The structure of \( A_{0.95}O \) is formed by \( A^{2+} \), \( A^{3+} \), and \( O^{2-} \) ions. The stoichiometry implies that 95 percentage of \( A \)-sites are occupied by \( A^{2+} \), while 5 percentage are substituted by \( A^{3+} \), and oxygen sites are fully occupied by \( O^{2-} \). In the diagrams:


- Diagram A represents this substitution correctly, where the lattice shows \( A^{2+} \) (green), \( A^{3+} \) (red), and \( O^{2-} \) (black) ions distributed as per the stoichiometric requirements.

- Diagrams B and C do not align with the stoichiometric ratio or distribution. Diagram B over-represents \( A^{3+} \), while C shows irregularity in \( O^{2-} \) ion placement.


Thus, the correct representation of the lattice is Diagram A.
Quick Tip: To determine the correct lattice structure, check for the accurate stoichiometry and ion distribution as per the compound formula.

Sugar [X] is D-glucose, which can cyclize to form a six-membered pyranose ring. The structure depends on the orientation of the -OH group on the anomeric carbon (C-1):


- In Diagram 1, the -OH group on the anomeric carbon (C-1) is oriented downward, representing the α-D-glucopyranose form.

- In Diagram 2, the -OH group on the anomeric carbon (C-1) is oriented upward, representing the β-D-glucopyranose form.

- In Diagram 3, the HO is added and representing in downward form.

- In Diagram 4, the CH2 is added and representing in downward form.

Since the question asks for the α-D-glucopyranose form, Diagram 2 is the correct representation.
Quick Tip: To identify α- and β-anomers, check the orientation of the -OH group on the anomeric carbon. In α-anomers, it points downward, while in β-anomers, it points upward.

In \( Mn_2O_7 \), manganese (Mn) is in its highest oxidation state of \( +7 \). The molecular structure of \( Mn_2O_7 \) exhibits the following features:

1. Each Mn atom is tetrahedrally surrounded by oxygen atoms, as the \( MnO_4^- \) tetrahedral unit is a part of the molecule.

2. The molecule contains an \( Mn-O-Mn \) bridge linking the two tetrahedral units.

3. There is no \( Mn-Mn \) bond in \( Mn_2O_7 \).

4. Mn is not octahedrally surrounded by oxygen atoms.


Based on these points, the correct statements are:

% Option
(A) Mn is tetrahedrally surrounded by oxygen atoms.

% Option
(C) Contains \( Mn-O-Mn \) bridge.


Conclusion: The correct answer is \( \boxed{1} \) (A and C only).
Quick Tip: In \( Mn_2O_7 \), manganese achieves its highest oxidation state of \( +7 \). The structure consists of two tetrahedral \( MnO_4^- \) units linked by an \( Mn-O-Mn \) bridge, with no \( Mn-Mn \) bond.

The dehydration of alcohols follows the order of their stability as carbocations formed during the reaction. The stability of the carbocation depends on the degree of substitution:


- Alcohol \( b \): Forms a tertiary carbocation, which is the most stable due to maximum alkyl group stabilization.

- Alcohol \( d \): Forms a secondary carbocation, which is moderately stable.

- Alcohol \( c \): Forms another secondary carbocation, similar to \( d \), but slightly less stable due to fewer substituents.

- Alcohol \( a \): Forms a primary carbocation, which is the least stable.


Thus, the order of dehydration follows the carbocation stability: \( b > d > c > a \).
Quick Tip: Dehydration of alcohols is faster when the resulting carbocation is more stable. The order of stability is: tertiary > secondary > primary.

1. Assertion (A): Among the noble gases, Krypton (Kr) is adsorbed more by activated charcoal compared to He, Ne, and Ar. This is because the extent of adsorption increases with an increase in molecular size and van der Waals forces, both of which are higher for Krypton. Thus, Assertion (A) is true.


