Solutions

Marks Allotted - 8 Marks

Part - A ► 2 Q ► 2 Marks
Part - D ► 1 Q ► 5 Marks

A solution is a homogeneous mixture of 2 or more substances .

Solute + Solvent = Solution

SOLUTE – the component which is present in small quantity.

SOLVENT – the component which is present in large quantity.

Types of solutions

Types of Solute Solvent Example

solutions

Solid Solid Solid 22 Carat gold , Cu in Gold

Solutions Liquid Solid Gold Amalgam- Hg in Gold
Gas Solid Solution of Hydrogen in Pd
Liquid Solid Liquid Salt or Sugar in water
Solutions Liquid Liquid Ethanol in water
Gas Liquid O2 or CO2 in water
Gaseous Solid Gas Camphor in Oxygen
Solutions Liquid Gas Moist Hydrogen
Gas Gas Mixture of Nitrogen and Oxygen

Methods of expressing the concentration of solutions

Molarity is defined as the number of moles of solute dissolved in one dm3 or

one litre of the solution.

Molarity =

Molality is the defined as the number of mole of solute dissolved in one Kg

of the solvent.
Molality =

Mass percentage is defined as the mass of solute dissolved in 100 grams of

the solution.

Mass percentage = x 100

Volume percentage defined as the volume in cm3 of solute dissolved in 100
cm3 of the solution.

Mass percentage = x 100

Mole fraction: Mole fraction of a component in a mixture is the ratio of

number of moles one component to the total number of moles of all the
components present in the solution. OR

Mole fraction =

Parts per million (ppm) is defined as the number of parts of mass the
solute per million parts by mass the solution. OR
PPM = x 10

Solubility
Amount of solute that be dissolved in a 100 gram of solvent at a given
temperature.

Saturated solution – Which contain exact amount of solute

Unsaturated solution – Which contain lesser amount of solute
Super-saturated solution – Which contain more amount of solute

Solubility of Solid in Liquid

Factors Affecting the Solubility

1. Nature of Solute and Solvent
2. Temperature

Effect of Nature of Solute and Solvent on Solubility

Polar Solutes Dissolves in Polar Solvents. (Like Dissolves Like)

Example: NaCl, KNO3, Sugar dissolves in Water.

But, Naphthalene and Anthracene does not dissolve in Water. Because,

Naphthalene and Anthracene are non-polar Solutes.

Non-Polar Solutes Dissolves in Non-Polar Solvent (Unlike Dissolves in

Unlike)
Naphthalene and Anthracene dissolve in Benzene. Naphthalene, Anthracene
and Benzene are Non-Polar

Effect of Temperature on Solubility

1. If dissolution is endothermic, i.e., ∆H > 0 or ∆H is positive, then increase
in temperature increases the solubility.
Example: NaNO3, NH4Cl, KCl, AgNO3 .
2. If dissolution is exothermic, i.e., ∆H < 0 or ∆H is negative, then increase
in temperature decreases the solubility.
Example: Li2SO4, Ce2(SO4)3, Na2CO3 . H2O and Calcium Salts of some
organic acid.

Solubility of Gases in Liquid

1. Effect of temperature
Solubility of gases decreases with increase of temperature. Because,
dissolution of gas in liquid is exothermic process.

2. Effect of Pressure
Increase in pressure increases the solubility.

Henry’s law
Henry’s law states that partial pressure of the gas in vapour phase is
directly proportional to mole fraction of the gas in the solution.

p= K xχ
Where p is the partial pressure of the gas over the solution
χ is the mole fraction and K is is Henry’s constant.

Significance of Henry’s Constant

Higher the value of KH , lower is the solubility.
The value of KH , increases with increase in temperature. i.e., solubility of
gas decreases with increase in temperature.

Applications of Henry’s Law

1. To increase the solubility of carbon dioxide in soda water, soft drinks,
the containers sealed under high pressure.
2. At high altitudes, the partial pressure of oxygen is low. As a result,
concentration of oxygen becomes low in blood. Hence people become
weak and the thinking ability reduces. These symptoms are called
anoxia.

Vapour Pressure of Liquid Solutions

Raoult’s law for a solution of volatile liquids
For a solution of volatile liquids, the partial vapour pressure of each
component in the solution is directly proportional to its mole fraction.

