Step 1: Analyze the Reaction from B to C (Reagent 'X')
The transformation from B to C involves the esterification of the phenolic OH group. This can be achieved using acetic anhydride ((CH₃CO)₂O) in the presence of an acid catalyst (H⁺). This reaction is known as Fischer esterification.

Step 2: Analyze the Reaction from B to A (Reagent 'Y')
The transformation from B to A involves the esterification of the carboxylic acid group (COOH) with methanol (CH₃OH) in the presence of an acid catalyst (H⁺) and heat (Δ). This is also a Fischer esterification.

Conclusion: The reagents X and Y are (CH₃CO)₂O/H⁺ and CH₃OH/H⁺, Δ, respectively, which corresponds to option (1).

Step 1: Consider the Trend Down the Group
Bond enthalpy generally decreases down a group in the periodic table. This is because as the atomic size increases, the bond length increases, and longer bonds are weaker.

Step 2: Analyze the Given Elements
The elements in question are C, Si, Ge, and Sn. They all belong to Group 14. Their atomic size increases down the group in the order C < Si < Ge < Sn.

Step 3: Determine the Bond Enthalpy Order
Since bond enthalpy decreases with increasing atomic size, the correct order of bond enthalpy is: C - C > Si - Si > Ge - Ge > Sn - Sn

Conclusion: The correct order is given in option (4).

Step 1: Analyze the Structures
All four structures represent ascorbic acid (vitamin C), but they differ in the position of the double bond within the ring and the configuration of the hydroxyl groups.

Step 2: Consider Resonance Stabilization
The most stable structure will be the one with the greatest resonance stabilization. Structure (1) has the most resonance structures possible because the double bond is conjugated with the carbonyl group, allowing for delocalization of electrons. This delocalization stabilizes the structure more than the other structures. Also, the hydroxyl group on C2 will donate electron density to the carbonyl group at C1, whereas in structure 2 the carbonyl group will pull electrons making it unstable.

Conclusion: Structure (1) is the most stable due to resonance stabilization and intramolecular hydrogen bonding possibility.

Step 1: Identify the Reaction Mechanism
The given reaction is a nucleophilic substitution reaction, specifically an S_N1 reaction. The S_N1 mechanism proceeds in two steps:

1. Ionization: The C-Cl bond breaks to form a carbocation intermediate ((C₆H₅)₃C⁺) and a chloride ion (Cl^-). This step is slow and rate-determining.
2. Nucleophilic Attack: The hydroxide ion (OH^-) attacks the carbocation to form the product ((C₆H₅)₃C-OH). This step is fast.

Step 2: Determine the Rate Law
Since the first step is rate-determining, the rate of the reaction depends only on the concentration of the alkyl halide ((C₆H₅)₃C-Cl): Rate = k[(C₆H₅)₃C-Cl] where k is the rate constant. The rate is independent of the concentrations of hydroxide ion (OH⁻) and pyridine.

Step 3: Analyze the Graphs
The correct graph should show a linear relationship between the rate and the concentration of (C₆H₅)₃C-Cl. This is represented by graph (3).

Conclusion: Graph (3) correctly represents the reaction.

Step 1: Identify the Reactants
The reactants are tetrahydrofuran (THF) and 2-methylpropene. The reaction is catalyzed by HF and takes place under heat.

Step 2: Determine the Reaction Mechanism
This reaction is an electrophilic addition of THF to the alkene. HF protonates the alkene to form a carbocation. The oxygen in THF acts as a nucleophile and attacks the carbocation. Finally, deprotonation occurs to yield the product.

Step 3: Determine the Major Product
The major product is determined by Markovnikov's rule, which states that the proton adds to the carbon of the double bond with more hydrogens. In this case, the carbocation will form on the more substituted carbon of 2-methylpropene, leading to product (1).

Conclusion: The major product 'X' is represented by structure (1).

Step 1: Analyze the Complexes for Isomerism
We are looking for a complex cation that exhibits two isomers.

1. [Co(H₂O)₆]³⁺: This complex has only one possible structure, as all ligands are the same (H₂O). So, it does not have isomers.
2. [Co(NH₃)₅Cl]²⁺: This complex can only exhibit ionization isomerism if another counter ion is present within the complex to exchange positions with the Cl ligand. With only one Cl, there's no possibility of isomerism here.
3. [Co(NH₃)₅NO₂]²⁺: This complex can exhibit linkage isomerism because the NO₂ ligand can coordinate to the metal ion through either the nitrogen atom (nitro isomer) or the oxygen atom (nitrito isomer). Thus, it has two isomers.
4. [Co(NH₃)₅Cl]⁺: Same explanation as (2), no isomerism is possible.

