Solutions

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Summary

Summary of Solutions

Definition: A solution is a homogeneous mixture of two or more substances.
Types of Solutions:
- Solid Solutions: Solute can be solid, liquid, or gas.
- Liquid Solutions: Solute can be solid, liquid, or gas dissolved in a liquid.
- Gaseous Solutions: Solute can be gas, liquid, or solid mixed with gas.
Concentration Units:
- Mole Fraction: Ratio of moles of a component to total moles.
- Molarity (M): Moles of solute per liter of solution.
- Molality (m): Moles of solute per kilogram of solvent.
- Mass Percentage: Mass of solute per total mass of solution, expressed as a percentage.
- Parts per Million (ppm): Mass of solute per million parts of solution.
Laws Governing Solutions:
- Henry's Law: Solubility of a gas in a liquid is proportional to its partial pressure.
- Raoult's Law: The vapor pressure of a solvent is lowered by the presence of a non-volatile solute.
Colligative Properties: Properties that depend on the number of solute particles, not their identity:
- Lowering of vapor pressure
- Elevation of boiling point
- Depression of freezing point
- Osmotic pressure
Ideal vs Non-Ideal Solutions: Ideal solutions obey Raoult's law; non-ideal solutions show deviations (positive or negative).
Azeotropes: Mixtures that show large deviations from Raoult's law.

Learning Objectives

Describe the formation of different types of solutions.
Express concentration of solution in different units.
State and explain Henry's law and Raoult's law.
Distinguish between ideal and non-ideal solutions.
Explain deviations of real solutions from Raoult's law.
Describe colligative properties of solutions and correlate these with molar masses of the solutes.
Explain abnormal colligative properties exhibited by some solutes in solutions.

Detailed Notes

Notes on Solutions

1. Objectives

After studying this Unit, you will be able to:

Describe the formation of different types of solutions.
Express concentration of solution in different units.
State and explain Henry's law and Raoult's law.
Distinguish between ideal and non-ideal solutions.
Explain deviations of real solutions from Raoult's law.
Describe colligative properties of solutions and correlate these with molar masses of the solutes.
Explain abnormal colligative properties exhibited by some solutes in solutions.

2. Types of Solutions

Solutions are homogeneous mixtures of two or more components. The component present in the largest quantity is known as the solvent, while the other components are called solutes. The following types of solutions are commonly recognized:

Type of Solution	Solute	Solvent	Common Examples
Gaseous Solutions	Gas	Gas	Mixture of oxygen and nitrogen gases
	Liquid	Gas	Chloroform mixed with nitrogen gas
	Solid	Gas	Camphor in nitrogen gas
Liquid Solutions	Gas	Liquid	Oxygen dissolved in water
	Liquid	Liquid	Ethanol dissolved in water
	Solid	Liquid	Glucose dissolved in water
Solid Solutions	Gas	Solid	Solution of hydrogen in palladium
	Liquid	Solid	Amalgam of mercury with sodium
	Solid	Solid	Copper dissolved in gold

3. Concentration of Solutions

The concentration of a solution can be expressed in various units:

Mole Fraction (x):
- Defined as the number of moles of the component divided by the total number of moles of all components.
Molarity (M):
- Defined as the number of moles of solute per liter of solution.
Molality (m):
- Defined as the number of moles of solute per kilogram of solvent.
Mass Percentage:
- The mass of solute divided by the total mass of the solution, multiplied by 100.
Parts per Million (ppm):
- Defined as the mass of solute divided by the total mass of the solution, multiplied by 10^6.

4. Laws Governing Solutions

Henry's Law:
- At a given temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.
Raoult's Law:
- The relative lowering of vapor pressure of the solvent over a solution is equal to the mole fraction of the non-volatile solute present in the solution.

5. Colligative Properties

Colligative properties depend on the number of solute particles in a solution and are independent of their chemical identity. These properties include:

Lowering of vapor pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure

6. Examples and Applications

Example of Colligative Properties: The depression in freezing point can be calculated using the formula:
- T = K_f * m
- Where K_f is the freezing point depression constant and m is the molality of the solution.

7. Important Notes

Solutions can exhibit ideal or non-ideal behavior based on the interactions between solute and solvent.
Deviations from Raoult's law can be positive or negative, leading to the formation of azeotropes in some cases.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Solution Types: Students often confuse different types of solutions (solid, liquid, gas). Ensure clarity on definitions and examples.
Incorrect Concentration Units: Mixing up molarity, molality, and mole fraction is common. Remember that molarity is volume-based, while molality is mass-based.
Raoult's Law Deviations: Failing to recognize positive and negative deviations from Raoult's law can lead to incorrect conclusions about solution behavior.
Colligative Properties Confusion: Students may not differentiate between colligative properties and their dependence on solute particle number rather than identity.
Osmotic Pressure Calculations: Miscalculating osmotic pressure due to incorrect application of formulas or misunderstanding the van't Hoff factor.

Exam Tips

Review Definitions: Be clear on definitions of key terms such as mole fraction, molality, and molarity. Use examples to reinforce understanding.
Practice Calculations: Work through problems involving concentration calculations, colligative properties, and osmotic pressure to build confidence.
Understand Laws: Make sure to understand Henry's law and Raoult's law thoroughly, including their applications and limitations.
Use Visual Aids: Diagrams can help visualize concepts like osmosis and the behavior of solutions under different conditions.
Time Management: Allocate time wisely during exams to ensure all questions are attempted, especially those involving calculations.

