Solid State - 50 Subjective Questions for HSC Revision

Chapter 1: Solid State - 50 Subjective Questions for HSC Revision | Chemca.in

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Chapter 1: Solid State

Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions

Click on any question to reveal its answer.

1. Define crystalline solids. State any two characteristics.

Definition: A crystalline solid is a homogeneous solid in which the constituent particles (atoms, ions, or molecules) are arranged in a definite, repeating three-dimensional pattern over a long range.

Characteristics:

They have a sharp and well-defined melting point.
They exhibit anisotropy (physical properties like refractive index vary in different directions).

2. What are amorphous solids? Give two examples.

Definition: Amorphous solids are those in which the constituent particles are not arranged in any regular, repeating pattern over a long range. They possess short-range order.

Examples: Glass, rubber, plastics.

3. Distinguish between crystalline and amorphous solids. (Any 4 points)

Property	Crystalline Solids	Amorphous Solids
Arrangement	Regular, long-range order	Irregular, short-range order
Melting Point	Sharp melting point	Melt over a range of temperatures
Isotropy	Anisotropic	Isotropic
Nature	True solids	Pseudo solids or supercooled liquids

4. Explain the term 'Isomorphism' with an example.

Isomorphism: When two or more different chemical substances crystallize in the same crystalline structural form, the phenomenon is called isomorphism. Such substances have the same atomic ratio.

Example: Sodium fluoride ($NaF$) and Magnesium oxide ($MgO$) are isomorphous (atomic ratio 1:1).

5. What is 'Polymorphism'? Give an example.

Polymorphism: The phenomenon where a single substance can exist in two or more different crystalline forms under different conditions of temperature and pressure is called polymorphism. In elements, it is called allotropy.

Example: Carbon exists as diamond, graphite, and fullerenes. Calcium carbonate ($CaCO_3$) exists as calcite and aragonite.

6. Define Anisotropy and Isotropy.

Anisotropy: The ability of a crystalline solid to exhibit different values of physical properties (like electrical resistance, refractive index, thermal expansion) when measured in different directions is called anisotropy.

Isotropy: The property of amorphous solids where physical properties have the same value regardless of the direction of measurement is called isotropy.

7. Why is glass considered a supercooled liquid?

Glass is an amorphous solid. Like liquids, it has a tendency to flow, though very slowly. This is observed in old buildings where glass panes are slightly thicker at the bottom due to gravity. Because of this sluggish flow property, glass is termed a pseudo solid or a supercooled liquid.

8. Classify crystalline solids based on intermolecular forces.

Based on intermolecular forces, crystalline solids are classified into four types:

Molecular Solids: Constituent particles are molecules held by Van der Waals forces. (e.g., Ice, solid $CO_2$)
Ionic Solids: Constituent particles are ions held by strong electrostatic forces. (e.g., $NaCl$)
Metallic Solids: Positive metal ions in a sea of delocalized electrons held by metallic bonds. (e.g., Copper, Iron)
Covalent (Network) Solids: Atoms held together by continuous covalent bonds forming a network. (e.g., Diamond, Quartz)

9. Write a short note on Ionic Solids.

Ionic Solids consist of positively (cations) and negatively (anions) charged ions. They are held together by strong coulombic or electrostatic forces of attraction.

Properties:

They are hard and brittle.
They have very high melting and boiling points due to strong ionic bonds.
They are electrical insulators in the solid state but become good conductors when melted (molten state) or dissolved in water because the ions become free to move.

10. Why are covalent network solids exceptionally hard and have very high melting points?

In covalent network solids (like diamond), atoms are linked to adjacent atoms by strong and highly directional covalent bonds throughout the entire crystal, forming a giant three-dimensional network structure. Breaking this massive network requires immense thermal energy, resulting in exceptional hardness and extremely high melting points.

11. Define: (a) Crystal lattice (Space lattice) (b) Unit cell.

(a) Crystal Lattice: A regular, three-dimensional arrangement of points in space representing constituent particles (atoms, ions, or molecules) of a crystal is called a crystal lattice.

(b) Unit Cell: The smallest repeating structural unit of a crystal lattice, which when repeated in different directions generates the entire crystal, is called a unit cell.

12. What are the parameters that characterize a unit cell?

A unit cell is characterized by 6 parameters:

Edge lengths: Three edge lengths $a, b, c$ along the three crystallographic axes. They may or may not be mutually perpendicular.
Interaxial angles: Three angles $\alpha, \beta, \gamma$ between the edges. ($\alpha$ between $b$ and $c$, $\beta$ between $a$ and $c$, $\gamma$ between $a$ and $b$).

13. What are Bravais lattices? How many are possible?

Bravais Lattices are the distinct three-dimensional arrangements of lattice points that describe the repeating structure of a crystal. The French mathematician Auguste Bravais proved that there are exactly 14 possible three-dimensional lattices, grouped into 7 crystal systems.

14. Differentiate between Simple, Body-Centered, and Face-Centered cubic unit cells.

Simple Cubic (SC): Particles are present only at the 8 corners of the cube.
Body-Centered Cubic (BCC): Particles are present at the 8 corners plus one particle exactly at the body center of the cube.
Face-Centered Cubic (FCC): Particles are present at the 8 corners plus one particle at the center of each of the 6 faces of the cube.

15. Calculate the total number of atoms per unit cell in a Simple Cubic (SC) lattice.

In a simple cubic unit cell, atoms are present only at the 8 corners.

