:-

A1. Sodium amalgam, copper dissolved in gold.

A2. Molarity of a solution decreases with increase in temperature because volume of

solution increases with increase in temperature.

A3. This is because mass of solvent does not affected by the temperature.

A4. Molality, Mole-fraction

A5. 88

A6. A supersaturated solution contains more of the solute is present in the saturated
solution. It is unstable.

A7. One

A8. Molarity is defined as number of moles of solute dissolved per litre of solution.

Molality is defined as number of moles solute dissolved per kilogram of solvent.

A9. 55.55M

A10. To increase the solubility of CO 2 .

A11. Ice

A12. To minimize the painful and dangerous effects accompanying decompression at

greater depth.

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A13. “The partial pressure of the gas in vapour phase ( ) is proportional to the mole

fraction of the gas( ) in the solution”

= ,

Here, is the Henry’s constant

Its application is

• To increase the solubility of CO 2 in the soda water, the bottle sealed under
high pressure.

A14. The temperature will decrease.

A15. The vapour pressure of liquid NH 3 is very high at room temperature. On cooling,
the vapour pressure of liquid NH 3 will decrease and hence, the liquid NH 3 will not
splash out.

A16. = 0, =0

A17. Minimum boiling azeotrope.

A18. < 0, < 0 i.e. negative,

A-B attraction > A-A and B-B solutions i.e. negative deviation from Raoult’s law.

A19. Pressure applied on the solution is more than the osmotic pressure.

A20. Five

A21. Reverse osmosis because the movement of solvent particle is higher

concentration to lower concentration on the application of pressure.

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A22. Due to the presence of H- bonding in between methanol and acetic acid, the
force of attraction increases, therefore energy is released.

A23. Molarity = = 0.8M

A24. B liquid contains higher boiling point, i.e. it having higher attraction among its
molecules posses less tendency for evaporation and thus has low vapour
pressure. Hence, A has higher vapour pressure than B at 50oC.

A25. = 1/2

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1 Mark: Very Short Answer Type Difficulty level: Difficult

Q1. Why do the gases always tend to be less soluble in liquids as rise in
temperature?

Q2. Calculate the molarity of a solution containing 4g of NaOH in 500ml solution.

Q3. Calculate the molarity of 0.006 moles of NaCl in 100 ml of solution.

Q4. What is the mole fraction of the solute in one molal aqueous solution?

Q5. What is the molar concentration of hydrogen, if 20 g hydrogen is present in 5 litre

vessel?

Q6. How many numbers of moles will be present in 2 litre of 0.5M NaOH?

Q7. For which of the following can Van’t Hoff factor not be greater than unity:

(a) NaCl (b) CaCl 2 (c) glucose (d) lunar caustic

Q8. Calculate the molality of 6 g of ethanoic acid in 80 g of benzene.

Q9. Sodium chloride is used to clear snow from the roads. Why?

Q10. Why does doctor advice gargles by saline water in case of sore throat?

Q11. State the underlying principle of purification of water by reverse osmosis.

Q12. Calculate the mass percentage of benzene and carbon tetra chloride if 18g of
benzene is dissolved in 120g of carbon tetra chloride.

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Answers:-

A1. With rise in temperature, the molecular kinetic energy of the molecules
increases. Due to the higher kinetic energy, molecules tend to break the bonds
with the liquids solvents and thus escape from the surface to pass into the
vapour phase. Thus, solubility of gases decreases in liquids as rise in
temperature.

A2. Number of moles of NaOH = 4/40 = 0.1 moles

Molarity = 0.1 × 1000/500 = 0.2 M

A3. Molarity of NaCl = = 0.06 M

A4. No. of moles of water = 1000/18 = 55.55

Mole fraction of solute = = 0.018

A5. Molar concentration of hydrogen = = =2

A6. No. of moles = molarity × volume in litre = 0.5 × 2 = 1.

A7. Glucose will have Van’t Hoff factor equal to one i.e. in unity.

A8. Molar mass of ethanoic acid (CH 3 COOH) = 60

Moles of ethanoic acid = 6/60 = 0.1 mol

Molality of ethanoic acid = = 1.25 m

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A9. Sodium chloride depresses the freezing point of water to such an extent that it
can not freeze to form ice and hence, it melts off easily at the prevailing
temperature.

A10. Saline water is a hypertonic solution of salt and water; therefore, fluids causing
irritation in throat will come out.