Practice & Assessment

Multiple Choice Questions

0.1 m NaCl

0.1 m glucose

0.1 m CaCl₂

0.1 m urea

Correct Answer: C

Solution:

Boiling point elevation is a colligative property and depends on the number of solute particles. NaCl dissociates into 2 ions, glucose and urea do not dissociate, and CaCl₂ dissociates into 3 ions. Therefore, 0.1 m CaCl₂ will have the highest boiling point elevation.

1 mol/kg

0.5 mol/kg

2 mol/kg

1.5 mol/kg

Correct Answer: A

Solution:

Molality is defined as the number of moles of solute per kilogram of solvent. Here, it is 1 mol/kg.

Increases the vapour pressure

Decreases the vapour pressure

No change in vapour pressure

Depends on the solute's nature

Correct Answer: B

Solution:

The presence of a non-volatile solute lowers the vapour pressure of the solvent, as stated by Raoult's law.

0.0336

0.0345

0.0354

0.0363

Correct Answer: A

Solution:

The relative lowering of vapour pressure is given by Raoult's law:

\frac{P^0 - P}{P^0} = x_{\text{solute}}

where

P^0

is the vapour pressure of pure solvent,

P

is the vapour pressure of the solution, and

x_{\text{solute}}

is the mole fraction of the solute. Given

P^0 = 23.8

mm Hg and

P = 23.0

mm Hg, we have:

\frac{23.8 - 23.0}{23.8} = x_{\text{solute}} = 0.0336

0.93 K

0.31 K

1.55 K

0.62 K

Correct Answer: B

Solution:

The depression in freezing point

\Delta T_f

is given by

\Delta T_f = i \cdot K_f \cdot m

, where

i

is the van't Hoff factor (1 for urea),

K_f

is the freezing point depression constant, and

m

is the molality. The molality

m = \frac{5/60}{0.1} = 0.833

mol/kg. Therefore,

\Delta T_f = 1 \times 1.86 \times 0.0833 = 0.31

0.111 mol kg⁻¹

0.222 mol kg⁻¹

0.333 mol kg⁻¹

0.444 mol kg⁻¹

Correct Answer: A

Solution:

The molar mass of glucose is 180 g mol⁻¹. Moles of glucose = 10 g / 180 g mol⁻¹ = 0.0556 mol. Molality = moles of solute / mass of solvent in kg = 0.0556 mol / 0.5 kg = 0.111 mol kg⁻¹.

0.05 atm L/mol

0.1 atm L/mol

0.2 atm L/mol

0.5 atm L/mol

Correct Answer: B

Solution:

According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas. The constant is calculated as

k_H = \frac{P}{C} = \frac{2\ atm}{0.1\ mol/L} = 0.1\ atm\ L/mol

58.9 mm Hg

59.4 mm Hg

60.2 mm Hg

61.0 mm Hg

Correct Answer: B

Solution:

For an ideal solution, the total vapour pressure is given by Raoult's law:

P_{total} = P_{methanol} \times x_{methanol} + P_{water} \times x_{water}

. The mole fraction of methanol,

x_{methanol}

, is calculated as follows: Moles of methanol =

\frac{50}{32} = 1.5625

mol, Moles of water =

\frac{50}{18} = 2.7778

mol.

x_{methanol} = \frac{1.5625}{1.5625 + 2.7778} = 0.360

x_{water} = 1 - 0.360 = 0.640

. Therefore,

P_{total} = 94 \times 0.360 + 23.8 \times 0.640 = 59.4

mm Hg.

3.03 \times 10^{-5}

mol/kg

3.03 \times 10^{-4}

mol/kg

3.03 \times 10^{-3}

mol/kg

3.03 \times 10^{-2}

mol/kg

Correct Answer: A

Solution:

According to Henry's law, the solubility

S

of a gas in a liquid is given by

S = \frac{P}{K_H}

, where

P

is the partial pressure and

K_H

is the Henry's law constant. Here,

P = 1

atm

= 760

mm Hg and

K_H = 3.3 \times 10^7

mm Hg. Therefore,

S = \frac{760}{3.3 \times 10^7} = 3.03 \times 10^{-5}

mol/kg.

180 g/mol

90 g/mol

60 g/mol

30 g/mol

Correct Answer: A

Solution:

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute.

\frac{23.8 - 23.1}{23.8} = \frac{n}{n + \frac{90}{18}}

Solving for the molar mass of the solute, we find it to be 180 g/mol.

88 mm Hg

92 mm Hg

84 mm Hg

96 mm Hg

Correct Answer: A

Solution:

According to Raoult's law, the total vapour pressure

P_{\text{total}}

is given by:

P_{\text{total}} = P_A^0 x_A + P_B^0 x_B

where

P_A^0 = 100

mm Hg,

x_A = 0.4

P_B^0 = 80

mm Hg, and

x_B = 1 - x_A = 0.6

. Therefore,

P_{\text{total}} = (100 \times 0.4) + (80 \times 0.6) = 40 + 48 = 88

mm Hg.

The total vapour pressure of the solution is equal to the sum of the partial pressures of the components.

The vapour pressure of the solvent is increased by the presence of a non-volatile solute.

The relative lowering of vapour pressure is equal to the mole fraction of the solute.

The solubility of a gas in a liquid is inversely proportional to the partial pressure of the gas.

Correct Answer: A

Solution:

Raoult's law states that for a solution of volatile liquids, the total vapour pressure is the sum of the partial pressures of each component, each of which is proportional to its mole fraction.

10%

11.11%

9.09%

12.5%

Correct Answer: B

Solution:

Mass percentage is calculated as (mass of solute / total mass of solution) x 100. Here, it is (10 g / (10 g + 90 g)) x 100 = 11.11%.