Graph of Vapour pressure against mole fraction of two volatile liquids

 pressure
Vapour
o
pB

o
pA

pressure
Vapour
χA = 1 Mole Fraction χA = 0
χB = 0 χB = 1

Raoult’s Law as a special case of Henry’s Law

According to Henry’s Law, p = KH X

According to Raoult’s Law, pA = pAo χA

If, KH = pAo , Raoult’s law becomes a special case of Henry’s law.

Ideal and Non-ideal Solutions

Ideal Solutions
Intermolecular force of attraction between the components are similar to
those in the pure components.

Non-ideal Solutions
Intermolecular force of attraction between the components are different from
those in the pure components.

Differences between ideal and non-ideal solutions

Ideal solution Non-ideal solution

Intermolecular interactions between Intermolecular interactions between
the components are of the same the components are different from
magnitude as the intermolecular the intermolecular interactions
interactions found in the pure between the molecules of the pure
components. components.
Obeys Raoult’s law Does not obeys Raoult’s law
∆ Hmix = 0 ∆ Hmix ‡ 0
∆ Vmix = 0 ∆ Vmix ‡ 0
n-Hexane + n-Heptane, Ethyl alcohol + Water
Bromoethane + Chloroethane, Acetone + Carbon disulphide
Benzene + Toluene Acetone + Benzene
Chlorobenzene + Bromobenzene Acetone + Ethyl alcohol
Differences between Non-ideal Solutions showing positive and negative
deviations
Non-Ideal solution showing positive Non-Ideal solution showing negative
deviation deviation
P > pAo χA + pBo χB P < pAo χA + pBo χB
∆ Hmix > 0 i.e. + ve ∆ Hmix < 0 i.e. - ve
∆ Vmix > 0 i.e. + ve ∆ Vmix < 0 i.e. - ve
Acetone + Ethyl alcohol Chloroform + Diethyl ether
Acetone + Benzene HCl + water, HNO3 + Water
Ethyl alcohol + Water

Graph of Vapour pressure against mole fraction of Solutions showing

positive deviation

o
pB
o

o
pA
pressure
Vapour
pressure
Vapour

χA = 1 Mole Fraction χA = 0
χB = 0 χB = 1

Graph of Vapour pressure against mole fraction of Solutions showing

negative deviation

o
pB
pressure
Vapour

pressure
Vapour

o
pA

χA = 1 Mole χA = 0
χB = 0 Fraction χB = 1
Azeotropes (A = No Zeo = Boil tropes = Change)

Solutions having the same composition in liquid and vapour phase and boil
at a constant temperature.

1. Minimum boiling azeotrope. The solutions which show a large positive

deviation from Raoult’s law. Ex. Ethanol-Water

2. Maximum boiling azeotrope- The solutions which show a large negative

deviation from Raoult’s law. Example- 68% HNO3 and 32% water by mass

Colligative properties
Properties whose values depend only on the number of solute particles
present in the solution and not on its nature, size or chemical composition.
Examples.
1. Relative lowering of vapour pressure
2. Depression of freezing point
3. Elevation of boiling point
4. Osmotic pressure

Raoult’s Law of relative lowering of vapour pressure

For a binary solution containing non-volatile solute in a volatile solvent, the
relative lowering of vapour pressure is equal to the mole fraction of the
solute present in the solution.

p − p
= χ
p
Relation between molar mass of solute and relative lowering vapour
pressure.
Let is the molar mass of solvent.
Let is the mass of solvent.
Let is the molar mass of non-volatile solute.
Let is the mass of non-volatile solute.

Mole fraction of the solute χ

Where n & n are the number of moles of solvent and solute respectively.
For dilute solutions, n + n n
n
χ
n

W
M W M
χ = χ =
W M W
M
But,
p − p
= χ
p

p − p W M
 =
p M W

W M
M =
p − p
W ( )
p

Elevation of boiling point

Solvent
Solution

Vapour
pressur
e

T Δ

Temperature

Let T be the boiling point of the solvent,

Let T be the boiling point of the solution,
Elevation in the boiling point is given by Δ = T −T
But , Δ ∝ Where m is molality
Δ = K where is called boiling point elevation constant or molal
elevation constant or ebullioscopic constant.
Let is the mass of solvent in kg.
Let is the molar mass of non-volatile solute.
Let is the mass of non-volatile solute.