A11. When pressure which is greater than osmotic pressure, is applied on the saline
water side of a semi-permeable membrane, water molecules move through the
membrane towards pure water side.

A12. Mass% of benzene = × 100 = 13.044%

Mass% of carbon tetra chloride = 100 – 13.044 = 86.956%

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2 Marks: Short Answer Type Difficulty level: Easy and Average

Q1. Define the vapour pressure of liquid. What happens to the vapour pressure
when:
(a) Dissolved solute is volatile.
(b) Dissolved solute is nonvolatile.

Q2. Distinguish between molarity and molality of a solution. When and why molality is
preferred over molarity handling solutions in experimental chemistry?

Q3. Define mole fraction. Calculate the mole fraction of water, if 95 mass percent
aqueous solution of ethanol further diluted with water, the mole fraction of
ethanol in diluted solution is 0.25. Is it an ideal solution?

Q4. Calculate molarity and molality of 18% solution (by weight) of sulphuric acid. Its
density is 1.020g cm–3. (At. Masses H = 1, O = 16, S = 32 amu)

Q5. Illustrate the depression of freezing point with the help of vapour pressure –
temperature curve of a solution. Show that depression of freezing point is a
Colligative property.

Q6. Explain why boiling point of solvent is increased on dissolving a non volatile
solute into it with the help of curve?

Q7. Explain why when fruits and vegetables are dried and placed in water, they
slowly swell and return to original form? Does a temperature increase accelerate
the process?

Q8. What will be molar concentration of solute in a human blood, if osmotic pressure
is 7.6 atm at body temperature of 37° C?

Q9. Calculate the mass of glucose required to making 2.5 kg of 0.2 molal aqueous
solutions.
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Answers: -

A1. Vapour pressure of liquid is the pressure exerted by the liquid vapours on the
liquid surface.
(a) When a volatile solute is dissolved in a liquid, the vapour pressure of solute
and solvent become less, if they form ideal solution.
(b) When a non volatile solute is dissolved in a liquid, the vapour pressure of
the solution becomes lowered.

A2. Molarity is defined as number of moles of solute dissolved per litre of solution.
Molality is defined as number of moles of solute dissolved per kilogram of
solution.
Molarity is a function of temperature and changes with temperature while molality
of solution does not change with temperature. Therefore, molality is used in
experiments involving the elevation of melting point and depression in freezing
point.

A3. Mole fraction of solute is defined as the ration of number of moles of solute to the
total number of moles of solution (solute + solvent).
Mole fraction of water = 1 – 0.25 = 0.75

It is not an ideal solution because force of attraction between ethanol and water
solution is less than that of ethanol-ethanol and water-water molecules.

A4. Molarity =

Molality = 2.24m

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A5. When a non volatile solute is added to a pure solvent, its vapour pressure is
lowered and at lower temperature, the vapour pressure of solid and liquid solvent
will be equal, i.e. depression of freezing point take place.

= —

Since depends upon molality of solute and not on nature of solute, therefore
is Colligative property.

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A6. When a non volatile solute is added to a pure solvent, its vapour pressure is
lowered and at higher temperature, the vapour pressure of solution will be equal
to atmospheric pressure. Hence, the boiling point of a solution of non volatile
solute is higher than that of pure solvent.

A7. When fruits and vegetables are dried and placed in water, osmosis is occur, i.e.
water molecules enters through semipermeable membrane present in cell wall,
therefore they swell.
If temperature is increased, osmosis will faster because osmosis is directly
proportional to the concentration as well as temperature.

A8. ЛV = nRT Л= RT Л = CRT

7.6 = C × 0.0821 × 310K C = 0.3108 mol/litre

A9. 2.5 kg of 1 molal solution glucose = 2.5 mol

2.5 kg of 0.2 molal solution contain glucose = 2.5 × 0.2 = 0.5 mol

Molecular mass of glucose = 180 g mol–1

Mass of glucose required = 0.5 × 180 = 90 g

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2 Marks: Short Answer Type Difficulty level: Difficult

Q1. Calculate the volume of 75 % (by mass) of H 2 SO 4 (density = 1.80g mol–1)

required to prepare 1 litre of 0.2 molar H 2 SO 4 .
(At. Masses: H = 1, O = 16, S = 32)

Q2. The vapour pressure of the solution is found to be 192.5 mm Hg at room

temperature. When 2 g of non volatile solute is dissolve in 100 g of acetone
(molar mass = 58) at same temperature. Calculate the molar mass of solute.
(The vapour pressure of pure acetone is 195 mm Hg at room temperature)