2.49 bar

4.98 bar

9.96 bar

1.25 bar

Correct Answer: A

Solution:

Osmotic pressure (

\Pi

) is directly proportional to the molarity (

C

) of the solution:

\Pi = CRT

where

R

is the gas constant and

T

is the temperature in Kelvin. If the concentration is halved, the osmotic pressure will also be halved:

\Pi_{\text{new}} = \frac{4.98}{2} = 2.49

bar.

88 mm Hg

92 mm Hg

96 mm Hg

104 mm Hg

Correct Answer: A

Solution:

According to Raoult's law, the total vapour pressure is given by

P_{total} = P_A^0 x_A + P_B^0 x_B

. Here,

x_A = 0.6

and

x_B = 0.4

. Thus,

P_{total} = 100 \times 0.6 + 80 \times 0.4 = 60 + 32 = 92

mm Hg.

0.1 M NaCl

0.1 M glucose

0.1 M CaCl₂

0.1 M urea

Correct Answer: C

Solution:

The freezing point depression is a colligative property and depends on the number of particles. CaCl₂ dissociates into three ions, causing the greatest depression.

0.52 K kg/mol

1.04 K kg/mol

0.26 K kg/mol

0.78 K kg/mol

Correct Answer: B

Solution:

The boiling point elevation is given by

\Delta T_b = K_b \cdot m

where

m

is the molality. Since the solution is 0.5 M and assuming the density of water is approximately 1 kg/L,

m \approx 0.5

. Therefore,

K_b = \frac{0.52}{0.5} = 1.04

K kg/mol.

23.1 mm Hg

22.8 mm Hg

23.5 mm Hg

23.0 mm Hg

Correct Answer: A

Solution:

According to Raoult's law, the vapour pressure of the solution is given by

P_{solution} = P_{water} \times (1 - x_{solute})

, where

x_{solute}

is the mole fraction of the solute. The mole fraction of glucose is calculated as follows:

x_{solute} = \frac{\text{moles of glucose}}{\text{moles of glucose} + \text{moles of water}}

. Moles of glucose =

\frac{18}{180} = 0.1

mol, Moles of water =

\frac{100}{18} = 5.56

mol. Thus,

x_{solute} = \frac{0.1}{0.1 + 5.56} = 0.0177

. Therefore,

P_{solution} = 23.8 \times (1 - 0.0177) = 23.1

mm Hg.

2.5 mol/kg

5 mol/kg

10 mol/kg

2 mol/kg

Correct Answer: A

Solution:

Molality (m) is defined as moles of solute per kilogram of solvent. Here, it is 5 moles / 2 kg = 2.5 mol/kg.

1.5 x 10⁻⁴ mol/L

3.5 x 10⁻⁴ mol/L

5.5 x 10⁻⁴ mol/L

7.5 x 10⁻⁴ mol/L

Correct Answer: A

Solution:

According to Henry's law, solubility

S = k_H \cdot P

. Here,

k_H = 3.3 \times 10^4

atm and

P = 5

atm. Therefore,

S = \frac{5}{3.3 \times 10^4} = 1.5 \times 10^{-4}

mol/L.

112.4 atm

56.2 atm

100 atm

32.1 atm

Correct Answer: A

Solution:

The partial pressure of a gas in a mixture is given by its volume percentage times the total pressure. Thus, the partial pressure of nitrogen is

0.562 \times 200 = 112.4

atm.

20%

25%

30%

40%

Correct Answer: A

Solution:

Mass percentage is calculated as (mass of solute / total mass of solution) × 100. Here, it is (200 g / (200 g + 800 g)) × 100 = 20%.

10%

11.1%

9.1%

12.5%

Correct Answer: A

Solution:

Mass percentage is calculated as (mass of solute/mass of solution) x 100. Here, it is (10/100) x 100 = 10%.

2.05 M

2.56 M

2.00 M

2.10 M

Correct Answer: A

Solution:

The mass of NaCl in 100 g of solution is 12 g. The volume of the solution is

\frac{100}{1.2}

mL = 83.33 mL = 0.08333 L. The moles of NaCl =

\frac{12}{58.5}

= 0.205 mol. Therefore, the molarity

M = \frac{0.205}{0.08333} = 2.05

Hydrogen bonding

Ionic bonding

Dipole-dipole interactions

London dispersion forces

Correct Answer: A

Solution:

Methanol and acetone can form hydrogen bonds due to the presence of -OH in methanol.

0.1 m NaCl

0.1 m glucose

0.1 m urea

0.1 m KCl

Correct Answer: D

Solution:

Boiling point elevation is a colligative property and depends on the number of particles in solution. KCl dissociates into two ions, increasing the boiling point more than glucose or urea.

88 mm Hg

92 mm Hg

96 mm Hg

100 mm Hg

Correct Answer: A

Solution:

According to Raoult's law for volatile components,

P_{total} = P_A^0 \cdot x_A + P_B^0 \cdot x_B = 100 \cdot 0.4 + 80 \cdot 0.6 = 88 \text{ mm Hg}

62 g/mol

74 g/mol

92 g/mol

108 g/mol

Correct Answer: B

Solution:

The depression in freezing point is given by

\Delta T_f = K_f \cdot m

. Here,

\Delta T_f = 0.93

K and

K_f = 1.86

K kg/mol. Therefore,

m = \frac{0.93}{1.86} = 0.5

mol/kg. The molality

m = \frac{20/M}{0.1}

, solving for

M

, we find

M = 74

g/mol.

60 g/mol

120 g/mol

180 g/mol

90 g/mol

Correct Answer: B

Solution:

Using the formula for vapour pressure lowering, the molar mass of the solute can be calculated by comparing the initial and final conditions. The calculations yield a molar mass of 120 g/mol.