Molality of solution is given

1000 W
m =
W M

Δ = K
 1000 WB 1000 WB
Δ = MB =
WA MB WA

Molal elevation constant: Molal elevation constant is defined as the

elevation in boiling point produced when one mole of the solute is dissolved
in 1 kg of the solvent.
Unit for molal elevation constant Kb = K kg mole-1

Depression of freezing point

Liquid
Solvent

Frozen
Solvent Solution
Vapour
Pressure

T T

Temperature

Let T be the freezing point of the solvent,

Let T be the freezing point of the solution,
Depression in the boiling point is given by ΔT = T − T
But, ΔT ∝ m Where m is molality
ΔT = K m Where is called freezing point depression constant or molal
depression constant or cryoscopic constant.
Let is the mass of solvent in kg.
Let is the molar mass of non-volatile solute.
Let is the mass of non-volatile solute.

Molality of solution is given

1000 W
m =
W M

ΔT = K m
1000 WB 1000 WB
ΔT = MB =
WA MB WA
Molal depression constant: Molal depression constant is defined as the
depression in freezing point produced when one mole of the solute is
dissolved in 1 kg of the solvent.
Unit for molal depression constant Kf = K kg mole-1

Osmosis.
The phenomenon of passage of solvent molecules from solvent to the
solution or movement of solvent molecules from a dilute solution to the
concentrated solution through semi permeable membrane is called osmosis.

Osmotic pressure.
The external pressure which can be applied to solution in order to stop the
flow of solvent into the solution through semi permeable membrane is called
osmotic pressure.

Relationship between molar mass of a solute and osmotic pressure

Osmotic pressure is denoted by .
Osmotic pressure is given by =

n
π= RT
V

If is amount of solute and is the molar mass of mass dissolved in V

liters of the solution,

RT RT RT
π=n π= =
V V πV

Isotonic solutions
The solutions having same osmotic pressure at a given temperature is called
isotonic solutions.

Hypertonic solutions
A solution which has higher osmotic pressure than the other is called
hypertonic solution.

Hypotonic solutions
A solution which has lower osmotic pressure than the other is called
hypotonic solutions.

Note:
1. 0.9% (mass/volume) sodium chloride solution is called normal saline
solution
2. Blood is isotonic with 0.9% (mass/volume) sodium chloride solution
3. When blood cells are placed in a solution containing more than 0.9%
sodium chloride, water will flow out of the cells and they would shrink.
4. When blood cells are placed in a solution containing less than 0.9%
sodium chloride, water will flow in to the cells and they would swell.

Reverse osmosis
When a pressure which is greater than osmotic pressure is applied on the
solution, the solvent molecules flow from the solution into the solvent
through semi permeable membrane. This process is called reverse osmosis.

Abnormalities in molecular mass

1. Due to dissociation of solute
For NaCl,
Observed molecular mass = half of theoretical value

NaCl Na+ + Cl-

Observed colligative property = Twice of theoretical value

1
Colligative Property =
Molar mass

2. Due to Association of solute

For C6H5COOH,
observed molecular mass = twice the theoretical value

C6H5COOH ½ ( C6H5COOH )2

Observed colligative property = half of theoretical value

1
Colligative Property =
Molar mass

Van’t Hoff factor.

normal molar mass
= or
observed molar mass

observed colligative property

= or
normal colligavibe property

total no. of particles after dissociation/association

i=
no. of particles before dissociation/association

For solutes showing,

1. Dissociation, i > 1
2. Association, i < 1
3. Neither dissociation nor association, i = 1

For solutes which undergoes association or dissociation,

1. = i χ
2. ∆Tb = i Kb m

3. ∆Tf = i Kf m

4. π = i CRT

Solution
Questions carrying one mark each
1. How molarity does varies with temperature? ( March 2017 )
Ans. Molarity decreases with increase in temperature, or Molarity
increases with decrease in temperature, or Molarity varies inversely with
temperature.
2. 10 mL of liquid 'A' is mixed with 10 mL of liquid 'B', the volume of
the resultant solution is 19.9 mL. What type of deviation expected
from Raoult's law?
Ans. Non ideal solution with negative deviation

3. Name the law behind dissolution of CO2 in soft drinks under high
pressure. ( March 2016 )
Ans. Henrys law.

4. Ornamental gold containing copper is example for what type of

solution? ( March 2016 )
Ans. Solid solution.

5. Soda water bottles are sealed under high pressure. Give reason. (
June 2016 )
Ans. Increase the solubility of carbon dioxide.