2. Reason (R): The given reason states that the critical volume (\( V_c \)) and critical pressure (\( P_c \)) are highest for Krypton, and the compressibility factor (\( Z_c \)) at the critical point is lowest. However, this is incorrect because Krypton does not have the highest \( V_c \) or \( P_c \) among noble gases, and \( Z_c \) is not a determining factor for adsorption. Therefore, Reason (R) is false.


Conclusion: Assertion (A) is true, but Reason (R) is false. The correct answer is \( \boxed{1} \).
Quick Tip: The adsorption of gases on activated charcoal depends on molecular size and van der Waals forces, not directly on critical parameters like \( V_c, P_c, \) or \( Z_c \).

The given reaction involves a substitution of the hydroxyl group (-OH) with an ethyl carbamate (-COOEt) group to form the major product. Here’s how it works:


- The compound \( NH_2-CH_2OH \) reacts with ethyl chloroformate (\( EtO-C(=O)-Cl \)) under basic or catalytic conditions.

- The primary amine group (\(-NH_2\)) undergoes nucleophilic attack on the carbonyl carbon of ethyl chloroformate, resulting in the formation of a carbamate group (\( NH-C(=O)OEt \)).

- The major product 'A' is thus \( NH-C(=O)OEt \), as shown in option (2).
Quick Tip: Carbamates (\( R-NH-C(=O)OR' \)) are commonly formed by the reaction of primary amines with chloroformates (\( RO-C(=O)-Cl \)).

(A) Tranquilizers: These are drugs used to treat anxiety and depression. They belong to the class of antidepressant drugs. Therefore, (A) matches with (III).


(B) Aspirin: Aspirin is a well-known painkiller and also acts as an anti-blood-clotting agent. Therefore, (B) matches with (I).


(C) Antibiotics: These are drugs used to treat bacterial infections. Salvarsan, a drug used to treat syphilis, is an example. Therefore, (C) matches with (II).
% Option
(D) Antiseptic: These are substances used to prevent the growth of microorganisms. Soframicine is an antiseptic. Therefore, (D) matches with (IV).


Conclusion: The correct match is \( \boxed{3} \).
Quick Tip: Familiarize yourself with the uses and applications of common drugs and compounds to solve matching-type questions efficiently.

(A) Explanation of Statement I:
Chlorine forms oxides such as \( Cl_2O, ClO_2, Cl_2O_7 \), etc., when it reacts with oxygen. These oxides are highly reactive and unstable, with some having explosive tendencies. Hence, Statement I is true.

(B) Explanation of Statement II:
The chemical reactivity of an element can be assessed by studying its reactions with common elements such as oxygen and halogens. These reactions provide insights into the bonding behavior and stability of the compounds formed. Hence, Statement II is also true.


Conclusion: Both statements are true, making \( \boxed{1} \) the correct answer.
Quick Tip: Familiarize yourself with the properties of oxides and halides of nonmetals to understand their reactivity and stability.

The carbonate ion \( CO_3^{2-} \) exhibits resonance, where the true structure is a hybrid of the three resonance structures. These three structures are not separate, stable entities but are blended into one. The negative charge is delocalized over the three oxygen atoms, and there is no single structure that can be isolated.

(A) Option (1): It is not possible to isolate each individual resonance structure experimentally because the true structure is a resonance hybrid. Hence, Option 1 is false.


(B) Option (2): The three resonance structures do not exist in equilibrium with each other. Instead, they are blended into one hybrid structure. Hence, Option 2 is false.


(C) Option (3): The three resonance structures do not exist for equal amounts of time as separate entities. They are not time-dependent but are a blended average. Hence, Option 3 is false.


(D) Option (4): The true structure of \( CO_3^{2-} \) is the resonance hybrid of the three possible structures. Hence, Option 4 is correct.


Conclusion: The correct answer is \( \boxed{4} \). Quick Tip: In resonance structures, the true molecule or ion is a hybrid of the different structures, and none of the structures exist as distinct entities.

The question involves identifying the incorrect reaction among the given options:

1. Option (1): The reaction involves the substitution of \(-Br\) with \(-OH\) using aqueous KOH. This is a correct nucleophilic substitution reaction.