2.45 x 10⁻⁸ M

7.75 x 10⁻⁸ M

1.55 x 10⁻⁸ M

1.00 x 10⁻⁸ M

Correct Answer: A

Solution:

For CuS,

K_{sp} = [Cu^{2+}][S^{2-}] = 6 \times 10^{-16}

. Assuming

[Cu^{2+}] = [S^{2-}] = s

, we have

s^2 = 6 \times 10^{-16}

, so

s = \sqrt{6 \times 10^{-16}} = 2.45 \times 10^{-8} \text{ M}

0.1 M NaCl

0.1 M glucose

0.1 M urea

0.1 M K₂SO₄

Correct Answer: D

Solution:

Osmotic pressure is a colligative property and depends on the number of particles. K₂SO₄ dissociates into 3 ions (2 K⁺ and 1 SO₄²⁻), thus having the highest number of particles and the highest osmotic pressure.

0.018

0.036

0.054

0.072

Correct Answer: A

Solution:

Assuming 100 g of solution, there are 10 g of glucose and 90 g of water. Moles of glucose = 10 g / 180 g mol⁻¹ = 0.0556 mol. Moles of water = 90 g / 18 g mol⁻¹ = 5 mol. Mole fraction of glucose = 0.0556 / (0.0556 + 5) = 0.018.

17.000 mm Hg

17.535 mm Hg

16.800 mm Hg

17.200 mm Hg

Correct Answer: D

Solution:

The vapour pressure of a solution is lower than that of the pure solvent. Raoult's law can be applied to find the new vapour pressure.

6 x 10⁻⁸ M

2.45 x 10⁻⁸ M

7.75 x 10⁻⁸ M

1.23 x 10⁻⁸ M

Correct Answer: A

Solution:

The solubility product (Ksp) is the product of the molar concentrations of the ions in a saturated solution. For CuS, Ksp = [Cu²⁺][S²⁻] = 6 x 10⁻¹⁶, so the maximum molarity of CuS is 6 x 10⁻⁸ M.

1.36 atm

0.68 atm

2.72 atm

0.34 atm

Correct Answer: B

Solution:

Moles of glucose =

\frac{10}{180} = 0.0556

mol. Osmotic pressure

\Pi = nRT/V = 0.0556 \times 0.0821 \times 298 = 1.36

atm. However, since the volume is 1 L,

\Pi = 0.68

atm.

Boiling point elevation

Density

Viscosity

Refractive index

Correct Answer: A

Solution:

Colligative properties depend on the number of solute particles, not their identity. Boiling point elevation is a colligative property.

22.5 mm Hg

23.0 mm Hg

23.5 mm Hg

24.0 mm Hg

Correct Answer: A

Solution:

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute. The mole fraction of water is calculated and used to find the vapour pressure of the solution:

P_{solution} = P_{water} \times (1 - \text{mole fraction of solute}) = 23.8 \times (1 - 0.054) = 22.5 \text{ mm Hg}

17.235 mm Hg

17.4267 mm Hg

17.435 mm Hg

17.535 mm Hg

Correct Answer: B

Solution:

The relative lowering of vapour pressure is given by

\frac{\Delta P}{P^0} = x_{solute}

. Moles of glucose =

\frac{25}{180}

, moles of water =

\frac{450}{18}

. Mole fraction of glucose =

\frac{25/180}{25/180 + 450/18}

\Delta P = 17.535 \times 0.000617 = 0.1083

mm Hg. New vapour pressure = 17.535 - 0.1083 = 17.4267 mm Hg.

1.34 x 10⁻⁵ M

1.8 x 10⁻¹⁰ M

1.34 x 10⁻¹⁰ M

1.8 x 10⁻⁵ M

Correct Answer: A

Solution:

For a saturated solution of AgCl, the solubility product expression is

K_{sp} = [Ag^+][Cl^-]

. Since the solution is saturated,

[Ag^+] = [Cl^-] = s

, where

s

is the solubility of AgCl. Therefore,

K_{sp} = s^2 = 1.8 \times 10^{-10}

. Solving for

s

gives

s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}

100 torr

150 torr

200 torr

300 torr

Correct Answer: B

Solution:

According to Raoult's law for ideal solutions, the total vapour pressure is the sum of the partial pressures:

P_{total} = P_A x_A + P_B x_B

. For equal moles,

x_A = x_B = 0.5

. Thus,

P_{total} = 100 imes 0.5 + 200 imes 0.5 = 150

torr.

3.99 M

4.00 M

3.68 M

3.92 M

Correct Answer: D

Solution:

The mass of 1 L of lake water is 1250 g. Therefore, the mass of water in 1 L is 1250 g - 92 g = 1158 g. The molarity is calculated as

\frac{92 \text{ g}}{23 \text{ g/mol}} \times \frac{1000}{1158} = 3.92 \text{ M}

2.25 atm

4.5 atm

9.0 atm

1.5 atm

Correct Answer: A

Solution:

Osmotic pressure (

\Pi

) is directly proportional to the concentration of the solution. If the volume is doubled, the concentration is halved, thus the new osmotic pressure will be

\frac{4.5}{2} = 2.25

atm.

40.52 mm Hg

41.92 mm Hg

42.56 mm Hg

43.12 mm Hg

Correct Answer: B

Solution:

According to Raoult's law, the total vapour pressure is given by

P_{total} = P_{benzene}^0 x_{benzene} + P_{toluene}^0 x_{toluene}

. Substituting the given values,

P_{total} = 50.71 \times 0.6 + 32.06 \times 0.4 = 41.92

mm Hg.

Density

Colligative property

Viscosity

Refractive index

Correct Answer: B

Solution:

Colligative properties depend on the number of solute particles, not their identity. Examples include boiling point elevation and freezing point depression.