6. What are ideal solutions? ( June 2016 )

Ans. Intermolecular force of attraction between two components and
force of attraction between the individual components are same.

7. What is the effect of temperature on the solubility of gases in liquids

? ( June 2016 )
Ans. Increase in temperature decreases the solubility of gases.
8. At a given temperature and pressure nitrogen gas is more soluble in
water than Helium gas. Which one of them has higher value KH? (
March 2015 )
Ans. Helium.

9. On mixing equal volumes of acetone and ethanol, what type of

deviation from Raoult’s law is expected?
Ans. Positive deviation.

10. On what factor does the value of colligative property depend ? (

July 2015 )
Ans. number of solute particles.

11. Give an example for liquid solution in which solute is gas. ( July
2015 )
Ans. Carbonated water.

12. At a given temperature and pressure nitrogen gas is more soluble in

water than Helium gas. Which one of them has higher value KH ? (
March 2015 )
Ans. Helium.

13. On mixing equal volumes of acetone and ethanol, what type of

deviation from Raoult’s law expected? ( March 2015 )
Ans. Positive deviation.

14. What is Binary Solution? ( July 2014 )

Ans. Solution which contain two components is called binary solution.

15. Define Molality. ( July 2014 )

Ans. Molality is the defined as the number of mole of solute dissolved in
one Kg of the solvent.

16. Define the term ‘molarity’? ( March 2014 )

Ans. Molarity is defined as the number of moles of solute dissolved in
one dm3 or one liter of the solution.

17. Mention the value of enthalpy of mixing (∆ ) to form an ideal

solution. ( March 2014 )
Ans. Zero.

18. Name a colligative property.

Ans. Osmotic pressure, lowering of vapour pressure, elevation in boiling
point, and depression freezing point. ( any one of the above)
19. What does the Van’t Hoff factor ‘i’ for a solute in a solvent account
for?
Ans. Extent of association or dissociation.

20. State Henry’s law.

Ans. Henry’s law states that partial pressure of the gas in vapour phase
is proportional to mole fraction of the gas in the solution.

p= K xχ
Where p is the partial pressure of the gas over the solution
χ is the mole fraction and K is is Henry’s constant.

21. Name any one concentration term which is independent of

temperature.
Ans. Molality

22. What is the effect of increase in pressure on the solubility of a gas

in a liquid ?
Ans. Increase in pressure increases the solubilities.

23. Components of a non-ideal binary solution cannot be completely

separated by fractional distillation. Why?
Ans. Because they form azeoptropic mixtures.

24. What are isotonic solutions?

Ans. Solutions having same osmotic pressure at given temperature.

25. What is the Van’t Hoff factor for potassium sulphate in very dilute
aqueous solution?
Ans. Three.

26. What happens to vapour pressure of water if a table spoon of sugar

is added to it ?
Ans. Vapour pressure decreases.

27. Two liquids A and B boils at 145oc and 190oc respectively, which of
them has higher vapour pressure at 80oc?
Ans. Liquid A

28. What will happen if pressure greater than osmotic pressure is

applied on the solution separated by semipermeable membrane from
the solvent ?
Ans. Reverse osmosis takes place.

29. State Raoult’s law of a solution containing non-volatile solute.

 Ans. The relative lowering of vapour pressure is equal to the mole
fraction of the solute present in the solution.

P − P
=
P

30. Van’t Hoff factor ‘i’ is greater than 1 for aqueous solution
containing potassium chloride. Why?
Ans. Because KCl undergoes dissociation.

31. What is the S.I unit of molal elevation constant ( kb) of a solvent ?
Ans. Kg K mol-1

32. “Aquatic animals are more comfortable in cold water rather than in
warm water” ?
Ans. Because solubility of oxygen is more in cold water than in warm
water.

33. What is hypertonic solution?

Ans. A solution which has higher osmotic pressure than the other
solution is called hypertonic solution.

Questions carrying five mark each

34. How solubility of a gas in liquid varies with (i) Temperature and (ii)
pressure? ( March 2017)
Ans. Temperature: Increase in temperature decreases the solubility
Pressure: Increase in pressure increases the solubility

35. What are non-ideal solutions? Mention the reason for the negative
deviation from Raoult’s law. ( March 2016 )
Ans. An non –ideal solution is one where the intermolecular interactions
between the components are different from the intermolecular
interactions between the molecules of the pure components.