Each corner atom is shared by 8 adjacent unit cells. Thus, its contribution to one unit cell is $\frac{1}{8}$.

Total number of atoms ($Z$) = $8 \text{ corners} \times \frac{1}{8} \text{ atom per corner} = 1 \text{ atom}$.

16. Calculate the total number of atoms per unit cell in a Body-Centered Cubic (BCC) lattice.

A BCC unit cell has atoms at the 8 corners and 1 atom at the body center.

Contribution of 8 corner atoms = $8 \times \frac{1}{8} = 1$ atom.
Contribution of the atom at the body center (unshared) = $1 \times 1 = 1$ atom.

Total number of atoms ($Z$) = $1 + 1 = 2 \text{ atoms}$.

17. Calculate the total number of atoms per unit cell in a Face-Centered Cubic (FCC) lattice.

An FCC unit cell has atoms at 8 corners and 6 face centers.

Contribution of 8 corner atoms = $8 \times \frac{1}{8} = 1$ atom.
Each face center atom is shared by 2 unit cells. Contribution of 6 face atoms = $6 \times \frac{1}{2} = 3$ atoms.

Total number of atoms ($Z$) = $1 + 3 = 4 \text{ atoms}$.

18. What is Coordination Number in a crystal lattice?

Coordination Number: It is defined as the number of nearest neighboring particles (atoms, ions, or molecules) that are in direct contact with a given central particle in a crystal lattice.

Examples: In Simple Cubic, it is 6. In BCC, it is 8. In FCC/HCP/CCP, it is 12.

19. Differentiate between Hexagonal Close Packing (hcp) and Cubic Close Packing (ccp).

HCP (AB AB AB... type): The spheres of the third layer are aligned exactly above the spheres of the first layer. It generates a hexagonal unit cell. Tetrahedral voids of the second layer are covered by the third layer.
CCP (ABC ABC... type): The spheres of the third layer are not aligned with either the first or the second layer. It aligns with the first layer only at the fourth layer. It generates an FCC unit cell. Octahedral voids are covered by the third layer.

Both have a coordination number of 12 and a packing efficiency of 74%.

20. Define Packing Efficiency. State the packing efficiency of SC, BCC, and FCC.

Packing Efficiency: It is the percentage of total space in the unit cell that is occupied by the constituent particles (spheres).

$\text{Packing Efficiency} = \frac{\text{Volume occupied by particles}}{\text{Total volume of unit cell}} \times 100$

Simple Cubic (SC): $52.4\%$
Body-Centered Cubic (BCC): $68\%$
Face-Centered Cubic (FCC / ccp / hcp): $74\%$

21. Derive the packing efficiency of a Simple Cubic unit cell.

In a simple cubic cell, particles touch each other along the edges.

Let edge length = $a$, radius of particle = $r$. Therefore, $a = 2r \Rightarrow r = a/2$.

Volume of the cubic unit cell = $a^3 = (2r)^3 = 8r^3$.

Number of atoms in SC cell ($Z$) = $1$.

Volume of 1 atom (sphere) = $\frac{4}{3}\pi r^3$.

$\text{Packing Efficiency} = \frac{\text{Volume of 1 atom}}{\text{Volume of unit cell}} \times 100$

$= \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100 = \frac{\pi}{6} \times 100 \approx 52.36\%$.

22. Derive the relationship between atomic radius (r) and edge length (a) for a BCC lattice.

In a BCC unit cell, atoms touch each other along the body diagonal.

Length of the body diagonal in a cube of edge '$a$' is $\sqrt{3}a$.

Along the body diagonal, there are two corner radii and the full diameter of the center atom, so the length equals $4r$.

Therefore, $4r = \sqrt{3}a$

$r = \frac{\sqrt{3}}{4} a$

23. Derive the relationship between atomic radius (r) and edge length (a) for an FCC lattice.

In an FCC unit cell, atoms touch each other along the face diagonal.

Length of the face diagonal in a cube of edge '$a$' is $\sqrt{2}a$.

Along the face diagonal, there is one full atom and two half atoms from the corners touching, which equals $4r$.

Therefore, $4r = \sqrt{2}a$

$r = \frac{\sqrt{2}}{4} a = \frac{1}{2\sqrt{2}} a$ or $a = 2\sqrt{2}r$.

24. Define Tetrahedral Void and Octahedral Void.

Tetrahedral Void: The empty space left between four mutually touching spheres, where their centers form a regular tetrahedron, is called a tetrahedral void.

Octahedral Void: The empty space surrounded by six touching spheres, formed when a triangular void of one layer overlaps with a reversed triangular void of another layer, is called an octahedral void.

25. What is the relation between the number of constituent particles and the number of tetrahedral and octahedral voids in a close-packed structure?

Let the number of close-packed spheres (constituent particles) in a lattice be $N$.

The number of Octahedral voids generated = $N$.
The number of Tetrahedral voids generated = $2N$.

Therefore, total voids = $3N$.