2. Option (2): The reaction involves alcoholic KOH, which promotes elimination (E2 mechanism). However, the product shown is incorrect because the major product should be a double bond formed due to elimination, not a substitution product.

3. Option (3): The reaction is a Friedel-Crafts acylation, which is correctly represented.

4. Option (4): This is a Dow process, correctly showing the conversion of chlorobenzene to phenol under high temperature and pressure.


Thus, option (2) is the incorrect reaction. Quick Tip: Alcoholic KOH generally promotes elimination reactions (E2 mechanism), while aqueous KOH favors nucleophilic substitution reactions (SN2 or SN1).

The Prussian blue precipitate is formed when ferric ions (\( Fe^{3+} \)) react with hexacyanoferrate (\( [Fe(CN)_6]^{4-} \)) ions. The product of this reaction is \( Fe_4[Fe(CN)_6]_3 \), which is an insoluble complex that imparts the characteristic deep blue color.


- Step 1: Ferric chloride (\( FeCl_3 \)) dissociates to produce \( Fe^{3+} \) ions in solution.

- Step 2: Potassium hexacyanoferrate (\( K_4[Fe(CN)_6] \)) dissociates to produce \( [Fe(CN)_6]^{4-} \) ions in solution.

- Step 3: \( Fe^{3+} \) ions combine with \( [Fe(CN)_6]^{4-} \) to form the insoluble complex \( Fe_4[Fe(CN)_6]_3 \), known as Prussian blue.

\[ 4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow Fe_4[Fe(CN)_6]_3 \downarrow \quad (Prussian blue precipitate) \] Quick Tip: The formation of Prussian blue is a qualitative test for the presence of ferric ions (\( Fe^{3+} \)) in solution.

Double salts are compounds that dissociate completely into their constituent ions when dissolved in water. They are formed by the combination of two salts and retain their identity only in the crystalline state. Among the given options:


- (A) \( FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O \): Known as Mohr's salt, this is a classic example of a double salt.

- (B) \( CuSO_4\cdot4NH_3\cdotH_2O \): This is a coordination compound, not a double salt.

- (C) \( K_2SO_4\cdotAl_2(SO_4)_3\cdot24H_2O \): Known as potash alum, this is a double salt.

- (D) \( Fe(CN)_2\cdot4KCN \): This is a coordination compound (ferrocyanide complex), not a double salt.


Thus, the correct examples of double salts are (A) and (C).
Quick Tip: Double salts completely dissociate into their constituent ions in water, while coordination compounds do not dissociate fully due to the formation of coordination complexes.

The splitting of d-orbitals in a complex depends on the strength of the ligand according to the spectrochemical series. Strong field ligands cause greater splitting, while weak field ligands result in smaller splitting. Among the given complexes:


1. (1) \([ Fe(C_2O_4)_3 ]^{3-}\): Oxalate (\( C_2O_4^{2-} \)) is a moderately strong field ligand, causing moderate splitting.

2. (2) \([ FeF_6 ]^{3-}\): Fluoride (\( F^- \)) is a weak field ligand, causing minimal splitting.

3. (3) \([ Fe(CN)_6 ]^{3-}\): Cyanide (\( CN^- \)) is a very strong field ligand, causing the largest splitting of d-orbitals.

4. (4) \([ Fe(NH_3)_6 ]^{3+}\): Ammonia (\( NH_3 \)) is a moderate field ligand, causing less splitting compared to cyanide.


Thus, the complex \([ Fe(CN)_6 ]^{3-}\) will show the largest splitting of d-orbitals due to the presence of cyanide ligands.
Quick Tip: The strength of ligand field splitting follows the spectrochemical series: \[ I^- < Br^- < Cl^- < F^- < H_2O < NH_3 < C_2O_4^{2-} < CN^-. \]

Photochemical smog is a type of air pollution that occurs when sunlight reacts with pollutants such as nitrogen oxides (NOx) and volatile organic compounds (VOCs). These pollutants are primarily emitted from automobiles and industrial processes. The smog contains harmful substances like ozone (O\(_3\)) and peroxyacetyl nitrates (PANs), which can harm human health and the environment.