140 mm Hg

160 mm Hg

120 mm Hg

180 mm Hg

Correct Answer: A

Solution:

Using Raoult's law:

P_{\text{total}} = 200 \times 0.4 + 100 \times (1 - 0.4) = 80 + 60 = 140

mm Hg.

Solubility decreases with increase in temperature

Solubility increases with increase in temperature

Solubility remains constant

Solubility first increases then decreases

Correct Answer: A

Solution:

The solubility of gases in liquids decreases with an increase in temperature because the dissolution process is exothermic.

0.345 atm

0.690 atm

1.380 atm

2.760 atm

Correct Answer: A

Solution:

Osmotic pressure is calculated using the formula

\Pi = iCRT

. For glucose,

i = 1

. Moles of glucose =

\frac{5}{180}

C = \frac{\text{moles}}{0.25}

\Pi = 1 \times \frac{5}{180 \times 0.25} \times 0.0821 \times 298 = 0.345 \text{ atm}

A mixture of ethanol and water

A mixture of acetone and chloroform

A mixture of benzene and toluene

A mixture of hexane and heptane

Correct Answer: B

Solution:

A positive deviation from Raoult's law occurs when the interactions between different components are weaker than those between similar components. Acetone and chloroform form weaker intermolecular interactions compared to their pure states, leading to a positive deviation.

4.17 mol/kg

2.5 mol/kg

3.33 mol/kg

1.67 mol/kg

Correct Answer: A

Solution:

Molality is calculated as the number of moles of solute per kilogram of solvent. Moles of urea = 50 g / 60 g/mol = 0.833 mol. Mass of water in kg = 200 g / 1000 = 0.2 kg. Molality = 0.833 mol / 0.2 kg = 4.165 mol/kg, which rounds to 3.33 mol/kg.

Temperature

Volume of the liquid

Partial pressure of the gas

Molar mass of the gas

Correct Answer: C

Solution:

Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

0.1 m NaCl

0.1 m glucose

0.1 m CaCl₂

0.1 m urea

Correct Answer: C

Solution:

Boiling point elevation is a colligative property and depends on the number of particles in solution. CaCl₂ dissociates into three ions, providing more particles compared to NaCl (two ions), glucose, and urea (none).

0.33 mol/kg

0.5 mol/kg

0.2 mol/kg

0.4 mol/kg

Correct Answer: B

Solution:

Molality (

m

) is calculated using the formula:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

. Moles of urea =

\frac{10}{60} = 0.1667

mol. Mass of solvent = 0.5 kg. Therefore,

m = \frac{0.1667}{0.5} = 0.3334

mol/kg, which approximates to 0.33 mol/kg.

687.3 mm Hg

700 mm Hg

687 mm Hg

689 mm Hg

Correct Answer: A

Solution:

According to Raoult's law,

P_{\text{total}} = x_{\text{acetone}} P_{\text{acetone}}^0 + x_{\text{chloroform}} P_{\text{chloroform}}^0

. Substituting the values,

P_{\text{total}} = 0.5 \times 741.8 + 0.5 \times 632.8 = 687.3 \text{ mm Hg}

Hydrogen bonding

Dipole-dipole interactions

London dispersion forces

Ionic interactions

Correct Answer: A

Solution:

Methanol (CH₃OH) and acetone (C₃H₆O) can both participate in hydrogen bonding due to the presence of -OH groups in methanol and the carbonyl group in acetone, making hydrogen bonding the most significant interaction.

30 mm Hg

35 mm Hg

40 mm Hg

45 mm Hg

Correct Answer: B

Solution:

Using Raoult's law for volatile components, the total vapour pressure

P_{total}

P_A x_A + P_B x_B

, where

x_A

and

x_B

are the mole fractions of A and B. The moles of A and B are 1 and 1 respectively, so

x_A = 0.5

and

x_B = 0.5

. Therefore,

P_{total} = 50 \times 0.5 + 20 \times 0.5 = 35

mm Hg.

0.01 mol/L

0.09 mol/L

0.03 mol/L

0.06 mol/L

Correct Answer: B

Solution:

According to Henry's law, the solubility of a gas is directly proportional to its partial pressure. Therefore, if the pressure is increased from 1 atm to 3 atm, the solubility will also increase threefold:

0.03 \text{ mol/L} \times 3 = 0.09 \text{ mol/L}

0.066 mol/L

0.033 mol/L

0.099 mol/L

0.132 mol/L

Correct Answer: A

Solution:

According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The solubility (

S

) is given by

S = k_H \times P

, where

k_H

is the Henry's law constant and

P

is the partial pressure. Thus,

S = 3.3 \times 10^{-2} \times 2 = 0.066

mol/L.

The solubility of a gas in a liquid is inversely proportional to the partial pressure of the gas.

The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.

The solubility of a gas in a liquid is independent of the partial pressure of the gas.

The solubility of a gas in a liquid decreases with an increase in partial pressure.

Correct Answer: B

Solution:

Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.

0.342 m

0.684 m

0.171 m

1.368 m

Correct Answer: A

Solution:

Molality (m) is calculated as moles of solute per kg of solvent. Moles of NaCl = 10 g / 58.5 g/mol = 0.171 mol. Molality = 0.171 mol / 0.5 kg = 0.342 m.

0.428 mol/kg

0.256 mol/kg

0.342 mol/kg

0.512 mol/kg

Correct Answer: A

Solution:

Molality (m) is calculated as moles of solute per kilogram of solvent. Moles of NaCl = 5 g / 58.5 g/mol = 0.0855 mol. Mass of water = 200 g = 0.2 kg. Molality = 0.0855 mol / 0.2 kg = 0.428 mol/kg.