26. State the formula to calculate the density of a unit cell. Explain the terms involved.

The density ($\rho$) of a unit cell is given by:

$$ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} $$

Where:

$Z$ = Number of atoms per unit cell (1 for SC, 2 for BCC, 4 for FCC)
$M$ = Molar mass of the substance (g/mol)
$a$ = Edge length of the unit cell (in cm)
$N_A$ = Avogadro's number ($6.022 \times 10^{23} \text{ mol}^{-1}$)
$\rho$ = Density (g/cm³)

27. What are crystal defects (imperfections)? Mention the main types of point defects.

Crystal Defects: Any deviation from the perfectly ordered arrangement of constituent particles in a crystal is called a crystal defect or imperfection. These occur naturally during crystallization, especially when the process is fast.

Main types of Point Defects:

Stoichiometric Defects (Vacancy, Interstitial, Schottky, Frenkel)
Non-Stoichiometric Defects (Metal excess, Metal deficiency)
Impurity Defects

28. Explain Schottky Defect with an example.

Schottky Defect: It is a type of point defect in ionic solids that occurs when equal numbers of cations and anions are missing from their regular lattice sites, creating vacancies. This maintains the electrical neutrality of the crystal.

Conditions: Observed in highly ionic compounds with high coordination numbers and where cation and anion are of similar size.

Consequence: Decreases the density of the crystal.

Example: $NaCl, KCl, CsCl, AgBr$.

29. Explain Frenkel Defect with an example.

Frenkel Defect: It occurs when an ion (usually the smaller cation) is displaced from its normal lattice site and occupies an interstitial site between the lattice points.

Conditions: Observed in ionic compounds with low coordination numbers and a large difference in the size of cations and anions.

Consequence: Does not change the density of the crystal (mass and volume remain same).

Example: $ZnS, AgCl, AgBr, AgI$.

30. Distinguish between Schottky and Frenkel defects.

Schottky Defect	Frenkel Defect
Equal number of cations and anions are missing from lattice sites.	An ion leaves its site and occupies an interstitial site.
Decreases the density of the crystal.	Density of the crystal remains unchanged.
Occurs when sizes of cation and anion are almost similar.	Occurs when there is a large difference in size between ions.
Found in high coordination number crystals.	Found in low coordination number crystals.

31. Name a compound that exhibits both Schottky and Frenkel defects.

Silver Bromide ($AgBr$) exhibits both Schottky and Frenkel defects.

32. What are F-centers? How do they affect the properties of a crystal?

F-centers (Farbenzemtrum or Color Centers): In a metal excess defect caused by anion vacancies, an electron is trapped in the vacant anion lattice site to maintain electrical neutrality. This electron-occupied site is called an F-center.

Effect: They impart color to the crystal by absorbing visible light to excite the unpaired electron. For example, excess Sodium makes $NaCl$ yellow, excess Potassium makes $KCl$ violet/lilac. They also make the crystal paramagnetic and slightly electrically conductive.

33. Explain why ZnO turns yellow on heating.

Zinc oxide ($ZnO$) is white at room temperature. On heating, it loses oxygen gas and leaves behind $Zn^{2+}$ ions and electrons:

$$ ZnO \xrightarrow{\text{heat}} Zn^{2+} + \frac{1}{2}O_2 + 2e^- $$

The $Zn^{2+}$ ions move to interstitial sites, and the electrons move to neighboring interstitial sites. This creates a metal excess defect. The trapped electrons absorb radiation in the visible region, making the crystal appear yellow.

34. What is a metal deficiency defect? Give an example.

This defect occurs when the metal shows variable valency. A metal cation is missing from its lattice site, and to maintain electrical neutrality, an adjacent metal ion acquires a higher positive charge.

Example: In Ferrous oxide ($FeO$), some $Fe^{2+}$ ions are missing. To balance the charge, two other $Fe^{2+}$ ions become $Fe^{3+}$ ions. As a result, its actual formula ranges from $Fe_{0.93}O$ to $Fe_{0.96}O$.

35. Explain impurity defect with the example of $NaCl$ doped with $SrCl_2$.

When molten $NaCl$ containing a little amount of $SrCl_2$ is crystallized, some of the sites of $Na^+$ ions are occupied by $Sr^{2+}$ ions. Each $Sr^{2+}$ ion replaces two $Na^+$ ions (to maintain electrical neutrality). It occupies the site of one $Na^+$ ion, leaving the other site vacant. Hence, the number of cation vacancies produced is equal to the number of $Sr^{2+}$ ions added.

36. Explain Band Theory for classifying solids as conductors, insulators, and semiconductors.

Band theory describes the energy levels in a solid as continuous bands: the Valence Band (VB, highest occupied band) and the Conduction Band (CB, lowest unoccupied band). The energy gap between them is the Forbidden Energy Gap ($E_g$).

Conductors: The VB and CB overlap, or the VB is partially filled. Electrons move freely, hence excellent conductivity ($E_g = 0$).
Insulators: Large energy gap ($E_g > 3$ eV) between full VB and empty CB. Electrons cannot cross, so they do not conduct.
Semiconductors: Small energy gap ($E_g < 3$ eV). At absolute zero, they are insulators. At room temperature, some electrons gain enough thermal energy to jump from VB to CB, causing slight conductivity.

37. What is doping? Why is it done?

Doping: The process of intentionally adding an appropriate amount of a suitable impurity to a pure intrinsic semiconductor to increase its electrical conductivity is called doping.

Reason: Intrinsic semiconductors (like pure Si or Ge) have very low conductivity at room temperature, making them useless for practical electronic devices. Doping creates extrinsic semiconductors with highly tunable and increased conductivity.