The most effective way to control photochemical smog is to reduce the emissions of these harmful pollutants. The use of catalytic converters in automobiles and industrial exhaust systems helps to reduce the emissions of nitrogen oxides (NOx) and hydrocarbons (VOCs) that contribute to photochemical smog formation. Catalytic converters work by promoting chemical reactions that convert these pollutants into less harmful substances, such as nitrogen (N\(_2\)), carbon dioxide (CO\(_2\)), and water (H\(_2\)O).


(A) Option (1): Using tall chimneys may help disperse pollutants over a larger area, but it does not address the root cause of photochemical smog. Hence, this option is ineffective.


(B) Option (2): Complete combustion of fuel can reduce the production of particulate matter and some gases, but it does not fully prevent the formation of nitrogen oxides and volatile organic compounds that cause photochemical smog.


(C) \text bf{Option (3): Catalytic converters are specifically designed to reduce the emissions of harmful pollutants such as nitrogen oxides and hydrocarbons, which are major contributors to photochemical smog. Hence, this option is the most effective.


(D) Option (4): While catalysts are used in catalytic converters, using a catalyst alone without addressing the emissions is insufficient to control photochemical smog.


Conclusion: The correct answer is \( \boxed{3} \). Quick Tip: To control photochemical smog, reducing emissions of nitrogen oxides (NOx) and volatile organic compounds (VOCs) using catalytic converters is the most effective solution.

Let's analyze each compound:

- Slaked lime is calcium hydroxide, Ca(OH)\(_2\). This corresponds to option (II).

- Dead burnt plaster is calcium sulfate, CaSO\(_4\), which corresponds to option (IV).

- Caustic soda is sodium hydroxide, NaOH. This corresponds to option (I).

- Washing soda is sodium carbonate decahydrate, Na\(_2\)CO\(_3\)·10H\(_2\)O. This corresponds to option (III).


Thus, the correct matching is: \[ (A) – II, (B) – IV, (C) – I, (D) – III. \]

Conclusion: The correct answer is \( \boxed{3} \). Quick Tip: Remember that slaked lime is calcium hydroxide (Ca(OH)\(_2\)), dead burnt plaster is calcium sulfate (CaSO\(_4\)), caustic soda is sodium hydroxide (NaOH), and washing soda is sodium carbonate decahydrate (Na\(_2\)CO\(_3\)·10H\(_2\)O).

- A is incorrect: Beryllium oxide is amphoteric, not purely acidic.

- B is correct: Beryllium carbonate is unstable and decomposes in the atmosphere of CO\(_2\).

- C is correct: Beryllium sulfate is soluble in water.

- D is correct: Beryllium exhibits anomalous behavior due to its small size and high charge density, which makes it behave differently from other alkaline earth metals.


Conclusion: The correct answer is \( \boxed{2} \). Quick Tip: Beryllium (Be) exhibits anomalous behavior because of its small atomic size and high charge density, which results in different chemical properties compared to other elements in Group 2.

- Assertion A is correct: In an Ellingham diagram, the oxidation of carbon to carbon monoxide does indeed show a negative slope, indicating that the reaction becomes more favorable as temperature increases.

- Reason R is incorrect: While CO is a reducing agent, it does not tend to decompose at higher temperatures; rather, the decomposition of CO to C and O\(_2\) is unfavorable at high temperatures, as shown by its position on the Ellingham diagram.


Conclusion: The correct answer is \( \boxed{4} \). Quick Tip: In an Ellingham diagram, a negative slope indicates that the reaction becomes more favorable at higher temperatures, while a positive slope indicates that the reaction becomes less favorable.

The reaction involves the partial hydrogenation of but-2-yne to form cis-but-2-ene (A):


1. Statement A: Incorrect. A (cis-but-2-ene) is less soluble than B (trans-but-2-ene) due to the molecular geometry. Trans isomers are generally more symmetric and pack better, leading to higher solubility.

2. Statement B: Correct. Cis isomers (A) generally have higher boiling points due to stronger intermolecular forces and lower melting points compared to trans isomers (B).