Conclusion: The complex cation [Co(NH₃)₅NO₂]²⁺ exhibits linkage isomerism and has two isomers. Therefore, the correct answer is (3).

Step 1: Analyze Statement I
Sulfanilic acid contains an amine group (-NH₂) which is attached to the benzene ring, and a sulfonic acid group (-SO₃H). It does not contain a carboxyl group (COOH). Esterification is a characteristic reaction of carboxylic acids. Therefore, Statement I is incorrect.

Step 2: Analyze Statement II
Lassaigne's test is used to detect the presence of nitrogen, sulfur, halogens, and phosphorus in organic compounds. Sulfanilic acid contains sulfur and nitrogen. The red color in Lassaigne's test is due to the formation of ferric thiocyanate [Fe(SCN)₃] when sulfur is present. Thus, Statement II is correct.

Conclusion: Statement I is incorrect, but Statement II is correct. The correct answer is option (4).

Step 1: Analyze Assertion (A)
Gypsum (CaSO₄·2H₂O) is used for making fireproof wall boards. When heated, gypsum loses water and forms plaster of Paris, which is a good fire-resistant material. Hence, assertion (A) is correct.

Step 2: Analyze Reason (R)
Gypsum is unstable at high temperatures as it loses water molecules upon heating. Hence, Reason (R) is also correct.

Step 3: Determine the Relationship between (A) and (R)
While both statements are correct, the reason gypsum is used in fireproof wall boards is not simply because it's unstable at high temperatures. It's because the water molecules present in gypsum act as a fire retardant. When exposed to fire, the water molecules are released as steam, absorbing a significant amount of heat and preventing the spread of the fire. This process makes the wall board fire resistant. Therefore, (R) is not the correct explanation for (A).

Conclusion: Both (A) and (R) are correct, but (R) is not the correct explanation of (A). The correct option is (1).

Nessler's reagent is an alkaline solution of potassium tetraiodomercurate(II) (K₂[HgI₄]). Its chemical formula indicates the presence of potassium (K), mercury (Hg), and iodine (I). Oxygen is not present in the reagent itself but might be involved in the reaction when it's used to test for ammonia.

Conclusion: Oxygen is not present in Nessler's reagent. The correct option is (4).

Step 1: Analyze Assertion (A)
α-halocarboxylic acids react with dilute ammonia (NH₃) to give a good yield of α-amino carboxylic acids. This is because the carboxyl group (-COOH) increases the reactivity of the α-halo group towards nucleophilic substitution. In contrast, the yield of amines from simple alkyl halides reacting with ammonia is low due to overalkylation, where the initially formed amine can react further with the alkyl halide. Therefore, Assertion (A) is correct.

Step 2: Analyze Reason (R)
Amino acids exist as zwitterions in aqueous solutions and in the solid state. A zwitterion has both positive and negative charges within the same molecule, resulting in a net charge of zero. This is due to the acidic carboxyl group and the basic amino group present in amino acids. Thus, Reason (R) is correct.

Step 3: Analyze the Relationship between (A) and (R)
While both statements are individually correct, Reason (R) doesn't explain Assertion (A). The higher yield of amino acids from α-halocarboxylic acids is due to the enhanced reactivity of the α-halo group, not the zwitterionic nature of amino acids. The zwitterionic form is a characteristic of the product (amino acid) but doesn't explain the higher yield compared to the reaction of alkyl halides with ammonia.

Conclusion: Both (A) and (R) are correct, but (R) is not the correct explanation of (A). Therefore, the correct answer is (2).

Manufacturing of cement, steel manufacturing, and electricity generation in thermal power plants are major contributors to greenhouse gas emissions, primarily CO₂, which is a significant driver of global warming.

Cement production releases CO₂ through the calcination of limestone.

Steel manufacturing uses coal, a carbon-intensive fuel, releasing CO₂.

Thermal power plants also burn fossil fuels to generate electricity, leading to substantial CO₂ emissions.