Q3. If 2.5 g urea is dissolved in 150 ml of water, then what will be the value of
freezing point for this solution?( for water = 1.86 K kg mol–1, molar mass of

urea = 60)

Q4. Calculate the osmotic pressure of a solution obtained from the mixing of 200 cm3
of 0.25 M solution of urea and 100cm3 of 0.2 M of glucose at 27°C.
(R = 0.082 L atm mol–1 K–1)

Q5. Arrange the following aqueous solutions (each of strength 0.2 M), in order of
increasing boiling points and freezing points. Justify your answer.
CH 3 OH, NaCl, Ca 3 (PO 4 ) 2 , Na 2 SO 4

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Answers:-

A1. 75% means 75 g H 2 SO 4 in 100g of solution.

Volume = mass/density = = 55.5 cm3

Molar mass of H 2 SO 4 = 98

Molarity = = 12.63

M1V1 = M2V2

12.63 × V 1 = 0.2 × 1000 V 1 = 15.83 ml

A2. =

= 90.48

A3. = = 0.5167

Depression in freezing point = 0.5167°C

Freezing point of solution = – 0.5167°C

A4. No. of moles of urea = 0.25 × 200/1000 = 0.05 mol

No. of moles of glucose = 0.2 × 100/1000 = 0.01 mol
Total volume = 200 + 100 = 300cm3 = 0.3 L
Osmotic pressure of solution, Л = Л 1 + Л 2

= RT = × 0.0821 × 300 = 4.926 atm

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A5. For CH 3 OH ( = 1), NaCl ( = 2), Ca 3 (PO 4 ) 2 ( = 5), Na 2 SO 4 ( = 3)

• Greater the value of i, greater will be elevation of boiling point i.e. greater will
be boiling point.

Thus, increasing order of boiling point is:

CH 3 OH < NaCl < Na 2 SO 4 < Ca 3 (PO4) 2

• Greater the value of i, greater will be depression of freezing point i.e., lesser
will be freezing point.

Thus, increasing order of freezing point is:

Ca 3 (PO4) 2 < Na 2 SO 4 < NaCl < CH 3 OH

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3 Marks: Short Answer Type II Difficulty level: Easy and Average

Q1. What are non ideal solutions? Explain as how to they deviate from Raoult’s’ law,
with suitable diagram.

Q2. The Henry law constant for dissolved oxygen in water is 4.30 × 104 atm at room
temperature. If the partial pressure of oxygen is 0.2 atm in air. Calculate the
concentration (in moles per litre) of dissolved oxygen in water in equilibrium with
air, under ordinary atmospheric condition.

Q3. An aqueous solution freezes at 271.8 K, while pure water freezes at 273K.

(Given = 1.86 K kg mol–1, = 0.512 K kg mol–1 and vapour pressure of

pure water = 23.757 mmHg)

Determine:
(a) The molality of solution
(b) Boiling point of solution
(c) Lowering of vapour pressure of pre water at 298 K.

Q4. A decimolar solution of Fe 2 (SO 4 ) 3 is 50% dissociated at 310K. Calculate the

osmotic pressure of solution. (R = 0.0821litre atm K–1mol–1)

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Answers: -

A1. Solutions which do not obey Raoult’s law over the entire range of concentration
are called non-ideal solutions.
They have shown positive and negative deviation from Raoult’s law.

Positive deviation from Raoult’s law

Non ideal solutions showing positive deviation have force of attraction less than
the pure components, therefore, their vapour pressure is higher and boiling point
is lowered, that is why they form minimum boiling azeotropes.

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Negative deviation from Raoult’s law

Non ideal solutions showing negative deviation have force of attraction greater
than the pure components, therefore, their vapour pressure is lower and boiling
point is higher, that is why they form maximum boiling azeotropes.

A2. By Henry’s law,

Mole fraction of oxygen = = = = 4.6 × 10–6

1 litre water contains 55.5 mol

If = No. of moles of O 2 in the solution

= ≈ 4.6 × 10–6 =

= 4.6 × 10–6 × 55.5 = 2.6 × 10–6 moles

i.e. 2.6 × 10–6 moles of O 2 are dissolved in 1 litre of water.
Hence, concentration of dissolved oxygen in water = 2.6 × 10–6 moles per litre.