38. Explain n-type semiconductors.

When a Group 14 element (like Si or Ge, which has 4 valence electrons) is doped with a Group 15 element (like P, As, Sb, which has 5 valence electrons), four out of five electrons form covalent bonds with the host lattice. The fifth extra electron becomes delocalized and serves to conduct electricity. Because conductivity is due to negative charge carriers (electrons), it is called an n-type (negative type) semiconductor.

39. Explain p-type semiconductors.

When a Group 14 element (like Si or Ge) is doped with a Group 13 element (like B, Al, Ga, which has 3 valence electrons), the dopant can only form three covalent bonds. This creates a missing electron in the fourth bond, leaving a 'hole' (electron vacancy). These holes act as positive charge carriers that move through the lattice. Hence, it is called a p-type (positive type) semiconductor.

40. What is Paramagnetism? Give examples.

Paramagnetism: Substances that are weakly attracted by a magnetic field are called paramagnetic substances. This occurs when atoms/ions/molecules contain one or more unpaired electrons. They are magnetized in a magnetic field in the same direction, but lose their magnetism in the absence of the magnetic field.

Examples: $O_2, Cu^{2+}, Fe^{3+}, Cr^{3+}$.

41. What is Diamagnetism? Give examples.

Diamagnetism: Substances that are weakly repelled by a magnetic field are called diamagnetic substances. This occurs when all electrons in the substance are paired. The pairing cancels out their magnetic moments.

Examples: $H_2O, NaCl$, Benzene ($C_6H_6$), $N_2$.

42. What is Ferromagnetism? Give examples.

Ferromagnetism: Substances that are very strongly attracted by a magnetic field and can be permanently magnetized are called ferromagnetic substances. In solid state, the metal ions of these substances are grouped together into small regions called "domains". Each domain acts as a tiny magnet. In an unmagnetized piece, domains are randomly oriented. When placed in a magnetic field, all domains align in the direction of the field permanently.

Examples: Iron ($Fe$), Cobalt ($Co$), Nickel ($Ni$), Gadolinium ($Gd$), $CrO_2$.

43. Define Anti-ferromagnetism. Give an example.

Anti-ferromagnetism: Substances like $MnO$ have domain structures similar to ferromagnetic substances, but their domains are oppositely oriented in exactly equal numbers. The opposite magnetic moments perfectly cancel each other out, resulting in a zero net magnetic moment.

Example: Manganese oxide ($MnO$).

44. Define Ferrimagnetism. Give an example.

Ferrimagnetism: This is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers. They have a small net magnetic moment and are weakly attracted to a magnetic field compared to ferromagnetic substances. They lose ferrimagnetism on heating and become paramagnetic.

Example: Magnetite ($Fe_3O_4$), Ferrites like $MgFe_2O_4$.

45. What is Curie Temperature?

Curie Temperature ($T_c$): It is the specific temperature above which a ferromagnetic substance loses its ferromagnetism and becomes paramagnetic due to the randomization of its domains by thermal agitation.

46. Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of the copper atom?

For an FCC lattice, the relationship between edge length ($a$) and radius ($r$) is:

$a = 2\sqrt{2}r \Rightarrow r = \frac{a}{2\sqrt{2}}$

Given $a = 361$ pm.

$r = \frac{361}{2 \times 1.414} = \frac{361}{2.828} \approx 127.6 \text{ pm}$.

The radius of the copper atom is approximately 127.6 pm.

47. Why are ionic solids hard and brittle?

Ionic solids are hard because they consist of positive and negative ions held together by strong electrostatic (coulombic) forces of attraction arranged in a fixed 3D lattice. High energy is needed to break these bonds.

They are brittle because when a shearing force is applied, layers of ions slide over one another. This brings ions of the same charge close together, causing strong repulsion which shatters the crystal.

48. What makes diamond a good thermal conductor but an electrical insulator?

In diamond, each carbon atom is covalently bonded to four other carbon atoms in a rigid tetrahedral network.

It is an electrical insulator because there are no free or delocalized electrons available to carry a charge; all valence electrons are tied up in strong $C-C$ bonds.

It is a good thermal conductor because its rigid, continuous covalent network rapidly transfers thermal vibrations (phonons) throughout the crystal without scattering.

49. How do we determine the number of atoms in an unknown cubic unit cell from its density?

Using the density formula: $\rho = \frac{Z \cdot M}{a^3 \cdot N_A}$

We can rearrange it to find $Z$ (number of atoms per unit cell):

$$ Z = \frac{\rho \cdot a^3 \cdot N_A}{M} $$

By measuring the density ($\rho$), molar mass ($M$), and finding edge length ($a$) using X-ray diffraction, we compute $Z$. If $Z \approx 1$, it's Simple Cubic; if $Z \approx 2$, it's BCC; if $Z \approx 4$, it's FCC.

50. Explain the effect of temperature on the conductivity of a semiconductor.

At absolute zero, semiconductors behave as perfect insulators because the valence band is completely full and the conduction band is completely empty. The thermal energy is too low to cross the energy gap.

As the temperature increases, electrons in the valence band gain enough thermal energy to jump across the small energy gap ($E_g$) into the conduction band. This creates free electrons in the conduction band and holes in the valence band, increasing the electrical conductivity exponentially with temperature.

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Solutions - 50 Subjective Questions for HSC Revision

Chapter 2: Solutions - 50 Subjective Questions for HSC Revision | Chemca.in

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Chapter 2: Solutions

Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions

Click on any question to reveal its answer.