While urea production does have an environmental footprint, its contribution to global warming is much less than the other three activities listed. The primary greenhouse gas emissions associated with urea production are nitrous oxide (N₂O) from fertilizer application and CO₂ from energy use in the production process. However, these emissions are considerably lower compared to those from cement, steel, and electricity generation.

Conclusion: Among the given options, industrial production of urea is the least responsible for global warming.

Step 1: Reaction with NaHSO₃ and Dilute HCl (Formation of A)
The starting compound is butan-2-one. The reaction with NaHSO₃ followed by dilute HCl results in the formation of a cyanohydrin. The CN^- ion from NaCN attacks the carbonyl carbon, and the oxygen picks up a proton. The major product A is 2-hydroxy-2-methylbutanenitrile.

Step 2: Reduction with LiAlH₄ (Formation of B)
Lithium aluminum hydride (LiAlH₄) is a strong reducing agent. It reduces the nitrile group (CN) to an amine group (NH₂). The product B is 1-amino-2-methylbutan-2-ol.

Step 3: Hydrolysis with HCl/H₂O and Heat (Formation of C)
The amine group in B is hydrolyzed with HCl/H₂O and heat (△) into carboxylic group. The major product C is 2-hydroxy-2-methylbutanoic acid.

Conclusion: The structures of A, B, and C correspond to option (4).

Step 1: Analyze Assertion (A)
Cu²⁺ is more stable than Cu⁺ in aqueous solution. This is due to the higher hydration enthalpy of Cu²⁺ compared to Cu⁺. The hydration enthalpy compensates for the second ionization energy of copper, making Cu²⁺ more stable in an aqueous medium. Thus, Assertion (A) is correct.

Step 2: Analyze Reason (R)
The enthalpy of hydration is the energy released when one mole of gaseous ions is dissolved in water. The hydration enthalpy is directly proportional to the charge density of the ion. Since Cu²⁺ has a smaller ionic radius and a greater charge than Cu⁺, its charge density is higher. Consequently, the enthalpy of hydration for Cu²⁺ is much more negative (meaning greater energy release and stronger interaction with water) than that of Cu⁺. Thus, Reason (R) is incorrect. *It should say Enthalpy of hydration is much MORE for Cu²⁺*

Conclusion: Both (A) and edited-(R) are correct and (R) is the correct explanation of (A).

Deuterated hydrogen peroxide (D₂O₂) can be conveniently prepared in the laboratory by reacting K₂S₂O₈ with deuterated sulfuric acid (D₂SO₄) in D₂O (heavy water). This method allows for the direct incorporation of deuterium into the hydrogen peroxide molecule.

Conclusion: K₂S₂O₈ is the starting material for convenient preparation of D₂O₂ (Option 1).

Step 1: Recall the Freundlich Adsorption Isotherm
The Freundlich adsorption isotherm is given by: ^x⁄_m = KP^1⁄_n where x is the mass of adsorbate, m is the mass of adsorbent, P is the pressure, K is the Freundlich constant, and n is a constant.

Step 2: Linearize the Equation
Taking the logarithm of both sides, we get: log(^x⁄_m) = log K + ¹⁄_n log P

Step 3: Compare with the Given Equation
The given equation is y = 3x + 2.505, where y = log(^x⁄_m) and x = log P. Comparing this with the linearized Freundlich equation, we have:¹⁄_n=3 and log K = 2.505

Conclusion: The value of ¹⁄_n is ¹⁄₃, and log K is 2.505 (Option 3).

Step 1: Analyze Assertion (A)
In volumetric analysis, the concentration of the solutions used must be known accurately. KOH solutions are commonly used as titrants in acid-base titrations. The concentration of a KOH solution can change over time due to various factors. Therefore, it's crucial to check and standardize its concentration before use. Assertion (A) is correct.

Step 2: Analyze Reason (R)
KOH solutions absorb atmospheric carbon dioxide (CO₂). The reaction between KOH and CO₂ forms potassium carbonate (K₂CO₃) and water: 2KOH + CO₂ → K₂CO₃ + H₂O This reaction consumes KOH, reducing its concentration in the solution. Hence, Reason (R) is correct.

Step 3: Analyze the Relationship between (A) and (R)
The absorption of atmospheric CO₂ by KOH solution directly affects its concentration. This is the primary reason why the concentration of a KOH solution needs to be checked before use, especially if it's an older solution. Thus, Reason (R) is the correct explanation for Assertion (A).