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A3. (a) =273 – 271.8

Molality, = 0.645

(b) Elevation of boiling point, = × = 0.512 × 0.645 = 0.33K

Boiling point, = 373 + 0.33 = 373.33 K

(c) = = = =

1– = =1–

= 23.484 mmHg

A4. Decimolar means, C = 1/10 M

Fe 2 (SO 4 ) 3 2Fe3+ + 3SO 4 2–

n = 5,

α= 0.5 = =3

Osmotic pressure of solution = = 3 × 1/10 × 0.0821 × 310 = 7.6353 atm

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3 Marks: Short Answer Type Difficulty level: Difficult

Q1. 2.5 g of benzoic acid dissolved in 30 g of benzene shows a depression in

freezing point is 1.68K. The molal depression constant is 4.9 K kg mol–1. What
will be the percentage association of acid if it exits as dimer in the solution?

Q2. Two elements X and Y form two compounds which are having molecular
formulae XY 2 and XY 4 . When these dissolved in 20 g of benzene, 1 g of XY 2
lowers the freezing point by 2.4 K whereas 1 g of XY 4 lowers it by 1.3 K. The
molar depression constant for benzene is 5.1 K kg mol–1. What will be the atomic
masses of elements X and Y?

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Answers:-

A1. = = = 243

Calculated molecular mass of benzoic acid = 122

Van’t Hoff factor, = = = 0.502

Let α, be degree of association,

2C 6 H 5 COOH (C 6 H 5 COOH) 2
1 0 (initial)

1–α α/2 (after association)

Total no. of moles after association = 1 – α + α/2 = 1 – α/2

= = 0.502

α = (1– 0.502) × 2 = 0.498 × 2 = 0.996 = 99.6%

A2. Let a and b the atomic masses of X and Y

Molecular mass of XY 2 = a + 2b,

Molecular mass of XY 4 = a + 4b

If =

For XY 2 , a + 2b = = 110.87

For XY 4 , a + 4b = = 196.15

2b = 196.15 – 110.87 = 85.28

b = 42.64

a = 110.87 – 2 × 42.64 = 25.89

Hence, atomic mass of X = 25.89, atomic mass of Y = 42.64

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5 Marks: Long Answer Type

Q1. (a) Why is freezing point depression of 0.1M MgCl 2 solution nearly thrice that of
0.1M urea solution?
(b) Suggest the value of Van’t Hoff factor for MgCl 2 .
(c) The freezing point of pure nitrobenzene is 278.8 K. When 2 g of an
unknown nonvolatile substance is dissolved in 100 g of nitrobenzene, the
freezing point of solution is found to be 276.4 K. The freezing point
depression constant of nitrobenzene is 8.0K kg mol–1. Calculate the molar
mass of unknown substance.

Q2. (a) Explain why benzene and water are immiscible while ethyl alcohol and
water are miscible in all proportions?
(b) What is the value of for MgSO 4 .7H 2 O?
(c) Give the relations between
(i) Cryoscopic constant and enthalpy of fusion.
(ii) Ebullioscopic constant and enthalpy of vaporization.
(d) Calculate the osmotic pressure in Pascal exerted by a solution which is
prepared by dissolving 1 g of macro molecule of molar mass 195,000 in 500
ml of water at 310 K.

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Answers:

A1. (a) MgCl 2 dissociates into one Mg2+ and two Cl– i.e. = 3, while urea does not

dissociate, for glucose, i.e. = 1. Thus, freezing point depression will be

triple in MgCl 2 than in urea solution.

(b) =3

(c) = 278.8 – 276.4 = 2.4K

= 2.4 =

Hence, = 66.67 g mol–1

A2. (a) It is the based on the concept “like dissolve like “ i.e. polar solutes dissolve
in polar solvents and non polar solutes dissolve in non polar solvents. Water
and ethyl alcohol are miscible both are polar and show intermolecular
H-bonding. On other hand, benzene is non polar and it can not form
H-bonding with water and hence, benzene is immiscible.

(b) =2

(c) (i) =

Here, R = gaseous constant, = freezing point, = molar mass of solvent.