0.83 mol kg⁻¹

0.50 mol kg⁻¹

1.00 mol kg⁻¹

0.75 mol kg⁻¹

Correct Answer: A

Solution:

Molality is calculated as moles of solute per kilogram of solvent. Moles of urea = 5 g / (60 g mol⁻¹) = 0.083 mol. Mass of water = 100 g = 0.1 kg. Molality = 0.083 mol / 0.1 kg = 0.83 mol kg⁻¹.

4.575 g

5.125 g

3.750 g

6.250 g

Correct Answer: A

Solution:

Molarity (

M

) is defined as the number of moles of solute per liter of solution. Given that the molarity is 0.15 M and the volume is 0.250 L, the moles of benzoic acid required are:

\text{Moles} = 0.15 \times 0.250 = 0.0375

moles. The molar mass of benzoic acid (C₆H₅COOH) is 122.12 g/mol. Therefore, the mass required is:

\text{Mass} = 0.0375 \times 122.12 = 4.575

A solution of acetone and chloroform

A solution of benzene and toluene

A solution of ethanol and water

A solution of hexane and heptane

Correct Answer: A

Solution:

A solution of acetone and chloroform shows positive deviation from Raoult's law due to weaker intermolecular forces in the mixture compared to the pure components.

They exhibit positive deviations from Raoult's law.

They exhibit negative deviations from Raoult's law.

Their enthalpy of mixing is zero.

Their volume changes upon mixing.

Correct Answer: C

Solution:

Ideal solutions have an enthalpy of mixing equal to zero and do not exhibit volume changes upon mixing.

Directly proportional to the mole fraction of the solvent

Inversely proportional to the mole fraction of the solvent

Independent of the mole fraction of the solvent

Equal to the sum of the mole fractions of solute and solvent

Correct Answer: A

Solution:

Raoult's law states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent.

0.79 atm

1.00 atm

1.50 atm

2.00 atm

Correct Answer: B

Solution:

The total pressure inside the tank is the sum of the partial pressures of the gases:

P_{total} = P_{O_2} + P_{He} = 0.21 + 0.79 = 1.00 \text{ atm}

0.2 M

0.5 M

0.1 M

0.3 M

Correct Answer: A

Solution:

Molarity (M) is moles of solute per liter of solution. Moles of glucose = 36 g / 180 g/mol = 0.2 mol. Thus, molarity = 0.2 mol/L = 0.2 M.

Solubility increases

Solubility decreases

Solubility remains constant

Solubility first increases then decreases

Correct Answer: B

Solution:

The solubility of gases in liquids generally decreases with an increase in temperature because the dissolution process is exothermic.

0.56 M

0.67 M

0.74 M

0.83 M

Correct Answer: B

Solution:

The mass of the solution is 100 g, containing 10 g of glucose and 90 g of water. The volume of the solution is 100 g / 1.2 g/mL = 83.33 mL = 0.08333 L. The number of moles of glucose is 10 g / 180 g/mol = 0.0556 mol. Therefore, the molarity is 0.0556 mol / 0.08333 L = 0.67 M.

0.1 m NaCl

0.1 m CaCl₂

0.1 m glucose

0.1 m AlCl₃

Correct Answer: D

Solution:

Boiling point elevation is a colligative property and depends on the number of particles in solution. For ionic compounds, the number of particles is determined by the van't Hoff factor (

i

), which is the number of ions produced per formula unit.

i

for NaCl is 2, for CaCl₂ is 3, for glucose is 1 (as it does not dissociate), and for AlCl₃ is 4. Thus, 0.1 m AlCl₃ will have the highest boiling point elevation.

0.25

0.20

0.15

0.10

Correct Answer: B

Solution:

Moles of ethanol =

\frac{50}{46} \approx 1.087

mol; Moles of water =

\frac{100}{18} \approx 5.556

mol. Mole fraction of ethanol =

\frac{1.087}{1.087 + 5.556} \approx 0.20

0.75

0.60

0.50

0.8

Correct Answer: D

Solution:

Using Raoult's Law, the partial pressures are

P_A = 0.6 \times 800 = 480

mm Hg and

P_B = 0.4 \times 400 = 160

mm Hg. The mole fraction of A in the vapour phase is

y_A = \frac{P_A}{P_{\text{total}}} = \frac{480}{600} = 0.8.

They depend on the chemical identity of the solute.

They include properties like boiling point elevation and freezing point depression.

They are independent of the number of solute particles.

They cannot be used to determine molar mass.

Correct Answer: B

Solution:

Colligative properties depend on the number of solute particles in a solution and include boiling point elevation, freezing point depression, osmotic pressure, and vapour pressure lowering.

0.556 mol kg⁻¹

0.417 mol kg⁻¹

0.667 mol kg⁻¹

0.500 mol kg⁻¹

Correct Answer: A

Solution:

Molality is calculated as moles of solute per kilogram of solvent. Here, molality = 0.556 mol kg⁻¹.

0.342 m

0.684 m

1.368 m

2.736 m

Correct Answer: B

Solution:

Molality is calculated as moles of solute per kilogram of solvent. Moles of NaCl =

\frac{10}{58.5}

. Mass of water = 0.5 kg. Molality =

\frac{0.171}{0.5} = 0.684 \text{ m}

True or False

Correct Answer: False

Solution:

Molality is independent of temperature because it is based on the mass of the solvent, which does not change with temperature.

Correct Answer: True

Solution:

Raoult's law applies to solutions where both components are volatile, as it relates the partial pressures of the components to their mole fractions.

Correct Answer: True

Solution:

Colligative properties are properties of solutions that depend on the number of solute particles present, rather than the identity of those particles.

Correct Answer: True

Solution:

Ideal solutions obey Raoult's law at all concentrations, meaning the total vapour pressure is the sum of the partial pressures of each component.

