Thermal Properties Of Matter

Step 1: {Understanding concentration units}
Molarity and normality depend on volume, which changes with temperature. However, molality is defined as: Since mass is unaffected by temperature changes, molality remains constant.
Step 2: {Why molality is temperature-independent}
Molality does not involve volume, which expands or contracts with temperature. Hence, it is the preferred unit in temperature-dependent studies.

Step 1: Understanding Water-Soluble Vitamins
Water-soluble vitamins (B-complex and C) dissolve in water and are excreted in urine.
However, Vitamin B₁₂ is an exception.
Step 2: Storage of Vitamin B₁₂
Unlike other B vitamins, B₁₂ is stored in the liver.
It plays a vital role in red blood cell formation and neurological functions.
Final Answer: Vitamin B₁₂ is water-soluble but not excreted easily.

Step 1: Understanding Peptide Sequence Formation
- A tetrapeptide consists of 4 amino acids.
- The C-terminal amino acid is fixed as alanine.
- The N-terminal must be chiral (valine or serine).
Step 2: Listing Possible Sequences
Since glycine is achiral, the N-terminal choices are valine or serine, leading to 4 possible arrangements:
1. Val-Gly-Ser-Ala
2. Val-Ser-Gly-Ala
3. Ser-Gly-Val-Ala
4. Ser-Val-Gly-Ala
Final Answer: The correct number of sequences is 4.

Step 1: Understanding the Rate Law
The general rate law is: where is the order of the reaction.
Step 2: Given Condition
When the concentration of is doubled, the rate remains unchanged.
Mathematically, Since , we equate:
Step 3: Solving for
Dividing both sides: Final Answer: The reaction follows zero order kinetics.

Step 1: Applying the Arrhenius Equation
The Arrhenius equation relates the rate constant to temperature : where: - (since the rate constant doubles), - K, - K, - J mol K .
Step 2: Substituting Values

Step 3: Solving for
Using ,
Final Answer: The activation energy is 3.45 kJ/mol.

Step 1: Understanding the Rate Law
The rate of a reaction is given by: where is the rate constant, is the concentration of reactant, and is the order of the reaction.
Step 2: Applying Given Conditions
If the concentration of is increased four times, the rate doubles. Mathematically, Using the original rate equation:
Step 3: Dividing the Equations

Step 4: Solving for
Taking logarithm on both sides,
Final Answer: The order of reaction is 0.5.

To find the temperature at which the half-life of a first-order reaction is 10 minutes, we use the Arrhenius equation and the half-life formula for first-order reactions.

The half-life ( ) for a first-order reaction is given by:

where is the rate constant. For minutes, we convert this to seconds:

s

Thus:

Solving for :

The Arrhenius equation is:

Where:

• s
• kJ/mol (or J/mol)
• J/(mol K) (universal gas constant)

Substitute into the Arrhenius equation:

Take the natural logarithm on both sides:

Solve for :

1. Calculate .
2. Calculate .
3. Rearrange to solve for .

Performing these calculations gives:

Solve:

Calculating yields K.

Thus, the temperature at which the half-life of the reaction is 10 minutes is 311.15 K.

To solve the problem, we need to determine the rate of appearance of SO₃ based on the given rate of disappearance of O₂ for the reaction:
2SO₂ + O₂ ⇌ 2SO₃.

The stoichiometry of the reaction provides the key relationships between the rates:
1 mol of O₂ produces 2 mol of SO₃.

Let:

• Rate of disappearance of O₂ = mol L^-1 s^-1
• The rate of appearance of SO₃ is equal to twice the rate of disappearance of O₂ based on stoichiometry (1:2 ratio for O₂:SO₃)

Thus, the rate of appearance of SO₃ = 2 × Rate of disappearance of O₂ = 2 × = mol L^-1 s^-1.

The correct answer is:
mol L^-1 s^-1.

Step 1: Understanding Faraday’s Laws of Electrolysis
According to Faraday’s first law of electrolysis, the mass of metal deposited on the cathode is directly proportional to the charge passed: where: is the mass of the deposited metal,
is the electrochemical equivalent of the metal,
is the charge passed (which is the same for all solutions),
is Faraday’s constant C/mol.
Step 2: Electrochemical Equivalent and Equivalent Weight
The electrochemical equivalent ( ) is given by: The more the electrochemical equivalent ( ), the greater the mass of metal deposited.
The equivalent weight is given by:

Step 3: Calculating Equivalent Weights for Given Options
Zn from ZnSO - Atomic weight of Zn = 65 - Valency = 2 - Equivalent weight =
Fe from FeCl - Atomic weight of Fe = 56 - Valency = 3 - Equivalent weight =
Ag from AgNO - Atomic weight of Ag = 108 - Valency = 1 - Equivalent weight =
Ni from NiCl - Atomic weight of Ni = 58.7 - Valency = 2 - Equivalent weight =
Step 4: Identifying the Metal Deposited in Maximum Amount
- Since the mass deposited is directly proportional to equivalent weight, the metal with the highest equivalent weight will be deposited in the maximum amount. - Silver (Ag) has the highest equivalent weight (108 g/mol), so Ag will be deposited in the greatest amount. Thus, the correct answer is (C) AgNO .