3. Statement C: Incorrect. A (cis-but-2-ene) has a nonzero dipole moment due to the asymmetric arrangement of substituents, making it more polar than B.

4. Statement D: Incorrect. Bromine (Br\(_2\)) adds more easily to B (trans-but-2-ene) because of its more accessible double bond, as the substituents are farther apart compared to A.


Thus, the incorrect statements are A, C, and D.
Quick Tip: In alkene chemistry, trans isomers are generally less polar, more stable, and more soluble compared to their cis counterparts due to their symmetrical geometry.

- Assertion A is true: Hydrogen is considered an environment-friendly fuel because, when used as a fuel (for example, in fuel cells), it produces only water as a by-product, making it clean and non-polluting.

- Reason R is true: The atomic number of hydrogen is indeed 1, and it is the lightest element in the periodic table.

- However, Reason R does not explain why hydrogen is environment-friendly. The reason behind hydrogen's environment-friendliness is due to its clean combustion, not because of its atomic number or lightness.


Conclusion: The correct answer is \( \boxed{2} \). Quick Tip: Hydrogen is environment-friendly because its combustion produces only water, making it a clean source of energy. The lightness and atomic number of hydrogen do not directly explain this property.

- Molisch's Test is used for the detection of carbohydrates. So, (A) – II.

- Biuret Test is used for detecting peptide bonds, thus identifying proteins. So, (B) – I.

- Carbylamine Test is used for detecting primary amines. So, (C) – III.

- Schiff's Test is used to detect aldehydes. So, (D) – IV.


Conclusion: The correct matching is \( \boxed{(3)} \). Quick Tip: Molisch’s test is used to detect carbohydrates, Biuret test identifies peptides or proteins, Carbylamine test identifies primary amines, and Schiff’s test detects aldehydes.

The molality of a solution can be calculated using the formula: \[ Molality (m) = \frac{Moles of solute}{Mass of solvent in kg} \]

1. Step 1: Calculate the mass of 1 L of solution

The density of the solution is given as 1.0 g/mL, so the mass of 1 L of the solution is: \[ Mass of solution = 1.0 \, g/mL \times 1000 \, mL = 1000 \, g \]

2. Step 2: Calculate the mass of NaCl in 1 L solution

The molarity of the solution is 3 M, which means there are 3 moles of NaCl in 1 L of the solution. The molar mass of NaCl is: \[ Molar mass of NaCl = 23 + 35.5 = 58.5 \, g/mol \] \[ Mass of NaCl = 3 \, mol \times 58.5 \, g/mol = 175.5 \, g \]

3. Step 3: Calculate the mass of the solvent (water)

The mass of the solvent is the total mass of the solution minus the mass of NaCl: \[ Mass of solvent = 1000 \, g - 175.5 \, g = 824.5 \, g = 0.8245 \, kg \]

4. Step 4: Calculate the molality
\[ Molality (m) = \frac{3 \, mol}{0.8245 \, kg} = 3.64 \, mol/kg \]
Expressing this in the form of \(\times 10^{-2}\): \[ Molality = 364 \times 10^{-2} \, m \] Quick Tip: Molality depends only on the mass of the solvent and is independent of temperature, unlike molarity.

1. Statement (A): The de Broglie wavelength (\( \lambda \)) is calculated as: \[ \lambda = \frac{h}{mv} \]
Substitute the given values: \[ \lambda = \frac{6 \times 10^{-34}}{9 \times 10^{-31} \times 1000} = 6.67 \times 10^{-7} \, m = 666.67 \, nm. \]
This statement is true.


2. Statement (B): The emission of electrons in a cathode ray tube depends on the material of the cathode, as the work function and the energy required to emit electrons vary with the material. This statement is true.


3. Statement (C): Cathode rays are streams of electrons that start from the cathode (negative electrode) and travel towards the anode (positive electrode) in a cathode ray tube. This statement is true.


4. Statement (D): The electrons themselves are not affected by the nature of the gas present in the cathode ray tube. The nature of the gas may affect the fluorescence or color of the emitted light, but not the nature of the electrons. This statement is false.