Conclusion: Both (A) and (R) are correct, and (R) is the correct explanation for (A). The correct answer is (3).

Step 1: Understand Isoelectronic Species
Isoelectronic species are atoms or ions that have the same number of electrons.

Step 2: Calculate the Number of Electrons in Each Ion

1. Li⁺(2 electrons), Na⁺(10 electrons), Mg²⁺(10 electrons), Ca²⁺(18 electrons)
2. Ba²⁺(54 electrons), Sr²⁺(36 electrons), K⁺(18 electrons), Ca²⁺(18 electrons)
3. N^3-(10 electrons), O^2-(10 electrons), F^-(10 electrons), S^2-(18 electrons)
4. K⁺(18 electrons), Cl^-(18 electrons), Ca²⁺(18 electrons), Sc³⁺(18 electrons)

Conclusion: The set of ions in option (4) (K⁺, Cl^-, Ca²⁺, Sc³⁺) all have 18 electrons and are therefore isoelectronic.

Adding an inert gas like helium at constant volume does not affect the equilibrium position. This is because the partial pressures of the reactants and products remain unchanged. However, if helium is added at constant pressure, the volume of the system will increase. This decrease in concentration affects all gaseous products equally and therefore it shifts the equilibrium towards the side with more gas molecules, according to Le Chatelier's principle. In this case, that is the forward direction, producing more Cl₂ and PCl₃.

Conclusion: Since the question does not specify if it was done at constant volume or constant pressure, the answer will be both (1) and (4).

Electron Gain Enthalpy Trend
Electron gain enthalpy generally becomes less negative (less exothermic) down a group due to increasing atomic size. However, there can be exceptions due to factors like electron-electron repulsion and shielding effects. Fluorine has a smaller atomic size and therefore greater electron density and experiences inter electronic repulsions more than chlorine and hence its magnitude is smaller than that of Cl.

Analyzing the Options

1. Cl > F (Correct, due to small size and inter electronic repulsions of F)
2. S > Se (Incorrect, electron gain enthalpy becomes less negative down the group; therefore, S should have a more negative electron gain enthalpy than Se. This is an exception).
3. At > I (Correct)
4. Po > Te (Correct)

Conclusion: The incorrect option is (2).

Step 1: Analyze the O-O bond length
In H₂O₂, the oxygen atoms are bonded to hydrogen atoms. In F₂O₂, the oxygen atoms are bonded to fluorine atoms. Fluorine is more electronegative than hydrogen. The higher electronegativity of fluorine in F₂O₂ leads to a greater pull of electron density towards the fluorine atoms, which weakens the O-O bond and increases its bond length. Therefore, the O-O bond length in H₂O₂ is *longer* than in F₂O₂. So, X is longer.

Step 2: Analyze the O-H and O-F bond lengths
The O-H bond is formed between oxygen and hydrogen, while the O-F bond is formed between oxygen and fluorine. Fluorine has a smaller atomic radius than hydrogen. Also, oxygen and fluorine have much closer electronegativities, leading to a shorter O-F bond compared to the O-H bond where there's a larger electronegativity difference. Therefore, the O-H bond length in H₂O₂ is *shorter* than the O-F bond length in F₂O₂. So, Y is shorter.

Conclusion: The O-O bond length in H₂O₂ is longer than in F₂O₂, and the O-H bond length is shorter than the O-F bond length. This corresponds to option (4).

The balanced chemical equation for the combustion of ethane is:
C₂H₆(g) + ⁷⁄₂ O₂(g) → 2CO₂(g) + 3H₂O(l)

Heat evolved at constant volume (q_v): The heat absorbed by the calorimeter is given by: q_v = CΔT Where C is the heat capacity of the calorimeter system and ΔT is the temperature change.

q_v = 20 kJ K⁻¹ × 0.5 K = 10 kJ
Since the combustion is exothermic, the heat evolved is -10 kJ. This is for 0.3 g of ethane.