Step 1: Understanding Colligative Properties
- Colligative properties depend on the number of particles in the solution rather than their nature.
- The van't Hoff factor ( ) represents the number of ions each molecule dissociates into.
Step 2: Calculating the Number of Ions
BaCl dissociates into 3 ions: Ba and 2Cl .
AgNO dissociates into 2 ions: Ag and NO .
Urea does not dissociate, so .
(NH ) PO dissociates into 4 ions: 3 NH and 1 PO .
Step 3: Identifying the Substance with the Highest Colligative Property
The greater the number of ions, the higher the colligative properties.
Since (NH ) PO produces 4 ions, it exhibits the highest colligative effect.
Thus, the correct answer is (D).

Step 1: Understanding the Reaction Mechanism
The given reaction involves sodium ethoxide (CH CH ONa) and tert-butyl chloride [(CH ) CCl] in ethanol. Tertiary alkyl halides undergo elimination (E2) rather than substitution due to steric hindrance.
Step 2: Identifying the Major Product
Since sodium ethoxide is a strong base, elimination occurs via the E2 mechanism, leading to the formation of an alkene. The product formed is isobutene ( ). Thus, the correct answer is (B).

Step 1: Understanding Bronsted Bases
A Bronsted base is a substance that can accept protons. The strength of a Bronsted base depends on: 1. The availability of a lone pair for protonation. 2. The electron-donating or withdrawing effects of the surrounding groups.
Step 2: Evaluating Each Structure
Structure 1 has electron-withdrawing effects, reducing basicity.
Structure 2 has delocalization of the lone pair, reducing its basicity.
Structure 3 is in a constrained ring, affecting electron availability.
Structure 4 has a localized lone pair, making it the most basic.
Thus, the correct answer is (D).

Step 1: Understanding IUPAC Nomenclature Rules
The longest chain in the benzene ring is chosen as the parent chain. The substituents are assigned numbers based on priority groups to ensure the lowest possible sum.
Step 2: Numbering the Benzene Ring
Methyl ( ) is given the highest priority and assigned position 1. Chlorine ( ) is then assigned position 2, and nitro ( ) is assigned position 4.
Step 3: Naming the Compound
The correct name following IUPAC rules is 2-Chloro-1-methyl-4-nitrobenzene. Thus, the correct answer is (C).

Step 1: Understanding Melting Point Trends in Ionic Compounds
The melting point of an ionic compound depends on: 1. Lattice Energy: Higher lattice energy means a higher melting point. 2. Size of the Anion: Smaller anions lead to stronger lattice energy.
Step 2: Analyzing the Given Salts
Fluoride (F ) is the smallest anion, followed by chloride (Cl ) and bromide (Br ).
The order of lattice energy is: Since lattice energy determines melting point, we get:
Step 3: Correct Answer
Thus, the correct order is: Thus, the correct answer is (B).

Step 1: Understanding Disproportionation and Redox Reactions
A disproportionation reaction is a special type of redox reaction in which a single element undergoes both oxidation and reduction. In contrast, a normal redox reaction involves oxidation and reduction of different elements.
Step 2: Oxidation States Analysis
Let's analyze the oxidation states of phosphorus (P) and sulfur (S) in the reaction: In elemental phosphorus ( ), P has an oxidation state of .
In phosphorus trichloride ( ), P has an oxidation state of .
In sulfur oxychloride ( ), S has an oxidation state of .
In disulfur dichloride ( ), S has an oxidation state of .
In sulfur dioxide ( ), S has an oxidation state of .
Step 3: Identify the Nature of the Reaction
Phosphorus ( ) is oxidized from to .
Sulfur ( ) is reduced from to .
Since phosphorus only undergoes oxidation and sulfur only undergoes reduction, this is a simple redox reaction, NOT a disproportionation reaction.
Step 4: Verify Other Options
Option (A) involves phosphorus undergoing both oxidation and reduction, making it a disproportionation reaction.
Option (B) involves oxygen undergoing disproportionation from to both and .
Option (C) also shows disproportionation of phosphorus.