Thus, the number of true statements is 2 (A and B).
Quick Tip: The de Broglie wavelength of a particle is inversely proportional to its momentum (\( p = mv \)), and it helps describe wave-particle duality.

Bromic acid is \( HBrO_3 \). Let the oxidation state of bromine be \( x \).
In \( HBrO_3 \), the sum of oxidation states is: \[ x + (-2 \times 3) + 1 = 0 \quad \Rightarrow \quad x - 6 + 1 = 0 \quad \Rightarrow \quad x = +5. \]
Thus, the oxidation state of bromine in bromic acid is \( +5 \).

Perbromic acid is \( HBrO_4 \). Let the oxidation state of bromine be \( x \).
In \( HBrO_4 \), the sum of oxidation states is: \[ x + (-2 \times 4) + 1 = 0 \quad \Rightarrow \quad x - 8 + 1 = 0 \quad \Rightarrow \quad x = +7. \]
Thus, the oxidation state of bromine in perbromic acid is \( +7 \).

Sum of oxidation states: \( 5 + 7 = 12 \). Quick Tip: The oxidation state of bromine in its oxoacids increases as the number of oxygen atoms increases.

The Nernst equation for the given half-cell reaction is: \[ E = E^\circ - \frac{0.059}{n} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^n} \]
Where:
- \( E \) is the electrode potential,
- \( E^\circ = 1.54 \, V \),
- \( n = 5 \) (number of electrons transferred),
- \( [Mn^{2+}] = 0.001 \, M \),
- \( [MnO_4^-] = 0.1 \, M \),
- \( [H^+] = 10^{-pH} \).

Substitute the values into the equation: \[ 1.282 = 1.54 - \frac{0.059}{5} \log \frac{0.001}{0.1 \cdot [H^+]^5}. \]

Simplify the terms: \[ 1.282 = 1.54 - 0.0118 \log \frac{0.001}{0.1 \cdot [H^+]^5}. \]

Rearranging: \[ 0.0118 \log \frac{0.001}{0.1 \cdot [H^+]^5} = 1.54 - 1.282 = 0.258. \]
\[ \log \frac{0.001}{0.1 \cdot [H^+]^5} = \frac{0.258}{0.0118} = 21.86. \]

Simplify the logarithmic term: \[ \frac{0.001}{0.1 \cdot [H^+]^5} = 10^{21.86}. \]

Taking \( [H^+]^5 \): \[ [H^+]^5 = \frac{0.1}{0.001 \cdot 10^{21.86}}. \]

Taking the fifth root and solving for pH: \[ pH = 3. \] Quick Tip: The Nernst equation relates the electrode potential to the concentrations of the reactants and products, including the hydrogen ion concentration for redox reactions involving \( H^+ \).

The given conditions indicate that the compound must:


1. Not dissolve in NaOH or HCl, which means it is a neutral compound, such as an ether.

2. Not give an orange precipitate with 2,4-DNP, meaning it is not a carbonyl compound (aldehyde or ketone).

3. On hydrogenation, give an identical compound, which implies it has an unsaturated bond (e.g., an aromatic ring) that does not change the functional group upon reduction.


For the molecular formula \( C_9H_{10}O \), possible isomeric structures satisfying the above conditions are:


1. \( p \)-methoxy toluene (\( C_6H_4(OCH_3)CH_3 \)), where the methoxy group is attached to the aromatic ring in the para position.

2. \( o \)-methoxy toluene (\( C_6H_4(OCH_3)CH_3 \)), where the methoxy group is attached to the aromatic ring in the ortho position.


These compounds are ethers, do not react with NaOH, HCl, or 2,4-DNP, and remain the same upon hydrogenation.


Thus, the number of isomeric compounds is \( 2 \).
Quick Tip: Ethers are neutral compounds that do not react with acids, bases, or 2,4-DNP, and aromatic ethers retain their structure upon hydrogenation of the benzene ring.