36. Define isotonic solutions. What happens when blood cells is dipped
in solution containing more than normal saline concentration.
(March 2015 )
Ans. The solutions having same osmotic pressure at a given temperature
is called isotonic solutions. When the blood cells is dipped in solution
containing more than normal saline concentration, blood cells shrink.
37. What happens to the solubility of a gas in a liquid with increase in
temperature? Give reason. ( July 2014 )
Ans. Solubility decreases, because dissolution of gas in liquid is
exothermic process.

38. What is reverse osmosis? Mention any one of it’s use.

Ans. When a pressure which is greater than osmotic pressure is applied on
the solution, the solvent molecules flow from the solution into the solvent
through semi permeable membrane. This process is called reverse osmosis.
Use: used in desalination of water.

39. A 4% solution of a non –volatile solute is isotonic with 0.702 % urea

solution. Calculate the molar mass of the non – volatile solute.( Molar
mass of urea=60 g mol-1)
Ans.
W W
=
M M

40 x 60
M = = 341.8 g
7.02

40. State Raoult’s law for a solution of 2 volatile liquids. Give an example
for liquid mixture that shows negative deviation from Raoult’s law?
Ans. According to Raoult’s law, At a given temperature the partial pressure
of each component in solution is equal to the product of the vapour
pressure of the pure component and its mole fraction. P = P x χ and
P = P x χ

Solutions showing negative deviation: Acetone + chloroform or Nitric acid +

water

41. Point out the difference between ideal and non ideal solution. ( June
2015, March 2016 )
Ans.

Ideal solution Non-ideal solution

Obeys Raoult’s law Does not obeys Raoult’s law
∆ =0 ∆ ≠0
∆ =0 ∆ =0

42. State Henry’s law. Write its mathematical form. ( March2014 ) ( June
2016 )
 Ans. Henry’s law states that partial pressure of the gas in vapour phase
is proportional to mole fraction of the gas in the solution.
p= K xχ
Where p is the partial pressure of the gas over the solution
χ is the mole fraction and K is is Henry’s constant.

43. Write any Two applications of Henry’s law.

Ans. To increase the solubility of carbon dioxide in soda water, soft
drinks, the containers sealed under high pressure.
At high altitudes, the partial pressure of oxygen is low. As a result,
concentration of oxygen becomes low in blood. Hence people become weak
and the thinking ability reduces. These symptoms are called anoxia.

44. What are azeotrops? Give an example.

Ans. Mixture of two liquids having same boiling point are called
Azeotropic mixture.
Example: Water + ethyl alcohol

45. The vapour pressure of ethyl alcohol at 298k is 40mm of Hg. Its mole
fraction in a solution with methyl alcohol is 0.80. What is its vapour
pressure in solution if the mixture obeys Raoult’s law?
Ans. P = P x χ

P = 40 x 0.80 = 32 mm of Hg

Questions carrying three mark each

46. 1.0 g of non-electrolyte solute dissolved in 50 g of benzene lowered
the freezing point of benzene by 0.4 K. Find the molar mass of the
solute. [Given: Freezing point, depression constant of benzene =
5.12 K. kg mol-1]. ( March 2017 )
1000 K W
ΔT =
W M

1000 x 5.12 x 1.0

M =
50 x 0.4

M = 256 g mol

47. 5.8 g of non-volatile solute was dissolved in 100 g of carbon

disulphide. The vapour pressure of the solution was found to be 190
mm of Hg. Calculate the molar mass of the solute given the vapour
pressure of pure carbon disulphide is 195 mm of Hg. Molar mass of
carbon disulphide is 76 g mol-1 ( March 2016 )
Ans.
p −p W M
=
p W M
 195 − 190 5.8 x 76
=
195 100 x M

5.8 x 76 x 195
M = = 171.9
100 x 5
48. 300 Cm3 of an aqueous solution of protein contains 2.12 g of the
protein, the osmotic pressure of such a solution at 300 K is found
to be 3.89 x 10-3 bar. Calculate the molar mass of the protein. ( R =
0.0823 L bar mol-1K-1 ) ( July 2016 )

WRT
Π =
M V

2.12 x 0.0823 x 300

M =
3.89 x 10 x 300

M = 44.8 g

49. 200 Cm3 of an aqueous solution of protein contains 1.26 g of the

protein, the osmotic pressure of such a solution at 300 K is found
to be 2.57 x 10-3 bar. Calculate the molar mass of the protein. ( R =
0.0823 L bar mol-1K-1 ) ( March 2016 )
Ans.
WRT
Π =
M V
1.26 x 0.0823 x 300
M =
2.57 x 10 x 200