1. What is a solution? Define solute and solvent.

Solution: A solution is a homogeneous mixture of two or more non-reacting substances whose composition can be varied within certain limits.

Solute: The component of the solution which constitutes the smaller part and is dissolved in the solvent is called the solute.

Solvent: The component of the solution which constitutes the larger part and acts as a dissolving medium is called the solvent.

2. Differentiate between saturated, unsaturated, and supersaturated solutions.

Unsaturated Solution: A solution in which more solute can be dissolved at the same temperature without raising it.
Saturated Solution: A solution in which no more solute can be dissolved at a specific temperature. It is in dynamic equilibrium with the undissolved solid solute.
Supersaturated Solution: A metastable solution containing more solute than the saturated solution at that temperature. It usually crystallizes out the excess solute upon scratching or seeding.

3. Define Solubility. State its unit.

Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature to form a saturated solution.

Unit: It is generally expressed in $\text{mol L}^{-1}$ or $\text{g L}^{-1}$.

4. What are the factors affecting the solubility of a solid in a liquid?

The solubility of a solid in a liquid depends on:

Nature of Solute and Solvent: "Like dissolves like." Polar solutes dissolve in polar solvents (e.g., NaCl in water), and non-polar solutes dissolve in non-polar solvents (e.g., iodine in benzene).
Temperature: If the dissolution is endothermic, solubility increases with a rise in temperature. If exothermic, solubility decreases.
Pressure: Pressure has no significant effect on the solubility of solids in liquids because solids and liquids are highly incompressible.

5. How does temperature affect the solubility of a gas in a liquid? Give a reason.

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

Reason: Dissolution of a gas in a liquid is an exothermic process (gas + solvent $\rightleftharpoons$ solution + heat). According to Le Chatelier's principle, increasing the temperature shifts the equilibrium backward, favoring the release of the dissolved gas.

6. State Henry's Law and give its mathematical expression.

Henry's Law: It states that the solubility of a gas in a liquid at constant temperature is directly proportional to the pressure of the gas above the solution.

Mathematical Expression: $S \propto P \Rightarrow S = K_H \cdot P$

Where $S$ is the solubility (in mol/L), $P$ is the partial pressure of the gas, and $K_H$ is Henry's law constant.

7. What are the exceptions (limitations) to Henry's law?

Henry's law is applicable only if:

The gas behaves like an ideal gas (pressure is not too high, temperature is not too low).
The gas does not undergo any chemical reaction with the solvent. (e.g., $NH_3$ in water reacts to form $NH_4OH$, so it doesn't obey Henry's law well).
The gas does not dissociate or associate in the solvent.

8. State any two applications of Henry's Law.

In carbonated beverages: Soft drinks and soda water bottles are sealed under high pressure to increase the solubility of $CO_2$.
In deep-sea diving: Scuba divers breathe compressed air, leading to high solubility of $N_2$ in blood. To prevent "the bends" (painful bubbles of $N_2$ forming in blood upon surfacing), diving tanks are diluted with Helium, which is less soluble.

9. Why do aquatic animals find it more comfortable to live in cold water rather than warm water?

The solubility of oxygen (a gas) in water decreases as the temperature increases. In cold water, the concentration of dissolved oxygen is higher. Aquatic animals rely on this dissolved oxygen for respiration. Hence, they feel more comfortable in cold water where oxygen is abundant compared to warm water.

10. Define Mole Fraction. Write its formula.

Mole Fraction ($x$): It is defined as the ratio of the number of moles of a particular component to the total number of moles of all the components present in the solution.

For a binary solution containing $n_1$ moles of solvent and $n_2$ moles of solute:

Mole fraction of solvent, $x_1 = \frac{n_1}{n_1 + n_2}$

Mole fraction of solute, $x_2 = \frac{n_2}{n_1 + n_2}$

Note: $x_1 + x_2 = 1$. It is a dimensionless quantity and independent of temperature.

11. Define Vapor Pressure of a liquid.

Vapor Pressure: The pressure exerted by the vapor of a liquid in dynamic equilibrium with its liquid phase at a given constant temperature in a closed container is called the vapor pressure of the liquid.

12. State Raoult's Law for a solution containing two volatile liquid components.

Raoult's Law: It states that for a solution of volatile liquids, the partial vapor pressure of any volatile component in the solution is directly proportional to its mole fraction in the solution.

Mathematically: $P_1 = P_1^\circ \cdot x_1$ and $P_2 = P_2^\circ \cdot x_2$

Where $P_1$ is partial pressure, $P_1^\circ$ is vapor pressure of pure component 1, and $x_1$ is its mole fraction.

13. What are Ideal Solutions? State their characteristics.

Ideal Solutions: Solutions that obey Raoult's law exactly over the entire range of concentrations and temperatures.

Characteristics:

Follows Raoult's law: $P_{total} = P_1 + P_2$.
Enthalpy of mixing is zero: $\Delta H_{mix} = 0$ (No heat evolved or absorbed).
Volume of mixing is zero: $\Delta V_{mix} = 0$ (Total volume is sum of individual volumes).
Intermolecular interactions between solute-solvent (A-B) are identical to solute-solute (B-B) and solvent-solvent (A-A) interactions.

Example: Benzene + Toluene.