Let the initial concentration of both \( X(g) \) and \( A(g) \) be \( C \) and the degree of dissociation be \( \alpha \) (same for both reactions).

1. For the reaction \( X(g) \rightleftharpoons Y(g) + Z(g) \):
At equilibrium,
\[ Total moles = C(1 - \alpha) + C\alpha + C\alpha = C(1 + \alpha), \]
and the total pressure \( P_1 = k(C)(1 + \alpha), \) where \( k \) is a proportionality constant.

The equilibrium constant \( K_{p1} \) is given as:
\[ K_{p1} = \frac{(\alpha C)^2}{C(1 - \alpha)} = 3. \]
Solve for \( \alpha \):
\[ \alpha = \frac{3}{4}. \]

2. For the reaction \( A(g) \rightleftharpoons 2B(g) \):
At equilibrium,
\[ Total moles = C(1 - \alpha) + 2C\alpha = C(1 + \alpha), \]
and the total pressure \( P_2 = k(C)(1 + \alpha). \)

The equilibrium constant \( K_{p2} \) is given as:
\[ K_{p2} = \frac{(2\alpha C)^2}{C(1 - \alpha)} = 1. \]
Solve for \( \alpha \):
\[ \alpha = \frac{1}{2}. \]

3. Calculate the ratio of total pressures:
\[ \frac{P_1}{P_2} = \frac{C(1 + \frac{3}{4})}{C(1 + \frac{1}{2})} = \frac{12}{1}. \]

Thus, the ratio \( x : 1 \) is \( 12 : 1 \). Quick Tip: The degree of dissociation (\( \alpha \)) can be calculated by relating the equilibrium constant to the initial and equilibrium concentrations.

Chirality occurs in a molecule if it has at least one chiral center, which is a carbon atom bonded to four different groups. Analyze each compound:


1. Compound 1 (\( Ph-CH(COOH) \)): The carbon attached to \( COOH \), \( H \), and \( Ph \) is chiral. This compound is chiral.


2. Compound 2 (A fused aromatic ring with an ester group): No carbon atom is bonded to four different groups. This compound is not chiral.


3. Compound 3 (A sugar-like compound with multiple hydroxyl groups): The molecule has several chiral centers (multiple asymmetric carbons). This compound is chiral.


4. Compound 4 (A cyclopropane derivative with two \( COOH \) groups): The cyclopropane carbon atoms are symmetrically substituted and do not have four different groups. This compound is not chiral.


Thus, the total number of chiral compounds is \( 2 \).
Quick Tip: A chiral center is a carbon atom with four different groups attached to it. Symmetry in a molecule often eliminates chirality.

The decay of a substance follows the formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}, \]
where \( N(t) \) is the concentration at time \( t \), \( N_0 \) is the initial concentration, and \( t_{1/2} \) is the half-life of the substance.

Let the initial concentration of A be \( N_A \), and the initial concentration of B be \( N_B = 4N_A \).

For substance A: \[ N_A(t) = N_A \left( \frac{1}{2} \right)^{\frac{t}{15}}. \]

For substance B: \[ N_B(t) = 4N_A \left( \frac{1}{2} \right)^{\frac{t}{5}}. \]

We are asked to find the time \( t \) when the concentrations of A and B are the same, i.e., when \( N_A(t) = N_B(t) \). Therefore, we set the equations equal to each other: \[ N_A \left( \frac{1}{2} \right)^{\frac{t}{15}} = 4N_A \left( \frac{1}{2} \right)^{\frac{t}{5}}. \]

Canceling \( N_A \) from both sides: \[ \left( \frac{1}{2} \right)^{\frac{t}{15}} = 4 \left( \frac{1}{2} \right)^{\frac{t}{5}}. \]

Simplifying: \[ \left( \frac{1}{2} \right)^{\frac{t}{15}} = \left( \frac{1}{2} \right)^{\frac{t}{5}} \times 4. \]

Since \( 4 = 2^2 \), we can rewrite the equation as: \[ \left( \frac{1}{2} \right)^{\frac{t}{15}} = \left( \frac{1}{2} \right)^{\frac{t}{5}} \times \left( \frac{1}{2} \right)^{-2}. \]