Moles of ethane:
Molar mass of ethane (C₂H₆) = 2 × 12 + 6 × 1 = 30 g mol^-1
Moles of ethane = ^{0.3 g} ⁄_{30 g mol^-1} = 0.01 mol

Heat evolved per mole at constant volume (ΔU):
ΔU = ^{-10 kJ} ⁄ _{0.01 mol} = -1000 kJ mol^-1

Heat evolved at constant pressure (ΔH): For the given reaction, the change in the number of gaseous moles is:
Δn_g = n_products - n_reactants = (2) - (1 + ⁷⁄₂) = -2.5

The relationship between ΔH and ΔU is:
ΔH = ΔU + Δn_gRT
ΔH = −1000 kJ mol⁻¹ + (-2.5) × 8.3 × 10^-3 kJ K⁻¹mol^-1 × 300 K
ΔH = −1000 – 6.225 = -1006.225 kJ mol^-1

The nearest integer is -1006 kJ/mol.

Copper matte is a molten mixture primarily composed of Cu₂S and FeS. It is an intermediate product in the smelting of copper ore. Of the given compounds, only Cu₂S is present in copper matte.

Conclusion: Only one of the listed compounds (Cu₂S) is present in copper matte.

Tranquilizers
Tranquilizers are drugs used to treat anxiety and mental disorders. They work by depressing the central nervous system.

Analyzing the Given Compounds

• Chlordiazepoxide: This is a benzodiazepine and is used as a tranquilizer.
• Veronal: This is a barbiturate and acts as a sedative-hypnotic, *not* a tranquilizer.
• Valium (Diazepam): This is also a benzodiazepine and is used as a tranquilizer.
• Salvarsan: This is an organoarsenic compound historically used to treat syphilis, *not* a tranquilizer.

Conclusion: Three of the given compounds (Chlordiazepoxide, Valium) are tranquilizers. (Sometimes, Veronal is loosely classified as a tranquilizer because of its sedative properties, but it is primarily a hypnotic, not an anxiolytic.)

For a zero-order reaction, the integrated rate law is given by:
[A]_t = [A]₀ - kt
where [A]_t is the concentration of A at time t, [A]₀ is the initial concentration of A, and k is the rate constant.

The half-life (t_1/2) of a zero-order reaction is given by:
t_1/2 = ^[A]₀⁄_2k

Given that t_1/2 = 50 min, we can find the rate constant k:
k = ^[A]₀⁄_{(2 × t1/2)} = ^[A]₀⁄_{(2 × 50)} = ^[A]₀⁄₁₀₀

We are asked to find the time taken for the concentration of A to reduce to one-fourth of its initial value. Let this time be t. So, [A]_t = ^[A]₀⁄₄. Substituting this into the integrated rate law:

^[A]₀⁄₄ = [A]₀ - kt
^3[A]₀⁄₄ = kt

Substituting the value of k we found earlier:
t = (^3[A]₀⁄₄) × (¹⁰⁰⁄_[A]0) = 3 × 25 = 75 min

Moles of acetic acid:
Molar mass of acetic acid (CH₃COOH) = 2 × 12 + 4 × 1 + 2 × 16 = 60 g/mol
Moles of acetic acid = ^5g⁄_{60 g/mol} = ¹⁄₁₂ mol

Molality of acetic acid:
Mass of water = Volume × Density = 500 mL × 1 g/mL = 500 g = 0.5 kg
Molality (m) = ^{moles of solute}⁄_{mass of solvent (kg)} = ^{(1⁄₁₂ mol)}⁄_{0.5 kg} = ¹⁄₆ mol/kg

van't Hoff factor (i):
Acetic acid dissociates as follows: CH₃COOH ⇌ CH₃COO^- + H⁺
Since 20% of acetic acid dissociates, the degree of dissociation (α) = 0.2. For dissociation, i = 1 + α(n − 1), where n is the number of particles formed after dissociation. Here, n = 2.
i = 1 + 0.2(2 - 1) = 1 + 0.2 = 1.2

Depression in freezing point (ΔT_f):
ΔT_f = iK_fm where K_f is the molal depression constant.
ΔT_f = 1.2 × 1.86 K kg mol⁻¹ × ¹⁄₆ mol/kg = 0.372 K

Since the change in temperature in Kelvin and Celsius are the same, ΔT_f = 0.372°C = 372 × 10^-3 °C

Let's assume we have 100 mL of the solution. Since it's a 10% v/v solution, the volume of Br₂ is 10 mL and the volume of CCl₄ is 90 mL.