14. Distinguish between ideal and non-ideal solutions.

Ideal Solution	Non-Ideal Solution
Obeys Raoult's law at all concentrations.	Does not obey Raoult's law.
$\Delta H_{mix} = 0$	$\Delta H_{mix} \neq 0$
$\Delta V_{mix} = 0$	$\Delta V_{mix} \neq 0$
A-B interactions are similar to A-A and B-B.	A-B interactions are stronger or weaker than A-A and B-B.

15. Explain positive deviation from Raoult's law with an example.

In solutions showing positive deviation, the solute-solvent (A-B) attractive forces are weaker than the pure solvent-solvent (A-A) and solute-solute (B-B) forces.

As a result, molecules escape more easily into the vapor phase, making the total vapor pressure higher than predicted by Raoult's law ($P_{total} > P_1^\circ x_1 + P_2^\circ x_2$). Also, $\Delta H_{mix} > 0$ and $\Delta V_{mix} > 0$.

Example: Ethanol + Acetone.

16. Explain negative deviation from Raoult's law with an example.

In solutions showing negative deviation, the solute-solvent (A-B) attractive forces are stronger than the pure solvent-solvent (A-A) and solute-solute (B-B) forces.

As a result, the escaping tendency of molecules decreases, making the total vapor pressure lower than predicted by Raoult's law ($P_{total} < P_1^\circ x_1 + P_2^\circ x_2$). Also, $\Delta H_{mix} < 0$ and $\Delta V_{mix} < 0$.

Example: Chloroform + Acetone (strong hydrogen bond forms between them).

17. Define Azeotropes. What are minimum and maximum boiling azeotropes?

Azeotropes: Binary mixtures having the same composition in liquid and vapor phase and which boil at a constant temperature are called azeotropes. They cannot be separated by fractional distillation.

Minimum Boiling Azeotrope: Formed by solutions showing large positive deviation from Raoult's law. They boil at a lower temperature than either pure component. (e.g., 95.5% Ethanol + Water).
Maximum Boiling Azeotrope: Formed by solutions showing large negative deviation from Raoult's law. They boil at a higher temperature than either pure component. (e.g., 68% Nitric acid + Water).

18. What are Colligative Properties? Name the four colligative properties.

Colligative Properties: The physical properties of solutions that depend entirely on the number of solute particles present in the given amount of solvent, and not on the nature of the solute particles, are called colligative properties.

The four properties are:

Relative lowering of vapor pressure.
Elevation in boiling point.
Depression in freezing point.
Osmotic pressure.

19. Why does the vapor pressure of a liquid decrease when a non-volatile solute is added?

Vapor pressure depends on the escape of solvent molecules from the surface of the liquid. When a non-volatile solute is added, some of the surface area is occupied by the solute particles, which do not vaporize. This decreases the fraction of the surface covered by solvent molecules, lowering the rate of evaporation and consequently lowering the vapor pressure of the solution.

20. Define lowering of vapor pressure and relative lowering of vapor pressure.

Lowering of Vapor Pressure ($\Delta P$): It is the difference between the vapor pressure of the pure solvent ($P_1^\circ$) and the vapor pressure of the solution ($P_1$).
$\Delta P = P_1^\circ - P_1$

Relative Lowering of Vapor Pressure: It is the ratio of the lowering of vapor pressure to the vapor pressure of the pure solvent.
$\text{RLVP} = \frac{P_1^\circ - P_1}{P_1^\circ}$

21. State Raoult's Law for a solution containing a non-volatile solute.

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

Mathematically: $$\frac{P_1^\circ - P_1}{P_1^\circ} = x_2$$

Where $x_2$ is the mole fraction of the non-volatile solute.

22. Derive the formula to determine the molar mass of a solute from the relative lowering of vapor pressure.

From Raoult's law: $\frac{P_1^\circ - P_1}{P_1^\circ} = x_2$

We know, $x_2 = \frac{n_2}{n_1 + n_2}$. For dilute solutions, $n_2 << n_1$, so $n_1 + n_2 \approx n_1$.

Therefore, $x_2 \approx \frac{n_2}{n_1} = \frac{W_2 / M_2}{W_1 / M_1} = \frac{W_2 \cdot M_1}{W_1 \cdot M_2}$

Substituting this in Raoult's law equation:

$$\frac{P_1^\circ - P_1}{P_1^\circ} = \frac{W_2 \cdot M_1}{W_1 \cdot M_2}$$

Where $W_2, M_2$ are mass and molar mass of solute, and $W_1, M_1$ are mass and molar mass of solvent.

23. Define Boiling Point of a liquid.

The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the external atmospheric pressure.

24. Why is the boiling point of a solution containing a non-volatile solute always higher than that of the pure solvent?

The addition of a non-volatile solute lowers the vapor pressure of the solvent. To make this lowered vapor pressure equal to the external atmospheric pressure and make the solution boil, the solution must be heated to a higher temperature than the pure solvent. This increase in temperature is called the elevation of boiling point ($\Delta T_b$).

25. Define Ebullioscopic constant (Molal elevation constant, $K_b$) and state its unit.

Ebullioscopic Constant ($K_b$): It is defined as the elevation in boiling point when one mole of a non-volatile solute is dissolved in 1 kg (1000 g) of the solvent. It is the elevation in boiling point of a 1 molal solution.

Unit: $\text{K kg mol}^{-1}$.