Step 1: {Balance the Redox Reaction}
The balanced ionic equation for the reaction is: Step 2: {Stoichiometric Relationship}
From the equation, it is clear that 1 mole of KMnO reacts with 5 moles of FeSO .
Step 3: {Apply Molarity Equation}
Use the molarity equation to balance the reaction: Step 4: {Solve for }
.

Step 1: {Henderson-Hasselbalch Equation}
For an acidic buffer, the pH is given by: Step 2: {Concentrations}
Let the concentration of potassium acetate solution be M. Step 3: {Substitute Values}
Given, pH = 4.8 where
Step 4: {Solve for }
Substitute : Thus, the correct answer is (B) 0.04 M.

Step 1: {Understanding the Strength of Acids}
Stronger the acid, lower is the pH, and weaker the acid, higher the pH.
Step 2: {Order of Acid Strength}
The strength of the given acids is in the order:
Step 3: {Order of pH}
Hence, their pH is in the order: Thus, the correct answer is (C).

Step 1: {Understanding Free Expansion of an Ideal Gas}
In free expansion, the gas expands without doing work and without heat exchange with the surroundings. Since the process is adiabatic, , meaning no heat is exchanged.
Step 2: {Work Done in Free Expansion}
Since the expansion is against a vacuum, no work is done on or by the system. Hence, .
Step 3: {Temperature Change in Free Expansion}
For an ideal gas undergoing free expansion, there is no change in temperature, so . Thus, the correct answer is (D).

Step 1: {Understanding the Cyclic Process}
In a cyclic process, the internal energy change ( ) is zero. According to the first law of thermodynamics: where is heat absorbed and is the work done.
Step 2: {Work Done in a Cyclic Process}
The work done in a cyclic process is equal to the area enclosed by the cycle in the diagram. For a process with a circular path, the work done is the area of the circle, which is given by: where is the radius of the circle.
Step 3: {Calculating the Work Done}
From the diagram, the radius . Therefore: Since there is no change in internal energy, the heat absorbed . Thus, the heat absorbed is J.

Step 1: {Analysis of Canonical Forms}
Canonical forms are the different resonance structures that a molecule can have, resulting in delocalization of electrons. The species with the maximum number of canonical forms has the most stable resonance structure.
Step 2: {Analysis of Compounds}
SO has three resonance structures but lacks a -bond between sulfur and oxygen in each structure.
O has two resonance structures with a single -bond between the two oxygen atoms, but it doesn’t have the maximum number of canonical forms.
SO has two resonance structures, but again, it doesn’t maximize the number of canonical forms.
CO has three resonance structures, each with a -bond and delocalized electrons between the carbon and the oxygen atoms.
The carbonate ion (CO ) has the maximum number of canonical forms.
Thus, the correct answer is (D).

Step 1: {Isoelectronic Species}
The ions , , and are isoelectronic species, meaning they all have the same number of electrons. Specifically, each of these ions has 10 electrons. Despite having the same number of electrons, their ionic radii will differ because of the varying nuclear charges. The nuclear charge increases from nitrogen to oxygen to fluorine as we move across the periodic table: - has atomic number , - has atomic number , - has atomic number .
Step 2: {Ionic Radii Order}
As the nuclear charge increases, the electrons are attracted more strongly by the nucleus, which results in a decrease in ionic radius. Therefore, the ionic radii will decrease as we move from to to . This gives the following order of ionic radii: Hence, the correct answer is (A).

Step 1: Understanding Ozonolysis
- Ozonolysis of an alkene breaks the double bond and forms aldehydes and ketones.
- Identifying the correct alkene requires reversing this process.
Step 2: Identifying the Structure of X
The products given are Propan-2-one ( ) and methanal ( ).
The only alkene that produces these fragments is 2-Methylpropene ( ).
Final Answer: The correct alkene is 2-Methylpropene.

Step 1: Understanding Aldol Condensation
- Aldol condensation occurs in aldehydes/ketones that contain at least one -hydrogen.
- The -hydrogen is removed by a base to form an enolate ion, which then attacks another carbonyl compound.
Step 2: Analyzing the Given Compounds
- (Propionaldehyde) contains an -hydrogen, allowing aldol condensation.
- (Acrolein) and do not easily form enolates due to resonance stabilization.
- (Benzaldehyde) lacks -hydrogen, preventing aldol condensation.
Final Answer: undergoes aldol condensation.

Step 1: Understanding Lucas Test
- Lucas test differentiates alcohols based on their reaction with ZnCl₂/HCl.
Step 2: Order of Reactivity
- Tertiary alcohols react fastest via the S_N1 mechanism.
Final Answer: Tertiary alcohol via S_N1.