(ii) =

Here, = boiling point

(d) Л= × = = 26.422 Pascal

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B. Previous Year Questions

2 Marks – Short Answer Type I

Q1. A 0.1539 molal aqueous solution of cane sugar (mol. Mass = 342 gmol–1) has a
freezing point 271 K while the freezing point of pure water is 273.15 K. what will
be the freezing point of an aqueous solution containing 5 g of glucose
(mol. Mass = 180 gmol-1) per 100g of solution. [CBSE 2007]

Q2. State Henry’s Law correlating the pressure of a gas and its solubility in a solvent
and mentions two applications for the law. [CBSE 2008]

Q3. Calculate the temperature at which a solution containing 54 g of glucose,

(C 6 H 12 O 6 ) in 250 g of water will freeze. ( )
[CBSE 2008]

Q4. Define the term osmotic pressure. Describe how the molecular mass of a
substance can be determined by a method based on measurement of osmotic
pressure? [CBSE 2008]

Q5. 100mg of protein is dissolved in just enough water to make 10 ml of solution. If

this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar
mass of protein? (R = 0.0821L atm mol–1K–1 and 760 mm Hg = 1 atm)
[CBSE 2009]

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Answers

A1. Mass of solute ( ) = 5g, molar mass of ( ) = 180g, mass of solution = 100g,

Mass of solvent ( ) = 95g

=273.15 – 271 = 2.15K

= ×

2.51 = × 0.1539

= 2.15/0.1539 = 13.97

= = = 4.08

= 273.15 – 4.08 = 269.07 K

A2. Henry’s law: the pressure of gas over a solution is directly proportional to the
mol fraction of the gas dissolved in the solution.

P = KH.

K H = Henry’s constant

Applications of Henry’s constant

(i) To increase the solubility of CO 2 in soft drinks and soda water, the bottle is
sealed under high pressure.

(ii) Deep sea divers use He and O 2 mixture for respiration in place of N 2 and O 2
because N 2 being more soluble in blood at high pressure creates painful
effect on the human body.

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A3. Mass of solute ( ) = 54g, molar mass of ( ) = 180g, mass of solvent ( )=

250g

= ×

= = = 2.23K

= 273.15 – 2.23 = 270.77 K

A4. Osmotic Pressure ( ):

The pressure which must be exerted on the solution to prevent any net
movement of solvent between the solution and its solvent, when these are
separated by a semi permeable membrane, is called osmotic pressure.

Here, = mass of solute, T = temperature, = osmotic pressure, V = volume

of solution,

R = constant

A5.

= 100 × 10–3g, T = 25 + 273 = 298K, = 13.3 mm Hg

=
= 13980 g mol–1 or 1.4 × 104 g mol–1

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3 Marks – Short Answer Type II

Q1. State’s law for solution of volatile liquid. Taking suitable examples explain the
meaning of positive and negative deviations from Raoult’s’ law.
[CBSE2008, 10]

Q2. A solution prepared by dissolving 1.25 g of oil of winter green in 99 g of benzene

has boiling point of 80.31oC. Determine the molar mass of this compound.

(B.P. of pure benzene = 80.31oC and for benzene = 2.53oC kg mol–1)

[CBSE 2010]

Q3. A solution of glycerol (molar mass = 92 g mol–1) in water was prepared by

dissolving in 500 g of water. This solution has a boiling point of 100.42oC. What
mass of glycerol was dissolved to make this solution?
( for water = 0.512K kg mol–1) [CBSE 2010]

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Answers

A1. Raoult’s law states that for any solution, partial vapour pressure of each volatile
component in the solution directly proportional to its mole fraction.

PA

PA= ×

Positive deviation from Raoult’s law

Non ideal solutions showing positive deviation have force of attraction less than
the pure components, therefore, their vapour pressure is higher and boiling point
is lowered, that is why they form minimum boiling azeotropes.

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Negative deviation from Raoult’s law

Non ideal solutions showing negative deviation have force of attraction greater
than the pure components, therefore, their vapour pressure is lower and boiling
point is higher, that is why they form maximum boiling azeotropes.

A2. = — = (80.31– 80.10) oC = 0.21 oC

0.21 =

= 152 g mol–1

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A3. = — = (100.42 – 100)oC = 0.42oC

0.42 =

= 37.7g

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C. Must Do Questions

These questions are frequently asked in the Board Examination from the topics/ sub-
topics, of this chapter.

Q1. Why does the molality of a solution remain unaffected with temperature?

Q2. Why are soft drinks and soda water bottles sealed under pressure?

Q3. What is the condition applied in reverse osmosis?

Q4. What temperature change is expected during the mixing of two liquids whose
solution shows a positive deviation from Raoult’s law?

Q5. Why do the gases always tend to be less soluble in liquids as rise in
temperature?

Q6. What is the enthalpy and volume change, when an ideal solution is formed?

Q7. Calculate the molality of 6 g of ethanoic acid in 80 g of benzene.