26. Derive the relationship between molar mass of a solute and elevation of boiling point.

Experimentally, elevation of boiling point ($\Delta T_b$) is directly proportional to molality ($m$).
$\Delta T_b = K_b \cdot m$

We know molality $m = \frac{n_2}{W_1(\text{in kg})} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 \cdot W_2}{M_2 \cdot W_1}$

Substituting $m$ in the equation:

$$\Delta T_b = \frac{1000 \cdot K_b \cdot W_2}{M_2 \cdot W_1}$$

Rearranging for molar mass ($M_2$): $$M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$$

27. Define Freezing Point.

The freezing point of a substance is the temperature at which its liquid phase and solid phase are in equilibrium, meaning they have the same vapor pressure.

28. Why does a solution of a non-volatile solute freeze at a lower temperature than the pure solvent?

Adding a non-volatile solute lowers the vapor pressure of the liquid solvent. Therefore, the vapor pressure of the solution will become equal to the vapor pressure of the solid solvent at a much lower temperature compared to the pure solvent. This results in a depression of the freezing point ($\Delta T_f$).

29. Define Cryoscopic constant (Molal depression constant, $K_f$) and state its unit.

Cryoscopic Constant ($K_f$): It is defined as the depression in freezing point produced when one mole of a non-volatile solute is dissolved in 1 kg (1000 g) of the solvent. It is the freezing point depression of a 1 molal solution.

Unit: $\text{K kg mol}^{-1}$.

30. Derive the formula for determining molar mass from depression in freezing point.

Depression of freezing point ($\Delta T_f$) is directly proportional to molality ($m$).
$\Delta T_f = K_f \cdot m$

Molality $m = \frac{1000 \cdot W_2}{M_2 \cdot W_1}$

Substituting $m$ in the equation:

$$\Delta T_f = \frac{1000 \cdot K_f \cdot W_2}{M_2 \cdot W_1}$$

Rearranging for molar mass ($M_2$): $$M_2 = \frac{1000 \cdot K_f \cdot W_2}{\Delta T_f \cdot W_1}$$

31. What is an anti-freeze? Give an example and state its application.

Anti-freeze: A substance added to a solvent (like water) to lower its freezing point significantly, preventing it from freezing in cold climates.

Example: Ethylene glycol is used as an anti-freeze in the radiators of cars.

Application: In cold countries, water in car radiators would freeze and burst the pipes. Adding ethylene glycol depresses the freezing point below $0^\circ \text{C}$, keeping the coolant in liquid form.

32. What is a semipermeable membrane (SPM)? Give two examples.

Semipermeable Membrane: A membrane that allows the passage of solvent molecules (like water) through its microscopic pores but prevents the passage of larger solute molecules.

Examples: Animal bladder, parchment paper, cellophane, synthetic membranes like cellulose acetate, copper ferrocyanide ($Cu_2[Fe(CN)_6]$).

33. Define Osmosis.

Osmosis: The spontaneous net flow of solvent molecules from a region of lower solute concentration (pure solvent or dilute solution) to a region of higher solute concentration (concentrated solution) through a semipermeable membrane.

34. Define Osmotic Pressure ($\pi$).

Osmotic Pressure: The excess hydrostatic pressure that must be applied to the solution side to just stop the inward flow of pure solvent across a semipermeable membrane is called osmotic pressure.

35. State van't Hoff equation for osmotic pressure and show its relation to molar mass.

van't Hoff equation states that osmotic pressure ($\pi$) is proportional to molarity ($C$) and absolute temperature ($T$).

$$\pi = C \cdot R \cdot T$$

Where $C = \frac{n_2}{V} = \frac{W_2}{M_2 \cdot V}$ ($R$ is the gas constant).

Substituting $C$:

$$\pi = \frac{W_2 \cdot R \cdot T}{M_2 \cdot V} \Rightarrow M_2 = \frac{W_2 \cdot R \cdot T}{\pi \cdot V}$$

36. Define Isotonic, Hypertonic, and Hypotonic solutions.

Isotonic Solutions: Two solutions having the same osmotic pressure at a given temperature. No net osmosis occurs between them.
Hypertonic Solution: A solution with a higher osmotic pressure (higher concentration) relative to another solution.
Hypotonic Solution: A solution with a lower osmotic pressure (lower concentration) relative to another solution.

37. What happens when red blood cells (RBCs) are placed in a 0.5% NaCl solution? Give a reason.

The fluid inside RBCs is isotonic with 0.9% (w/v) NaCl solution. A 0.5% NaCl solution is hypotonic compared to the RBC fluid.

When placed in a 0.5% NaCl solution, water will flow into the cell via endosmosis. As a result, the RBCs will swell and may even burst.

38. What is Reverse Osmosis (RO)? State its most important application.

Reverse Osmosis: If a pressure larger than the osmotic pressure is applied to the solution side, pure solvent flows out of the solution across the semipermeable membrane into the pure solvent side. This process reverses the natural direction of osmosis.

Application: Desalination of seawater. When high pressure is applied to seawater, pure water is squeezed through a cellulose acetate membrane, leaving salts behind, providing drinking water.

39. Why is osmotic pressure considered the best colligative property for determining the molar mass of macromolecules like proteins and polymers?

Osmotic pressure is preferred because:

It is measured at room temperature, preventing the degradation of temperature-sensitive biomolecules like proteins.
The magnitude of osmotic pressure is measurable even for very dilute solutions containing macromolecules with huge molar masses, whereas $\Delta T_b$ or $\Delta T_f$ values would be too small to measure accurately.
It uses molarity instead of molality, which is easier to prepare for polymeric solutions.