This simplifies to: \[ \left( \frac{1}{2} \right)^{\frac{t}{15}} = \left( \frac{1}{2} \right)^{\frac{t}{5} - 2}. \]

Equating the exponents: \[ \frac{t}{15} = \frac{t}{5} - 2. \]

Solving for \( t \): \[ \frac{t}{15} - \frac{t}{5} = -2 \quad \Rightarrow \quad \frac{t}{15} - \frac{3t}{15} = -2 \quad \Rightarrow \quad -\frac{2t}{15} = -2 \quad \Rightarrow \quad t = 15 min. \]

Conclusion: It will take 15 minutes for the concentrations of A and B to become the same. Quick Tip: When solving problems involving radioactive decay, it's important to use the formula for exponential decay and balance the exponents to find the required time.

The reaction to determine \( X \) is: \[ H_2O(g) \rightarrow H(g) + OH(g). \]
Using Hess's Law, we can derive the enthalpy change for this reaction by breaking it into steps:

1. Break \( H_2O(g) \) into its constituent atoms: \[ H_2O(g) \rightarrow 2H(g) + O(g). \]
From the given data, the enthalpy change for this reaction can be written as: \[ \Delta H^\circ = \Delta H^\circ_{H_2 \rightarrow 2H} + \Delta H^\circ_{\frac{1}{2}O_2 \rightarrow O} = 436 + 249 = 685 \, kJ mol^{-1}. \]

2. Combine 1 hydrogen atom (\( H(g) \)) and 1 oxygen atom (\( O(g) \)) to form an OH radical: \[ H(g) + O(g) \rightarrow OH(g). \]
The enthalpy change for this step can be calculated using the reverse of the reaction: \[ H_2(g) + O_2(g) \rightarrow 2OH(g), \, \Delta H^\circ = 78 \, kJ mol^{-1}. \]
Divide by 2 to get the enthalpy for forming 1 mole of OH: \[ \Delta H^\circ = \frac{78}{2} = 39 \, kJ mol^{-1}. \]

3. Combine the results: \[ \Delta H^\circ = 685 - 39 = 499 \, kJ mol^{-1}. \]

Thus, \( X = 499 \). Quick Tip: Hess's Law allows us to calculate the enthalpy change of a reaction by summing the enthalpy changes of individual steps that lead to the overall reaction.

1. Calculate the moles of AgNO\(_3\):
\[ Moles of AgNO_3 = Molarity \times Volume (L) = 1 \times 0.020 = 0.020 \, mol. \]

2. Determine moles of KCl:

From the reaction: \[ AgNO_3 + KCl \rightarrow AgCl + KNO_3, \]
1 mole of AgNO\(_3\) reacts with 1 mole of KCl. Therefore, the moles of KCl are: \[ Moles of KCl = 0.020 \, mol. \]

3. Calculate the molality of KCl solution:

The volume of the solution is 25 mL, and the density is \( 1 \, g mL^{-1} \), so the mass of the solution is: \[ Mass of solution = 25 \, g. \]
Since KCl solution is assumed to be dilute, the mass of the solvent is approximately: \[ Mass of solvent = 25 \, g = 0.025 \, kg. \]
The molality is: \[ Molality = \frac{Moles of solute}{Mass of solvent (kg)} = \frac{0.020}{0.025} = 0.8 \, mol kg^{-1}. \]

4. Calculate the depression in freezing point:

KCl dissociates completely into \( K^+ \) and \( Cl^- \), so the van 't Hoff factor (\( i \)) is 2. The depression in freezing point is: \[ \Delta T_f = i \cdot K_f \cdot Molality. \] \[ \Delta T_f = 2 \cdot 2.0 \cdot 0.8 = 3.2 \, K. \]
The nearest integer is: \[ \Delta T_f = 3 \, K. \] Quick Tip: The van 't Hoff factor (\( i \)) accounts for the number of particles into which a solute dissociates in solution. For ionic compounds like KCl, \( i \) is the number of ions formed.