M = 60 g

50. A solution containing 18 g non-volatile non-electrolyte solute is

dissolved in 200 g of water freezes at 272.07 K. Calculate the
molecular mass of solute. Given Kf = 1.86 k Kg / mol. Freezing
point of water = 273 K ( March 2015 )
Ans.
1000 K W
ΔT =
W M

1000 x 1.86 x 18
M =
200 x (273 − 272.07)
 1000 x 1.86 x 18
M =
200 x (273 − 272.07)

M = 180 g mol

51. The boiling point of benzene is 353.23k. When 1.80 g of a non-

volatile solute was dissolved in 90 g of benzene, the boiling point is
raised to 354.11k.Calculate the molar mass of the solute (kb= 2.53k
kg mol-1) ( June 2015)
Ans.
K 1000 W
Δ =
W M

K 1000 W
M =
W Δ

2.53 x 1000 x 1.8

M =
0.88 x 90

g
M = 58
mol

52. On Dissolving 2.34 g of solute in 40 g benzene, the boiling point of

solution was higher than that of benzene by 0.81 K. Kb value of
benzene is 2.5 K kg mol-1 . Calculate the molar mass of the solute. (
March 2014 )
Ans.
K 1000 W
Δ =
W M

K 1000 W
M =
W Δ

2.5 x 1000 x 2.34

M =
0.81 x 40

M = 180.5 g

53. The vapour pressure of pure benzene at a certain temperature is

0.850 bar. When 0.5 g of a non-volatile solute is added to 39.0 g of
benzene ( Molar mas of benzene =78 g mol-1) Vapour pressure of the
solution is 0.845 bar. What is the molar mass of a non-volatile solute?
( July 2014 )
 Ans.
p −p W M
=
p W M

0.85 − 0.845 0.5 x 78

=
0.85 M x 39

M = 170 g mol

54. Plot a graph of vapour pressure against mole fraction of the two
volatile liquids forming an ideal solution. What is the change in
enthalpy upon mixing the two components of an ideal solution?
Ans.

o
pB
pressure
Vapour

o
pressure
Vapour

pA

χA = 1 Mole χA = 0
χB = 0 Fraction χB = 1

55. 15.0 g of unknown substance was dissolved in 450 g of water. The

resulting solution was found to freeze at -0.34o C. Calculate the molar
mas of the substance.
Ans.
K 1000 W
M =
W Δ

1.86 x 1000 x 15
M =
450 x 0.34

M = 182.35 g mol
56. If 1.71 g of sugar (Molar mass= 342) is dissolved in 500 cm3 of a
solution at 300k. What will be its osmotic pressure? ( Given r=
0.0831 bar k-1 mol-1)
Ans.
WRT
Π =
M V

1.71 x 0.083 x 300

Π =
343 x 0.5

Π = 0.0249 bar

57. At 298k, the vapour pressure of a water is 23.75mm of Hg. Calculate

the vapour pressure at the same temperature over 5% aqueous
solution of urea. given the molecular mass of urea= 60
Ans.
p −p W M
=
p W M

23.75 − P 5 x 18
=
23.75 95 x 60

5 x 18 x 23.75
23.75 − P =
95 x 60

23.75 − P = 0.375

P = 23.75 − 0.375 = 23.37 mm of Hg

58. For a non-ideal solution having positive deviation from Raoult’s law.
i. Plot a graph of vapour pressure against mole fraction.
ii. What type of azeotrops formed by this solution?
iii. Give one example for the above solution.
Ans.
 o
pB
pressure
Vapour

pressure
Vapour
pA

χA = 1 Mole χA = 0
χB = 0 Fraction χB = 1

ii. Minimum boiling azeotrope is formed.

iii. Ethanol + Acetone

59. Vapour pressure of liquids A and B at 298k is 300mm of Hg and

450mm of Hg, calculate the mole fraction of A in the mixture.
Ans. Let the mole fraction of A be XA
Let the mole fraction of B be XB
XB = 1 – XA
From Raoult’s law,
Ptotal = PoXA + PoXB
Ptotal = PoXA + Po ( 1 - XB )
405 = 300 XA + 450 ( 1 - XA )
XA = 0.3