40. What is meant by abnormal molar mass? Why does it happen?

When the molar mass calculated using colligative properties differs from the theoretical (true) molar mass, it is called an abnormal molar mass.

Reason: This occurs when the solute undergoes either dissociation (e.g., $NaCl \rightarrow Na^+ + Cl^-$) or association (e.g., acetic acid dimerizing in benzene) in the solution. This changes the actual number of particles in the solution, changing the colligative property value and thus giving an incorrect molar mass.

41. Define van't Hoff factor (i).

van't Hoff factor ($i$): It is defined as the ratio of the observed (experimental) colligative property to the theoretical (calculated) colligative property.

$$i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}}$$

Since colligative property $\propto \frac{1}{\text{Molar Mass}}$, it can also be written as:

$$i = \frac{\text{Theoretical Molar Mass}}{\text{Observed Molar Mass}}$$

42. Write the modified equations for all four colligative properties incorporating the van't Hoff factor.

To correct for association or dissociation, the theoretical formulas are multiplied by '$i$':

RLVP: $\frac{P_1^\circ - P_1}{P_1^\circ} = i \cdot x_2$ (for dilute solutions, $i \cdot \frac{n_2}{n_1}$)
Elevation in B.P: $\Delta T_b = i \cdot K_b \cdot m$
Depression in F.P: $\Delta T_f = i \cdot K_f \cdot m$
Osmotic Pressure: $\pi = i \cdot C \cdot R \cdot T$

43. Relate the van't Hoff factor ($i$) with the degree of dissociation ($\alpha$).

For a solute that dissociates into $n$ ions:

$$\alpha = \frac{i - 1}{n - 1}$$

Where $\alpha$ is the degree of dissociation, $i$ is the van't Hoff factor, and $n$ is the number of moles of ions produced per mole of solute formula unit.

44. What is the value of the van't Hoff factor ($i$) for $K_2SO_4$ assuming complete dissociation?

$K_2SO_4$ dissociates as follows:

$$K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$$

One mole of $K_2SO_4$ yields $2 + 1 = 3$ moles of ions.

For complete dissociation ($\alpha = 1$), the van't Hoff factor $i = n$. Therefore, $i = 3$.

45. Explain why the van't Hoff factor for acetic acid dissolved in benzene is less than 1.

In a non-polar solvent like benzene, acetic acid molecules undergo association (dimerization) due to intermolecular hydrogen bonding.

$$2 CH_3COOH \rightleftharpoons (CH_3COOH)_2$$

Two molecules associate to act as a single particle. This decreases the total number of particles in the solution. Since $i = \frac{\text{Observed particles}}{\text{Theoretical particles}}$, and observed particles are fewer, $i < 1$ (it is approximately 0.5 for complete dimerization).

46. Define Molality ($m$). Why is it preferred over molarity for studying colligative properties involving temperature changes?

Molality ($m$): The number of moles of solute dissolved per kilogram (1000g) of the solvent.

Preference: Molarity involves the volume of the solution, which changes with temperature due to expansion or contraction. Molality involves the mass of the solvent, which is independent of temperature. Thus, molality is temperature-independent, making it ideal for boiling point and freezing point experiments.

47. What is Ebullioscopy and Cryoscopy?

Ebullioscopy: The method of determining the molar mass of a non-volatile solute by measuring the elevation in the boiling point of the solvent.
Cryoscopy: The method of determining the molar mass of a non-volatile solute by measuring the depression in the freezing point of the solvent.

48. Draw and explain the vapor pressure vs temperature graph showing elevation of boiling point.

In a graph of Vapor Pressure (y-axis) vs Temperature (x-axis):

There are two upward-curving lines: one for the pure solvent (upper curve) and one for the solution (lower curve, since VP is lowered).
A horizontal line is drawn at 1 atm (external atmospheric pressure).
The temperature where the pure solvent curve intersects 1 atm is $T_b^\circ$ (boiling point of pure solvent).
The temperature where the solution curve intersects 1 atm is $T_b$ (boiling point of solution).
Because the solution curve is lower, it intersects the 1 atm line at a higher temperature. Therefore, $T_b > T_b^\circ$, showing $\Delta T_b = T_b - T_b^\circ$.

49. Which will have a higher boiling point: 0.1 M NaCl or 0.1 M Glucose? Give reason.

0.1 M NaCl will have a higher boiling point.

Reason: Elevation in boiling point is a colligative property that depends on the number of particles. Glucose is a non-electrolyte and does not dissociate ($i=1$), providing 0.1 moles of particles. NaCl is a strong electrolyte and dissociates completely into $Na^+$ and $Cl^-$ ions ($i=2$), providing $0.1 \times 2 = 0.2$ moles of particles. More particles mean greater elevation in boiling point.

50. Calculate the mass percent of benzene in a solution containing 30g of benzene dissolved in 120g of carbon tetrachloride.

Mass of solute (benzene) = $30$ g.

Mass of solvent (carbon tetrachloride) = $120$ g.

Total mass of solution = $30 + 120 = 150$ g.

$$\text{Mass percent} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 100$$

$$\text{Mass percent of benzene} = \frac{30}{150} \times 100 = 20\%$$

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