Mass of Br₂ = Volume × Density = 10 mL × 3.2 g/mL = 32 g
Moles of Br₂ = ^Mass⁄_{Molar Mass} = ^{32 g}⁄_{160 g/mol} = 0.2 mol

Mass of CCl₄ = Volume × Density = 90 mL × 1.6 g/mL = 144 g
Molar mass of CCl₄ = 12 + (4 × 35.5) = 12 + 142 = 154 g/mol

Molality (m) = ^{Moles of solute}⁄_{Mass of solvent (in kg)}
m = ^{0.2 mol}⁄_{(144 g / 1000 g/kg)} = ^{0.2 mol}⁄_{0.144 kg} = 1.3888... mol/kg ≈ 1.39 mol/kg

Therefore, x = 139.

AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)
K_sp = [Ag⁺][Br⁻] = 4.9 × 10^-13

Let the solubility of AgBr be 's' mol/L. Then, [Ag⁺] = s and [Br⁻] = s.
s² = 4.9 × 10^-13
s = √(4.9 × 10^-13) = 7 × 10^-7 M

Since 1 L of saturated AgBr solution is taken, the concentration of Ag⁺ and Br⁻ from AgBr are both 7 × 10^-7 M. We are adding 1 × 10⁻⁵ M AgNO₃.

The Ag⁺ from AgNO₃ will be significantly greater than the Ag⁺ from AgBr, so we can approximate the total [Ag⁺] as 1 × 10⁻⁵ M.

The common ion effect will suppress the solubility of AgBr, so the [Br⁻] remains approximately 7 × 10^-7 M. The [NO₃⁻] will be 1 × 10^-5 M.

Conductivity (κ) = Σλ_ic_i
κ = λ_Ag⁺[Ag⁺] + λ_Br^-[Br⁻] + λ_NO3^-[NO₃⁻]
κ = (6 × 10⁻³)(1 × 10⁻⁵) + (8 × 10⁻³)(7 × 10⁻⁷) + (7 × 10^-3)(1 × 10⁻⁵)
κ = 6 × 10⁻⁸ + 5.6 × 10^-9 + 7 × 10^-8
κ ≈ 13.56 × 10⁻⁸ ≈ 14 × 10^-8 Sm^-1

An asymmetric carbon atom (chiral center) is a carbon atom that is bonded to four different groups. Examining the structure of testosterone reveals six such carbon atoms:

1. The carbon atom at the junction of the six-membered ring with the ketone group (C=O) and the five-membered ring.
2. The carbon atom in the five-membered ring attached to the methyl group (CH₃) and the hydroxyl group (OH).
3. The carbon atom at the top of the other six-membered ring that forms a bridge.
4. The carbon atom at the bottom of the other six-membered ring that forms a bridge.
5. The carbon atom at the top of the six-membered ring with the double bond.
6. The carbon atom at the bottom of the six-membered ring with the double bond.

Therefore, there are a total of six asymmetric carbon atoms in testosterone.

The electronic configuration of Mn is [Ar] 3d⁵ 4s².

In [Mn(H₂O)₆]²⁺, Mn is in +2 oxidation state. Water is a weak field ligand.

Electronic configuration of Mn²⁺ is [Ar] 3d⁵.

Since H₂O is a weak field ligand, there will be no pairing of electrons in the d orbitals.

Number of unpaired electrons (n) = 5

Spin only magnetic moment (µ_spin) = √n(n + 2) B.M. = √5(5+2) = √35 ≈ 5.92 B.M.

Nearest integer is 6.

Density (ρ) = ^{(Z × M)} ⁄ _{(NA × a³)}
where Z = number of atoms per unit cell, M = molar mass, N_A = Avogadro's number, a = edge length

For fcc, Z = 4, a = 2.0 Å = 2.0 × 10^-10 m
For bcc, Z = 2, a = 2.5 Å = 2.5 × 10^-10 m

Let M be the molar mass of the metal M. Then the density for fcc is:
ρ_fcc = ^4M ⁄ _{NA(2×10^-10)³} And for bcc is:
ρ_bcc = ^2M⁄_{NA(2.5×10^-10)³}

The ratio of densities is:
^ρ_fcc⁄_ρbcc= ^{(4M⁄_{NA(2×10-10)3})}⁄_{(^2M⁄NA(2.5×10^-10)³)} = ⁴⁄₂ × ^{(2.5×10-10)3}⁄_{(2×10^-10)³} = 2 × (^2.5⁄₂)³ ≈ 3.9

Nearest integer is 4.