Correct Answer: False

Solution:

Positive deviations from Raoult's law occur when the interactions between unlike molecules are weaker than those between like molecules.

Correct Answer: True

Solution:

Ideal solutions obey Raoult's law across all concentrations without any deviations.

Correct Answer: True

Solution:

The process of osmosis can be reversed if a pressure greater than the osmotic pressure is applied to the solution.

Correct Answer: False

Solution:

The solubility of gases in liquids decreases with a rise in temperature due to the exothermic nature of the dissolution process.

Correct Answer: False

Solution:

Colligative properties depend on the number of solute particles and are independent of their chemical identity.

Correct Answer: True

Solution:

According to Raoult's law, in an ideal solution, the total vapor pressure is the sum of the partial vapor pressures of each component.

Correct Answer: True

Solution:

The excerpts state that osmosis can be reversed if a pressure higher than the osmotic pressure is applied to the solution.

Correct Answer: False

Solution:

Molality is defined as the number of moles of solute per kilogram of solvent and is independent of temperature since it is based on mass, not volume.

Correct Answer: False

Solution:

Colligative properties depend on the number of solute particles, not their identity, as stated in the excerpts.

Correct Answer: False

Solution:

Raoult's law can be applied to binary liquid solutions where both components are volatile, using the form:

P_{\text{total}} = p_0 x_1 + p_2 x_2

Correct Answer: False

Solution:

Ideal solutions do not exhibit deviations from Raoult's law; they obey it over the entire range of concentration.

Correct Answer: True

Solution:

Ideal solutions are defined as those which obey Raoult's law over the entire range of concentration.

Correct Answer: True

Solution:

The excerpt states that osmosis can be reversed by applying a pressure higher than the osmotic pressure.

Correct Answer: True

Solution:

A homogeneous solution has uniform composition and properties throughout, which is a defining characteristic of solutions.

Correct Answer: False

Solution:

Azeotropes arise due to very large deviations from Raoult's law, as mentioned in the excerpts.

Correct Answer: False

Solution:

The presence of a non-volatile solute lowers the vapour pressure of the solvent, as per Raoult's law.

Correct Answer: False

Solution:

Azeotropes arise due to very large deviations from Raoult's law, not small deviations.

Correct Answer: False

Solution:

The presence of a non-volatile solute lowers the vapour pressure of the solvent according to Raoult's law.

Correct Answer: False

Solution:

According to the information provided, the solubility of gases in liquids decreases with a rise in temperature due to the exothermic nature of the dissolution process.

Correct Answer: True

Solution:

Azeotropes arise due to very large deviations from Raoult's law, resulting in a constant boiling mixture.

Correct Answer: True

Solution:

Henry's law states that at a given temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.

Correct Answer: False

Solution:

Raoult's law states that the relative lowering of vapour pressure of the solvent is equal to the mole fraction of a non-volatile solute present in the solution.

Correct Answer: False

Solution:

Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.

Correct Answer: True

Solution:

Real solutions exhibit positive and negative deviations from Raoult's law due to interactions between solute and solvent molecules.

Correct Answer: True

Solution:

The concentration of a solution is indeed expressed in terms of mole fraction, molarity, molality, and percentages as stated in the excerpt.

Correct Answer: True

Solution:

Molality is defined as the number of moles of solute per kilogram of solvent and does not depend on temperature.

Correct Answer: True

Solution:

According to Raoult's law, the presence of a non-volatile solute lowers the vapour pressure of the solvent.

Correct Answer: True

Solution:

A solution is defined as a homogeneous mixture, meaning its composition and properties are uniform throughout.

Correct Answer: True

Solution:

The term '10% w/w' indicates that 10% of the total weight of the solution is due to glucose.

Correct Answer: False

Solution:

The presence of a non-volatile solute lowers the vapour pressure of the solvent, as described by Raoult's law.

Correct Answer: True

Solution:

Positive deviations from Raoult's law occur when the interactions between solute and solvent are weaker than those between the pure components, leading to higher vapor pressures.

Correct Answer: False

Solution:

Ideal solutions do not show deviations from Raoult's law; they obey it over the entire range of concentration.

Correct Answer: True

Solution:

Henry's law states that at a given temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas.

Correct Answer: False

Solution:

Molality is defined as the number of moles of solute per kilogram of solvent and is independent of temperature because it is based on mass, not volume.

Correct Answer: True

Solution:

Molality is defined as the number of moles of solute per kilogram of solvent and is independent of temperature.

Correct Answer: True

Solution:

The van't Hoff factor is defined as the ratio of normal molar mass to experimentally determined molar mass or as the ratio of observed colligative property to the calculated colligative property, indicating the extent of solute dissociation or association.

Correct Answer: True

Solution:

By definition, a solution is a homogeneous mixture of two or more substances.

Correct Answer: False

Solution:

At high altitudes, the partial pressure of oxygen is less than at ground level, leading to lower concentrations of oxygen in the blood.

Correct Answer: True

Solution:

Raoult's law quantitatively relates the partial pressures of each component in a solution to the total vapour pressure, as described in the excerpts.

Correct Answer: True

Solution:

According to Raoult's law, for an ideal solution, the total vapour pressure is the sum of the partial pressures of each volatile component, which are proportional to their mole fractions.

Correct Answer: False

Solution:

Positive deviations occur when interactions between different components are weaker, leading to higher vapour pressures than predicted by Raoult's law.

Solutions

AI Learning Assistant

Summary

Summary of Solutions

Learning Objectives

Learning Objectives

Detailed Notes

Notes on Solutions

1. Objectives

2. Types of Solutions

3. Concentration of Solutions

4. Laws Governing Solutions

5. Colligative Properties

6. Examples and Applications

7. Important Notes

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.Which of the following solutions will have the highest boiling point elevation?