Q8. Sodium chloride is used to clear snow from the roads. Why?

Q9. Calculate the mass percentage of benzene and carbon tetra chloride if 18g of
benzene is dissolved in 120g of carbon tetra chloride.

Q10. Illustrate the depression of freezing point with the help of vapour pressure –
temperature curve of a solution. Show that depression of freezing point is a
Colligative property.

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Q11. Explain why boiling point of solvent is increased on dissolving a non volatile
solute into it with the help of curve?

Q12. What are non ideal solutions? Explain as how to they deviate from Raoult’s’ law
with suitable diagram.

Q13. Arrange the following aqueous solution (each of strength 0.2 M), in order of
increasing boiling points and freezing points. Justify your answer.

CH 3 OH, NaCl, Ca 3 (PO 4 ) 2 , Na 2 SO 4

Q14. 2.5 g of benzoic acid dissolved in 30 g of benzene shows a depression in

freezing point is 1.68 K. The molal depression constant is 4.9 K kg mol–1. What
will be the percentage association of acid if it exits as dimer in solution?

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D. HOTS (Higher Order Thinking Skills) questions.

Q1. Which Colligative property is preferred for the molar mass determination of
macromolecules? Why?

Q2. To get the hard boiled eggs, why common salt is added to water before boiling
the eggs?

Q3. How much ethyl alcohol must be added to 1 litre of water so that the solution will
freeze at 14°F? ( for water = 1.86 °C/mol)

Q4. How many ml. of a 0.2 M HCl are required to react completely with 1 g mixture of
K 2 CO 3 and KHCO 3 containing equimolar amounts of two?

Q5. Which of following, one has higher osmotic pressure?

(a) 200 ml of 2 M NaCl Solution and

(b) 200 ml of 1 M glucose Solution.

Q6. A sample of drinking water was found to be severely contaminated with

phosgene (COCl 2 ), supposed to be carcinogen. The level of contamination was
12 ppm (by mass).
(a) Express this in precent by mass.
(b) Determine the molality of phosgene in water sample.

Q7. Based on solute-solvent interactions, arrange the following in order of increasing

solubility in n-heptane. Give the explanation.

MgCl 2 , Methanol, Acetonitrile, n-hexane,

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Solutions CHEMISTRY

Answers: -

A1. Osmotic pressure is preferred over all other Colligative properties because:
(a) Even in dilute solution, the osmotic pressure values are appreciably high
and can be measured accurately.
(b) Osmotic pressure can be measured at room temperature. On the other
hand, elevation in boiling point is measured at high temperature whereas
the solute may decompose. The depression in freezing point is measured at
low temperature.

A2. Due to addition of common salt, the boiling point of the salt containing water
elevated. Hence, the egg becomes hard at high temperature.

A3. (°F – 32) / 9 = °C / 5

(14 – 32) / 9 = 0C / 5

in °C = 5 × (– 18) / 9

= – 10 °C

= (1 litre = 1000ml)

10 =

= 247.31 g

A4. + = 1g

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Solutions CHEMISTRY

= 0.58 g

= 0.58/138 = 0.0042 =

The basicity of is 2, while for is 1.

Thus, 1 g mixture of and would require

= (2 + 1) × 0.0042 = 0.0126 moles of HCl
0.0126 L of 1M HCl = 12.6 ml of 1M HCl

Using molarity equation,

M1V1 = M2V2

1 × 12.6 = 0.2 × V 2

V2 = 63 ml

A5. 200 ml of 2 M NaCl Solution has higher osmotic pressure.

NaCl is an electrolyte which dissolves in water to give Na+ and Cl– ions. Glucose
is non electrolytes. Thus, glucose has minimum concentration and hence, less
osmotic pressure.

A6. Molecular mass of phosgene = 99 g mol–1

12 ppm (by mass) = = 12 % by mass

To calculate molality, let us consider the solution weighing 1 kg.

Molality = =

= =

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Solutions CHEMISTRY

A7. MgCl 2 is an ionic compound while -heptane is non-polar. Therefore, they are
totally immiscible in all proportions.

Methanol and acetonitrile (CH 3 CN) both are polar, but acetonitrile is less polar
than methanol, while n-heptane is non- polar as a solvent. Therefore, Acetonitrile
will miscible more than methanol in n-heptane.
n-hexane and n-heptane both are non-polar. Therefore, they are miscible in all
proportions.

Hence, the order of increasing solubility will be

MgCl 2 < Methanol < Acetonitrile < n-hexane

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