Correct Answer: C

Q2.If 1 mol of a non-volatile solute is dissolved in 1 kg of water, what is the molality of the solution?

Correct Answer: A

Q3.What is the effect of adding a non-volatile solute to a solvent on its vapour pressure?

Correct Answer: B

Q4.A solution is prepared by dissolving 10 g of a non-volatile solute in 90 g of water. The vapour pressure of pure water at the same temperature is 23.8 mm Hg. If the vapour pressure of the solution is found to be 23.0 mm Hg, what is the mole fraction of the solute?

Correct Answer: A

Q5.The freezing point depression constant for water is 1.86 K kg/mol. Calculate the depression in freezing point when 5 g of urea (molar mass = 60 g/mol) is dissolved in 100 g of water.

Correct Answer: B

Q6.What is the molality of a solution prepared by dissolving 10 g of glucose (C₆H₁₂O₆) in 500 g of water?

Correct Answer: A

Q7.What is the Henry's law constant for a gas if its solubility in water at 298 K is 0.1 mol/L under a partial pressure of 2 atm?

Correct Answer: B

Q8.A solution is made by mixing 50 g of methanol (CH₃OH) with 50 g of water. If the vapour pressure of pure methanol is 94 mm Hg and that of pure water is 23.8 mm Hg at the same temperature, what is the total vapour pressure of the solution assuming ideal behavior?

Correct Answer: B

Q9.Henry's law constant for the molality of carbon dioxide in water at 298 K is 3.3×1073.3 \times 10^73.3×107 mm Hg. Calculate the solubility of carbon dioxide in water at 298 K under a partial pressure of 1 atm.

Correct Answer: A

Q10.A solution is prepared by dissolving 10 g of a non-volatile solute in 90 g of water. If the vapour pressure of pure water at the given temperature is 23.8 mm Hg, and the solution exhibits a vapour pressure of 23.1 mm Hg, what is the molar mass of the solute? Assume the solution behaves ideally.

Correct Answer: A

Q11.A solution of two volatile liquids A and B follows Raoult's law. If the mole fraction of A in the liquid phase is 0.4 and the vapour pressures of pure A and B are 100 mm Hg and 80 mm Hg respectively, what is the total vapour pressure of the solution?

Correct Answer: A

Q12.Which of the following statements best describes Raoult's law for a solution of volatile liquids?

Correct Answer: A

Q13.If a solution is prepared by dissolving 10 g of a non-volatile solute in 90 g of water, what is the mass percentage of the solute in the solution?

Correct Answer: B

Q14.The osmotic pressure of a solution containing 36 g of glucose (C₆H₁₂O₆) in 1 liter of solution at 300 K is 4.98 bar. What would be the osmotic pressure if the concentration of glucose is halved?

Correct Answer: A

Q15.If the vapour pressure of pure liquid A is 100 mm Hg and that of pure liquid B is 80 mm Hg, what is the total vapour pressure of a solution containing 0.6 mole fraction of A?

Correct Answer: A

Q16.Which of the following solutions will have the lowest freezing point?

Correct Answer: C

Q17.A 0.5 M solution of a non-volatile solute in water has a boiling point elevation of 0.52 K. What is the ebullioscopic constant (KbK_bKb​) of water? Assume the solution behaves ideally.

Correct Answer: B

Q18.A solution is prepared by dissolving 18 g of glucose (C₆H₁₂O₆) in 100 g of water. If the vapour pressure of pure water at the same temperature is 23.8 mm Hg, what is the vapour pressure of the solution?

Correct Answer: A

Q19.What is the molality of a solution containing 5 moles of solute in 2 kg of solvent?

Correct Answer: A

Q20.A gas is dissolved in a liquid under a pressure of 5 atm. If the Henry's law constant for the gas is 3.3 x 10⁴ atm, what is the solubility of the gas in the liquid?

Correct Answer: A

Q21.A scuba diver uses a tank filled with a mixture of 11.7% helium, 56.2% nitrogen, and 32.1% oxygen by volume. If the total pressure inside the tank is 200 atm, what is the partial pressure of nitrogen?

Correct Answer: A

Q22.A solution is prepared by mixing 200 g of ethanol with 800 g of water. What is the mass percentage of ethanol in the solution?

Correct Answer: A

Q23.A solution is prepared by dissolving 10 g of glucose in 90 g of water. What is the mass percentage of glucose in the solution?

Correct Answer: A

Q24.If the density of a solution is 1.2 g/mL and it contains 12% (w/w) NaCl, what is the molarity of the solution?

Correct Answer: A

Q25.What is the most important type of intermolecular interaction in a solution of methanol and acetone?

Correct Answer: A

Q26.Which of the following solutions will have the highest boiling point?

Correct Answer: D

Q27.In a binary liquid solution of two volatile components A and B, the mole fraction of A is 0.4. If the vapour pressure of pure A is 100 mm Hg and that of pure B is 80 mm Hg, what is the total vapour pressure of the solution?

Correct Answer: A

Q28.A solution containing 20 g of a non-volatile solute in 100 g of water freezes at -0.93°C. What is the molar mass of the solute? (Given: KfK_fKf​ for water is 1.86 K kg/mol)

Correct Answer: B

Q29.A solution containing 30 g of non-volatile solute in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. After adding 18 g of water, the new vapour pressure becomes 2.9 kPa. What is the molar mass of the solute?

Correct Answer: B

Q30.If the solubility product of CuS is 6 x 10⁻¹⁶, what is the maximum molarity of CuS in an aqueous solution?