Answers and Explanations: 1. JEE Main: Calculating Atomic Mass Atomic mass = (Mass of isotope 1 × Abundance of isotope 1) + (Mass of isotope 2 × Abundance of isotope 2)

Atomic mass = (12 amu × 98.9%) + (13 amu × 1.1%) Atomic mass = 11.868 amu + 0.143 amu Atomic mass ≈ 12 amu

2. JEE Main: Oxidation States of Sulfur

• In H2S, the oxidation state of sulfur is -2.
• In SO2, the oxidation state of sulfur is +4.
• In H2SO4, the oxidation state of sulfur is +6.

3. JEE Main: Balanced Chemical Equations

• Reaction of sodium with water: 2Na + 2H2O -> 2NaOH + H2

• Reaction of calcium carbonate with hydrochloric acid: CaCO3 + 2HCl -> CaCl2 + H2O + CO2

• Reaction of methane with chlorine: CH4 + 4Cl2 -> CCl4 + 4HCl

4. JEE Main: Neutralization of Hydrochloric Acid To neutralize 100 mL of 0.1 M hydrochloric acid, we need to use the same number of moles of sodium hydroxide (NaOH).

Moles of HCl = Concentration (M) × Volume (L) Moles of HCl = 0.1 M × 0.1 L = 0.01 moles

Since the reaction between HCl and NaOH is a 1:1 mole ratio, we also need 0.01 moles of NaOH.

Mass of NaOH = Moles of NaOH × Molar mass of NaOH Mass of NaOH = 0.01 moles × 40 g/mol = 0.4 grams

Therefore, 0.4 grams of sodium hydroxide is required to neutralize 100 mL of 0.1 M hydrochloric acid.

CBSE Board Exam:

1. Group and Period of Element with Atomic Number 17 The element with atomic number 17 is Chlorine (Cl). It belongs to Group 17 (Halogens) and Period 3.

2. Chemical Formula of Magnesium Oxide The chemical formula for the compound formed when magnesium reacts with oxygen is MgO.

3. Balancing Chemical Equation The balanced chemical equation for the reaction between Fe and HCl is: Fe + 2HCl -> FeCl2 + H2

4. Calculating Molarity Molarity (M) = Molar mass (g/mol) / Volume of solution (L) Moles of NaCl = Mass of NaCl / Molar Mass of NaCl Moles of NaCl = 10g/58.44g/mol =0.171mol Molarity=0.171mol/0.1L=1.71M

Therefore, the molarity of the prepared solution is 1.71 M.

Additional Numerical: 1. Calculating Electronegativity Electronegativity = (Ionization energy of valence electrons) / 2 Electronegativity = (520 + 7300) / 2 = 3910 kJ/mol

Therefore, the electronegativity of the element is 3910 kJ/mol.

2. Determining Lattice Energy Lattice energy = Enthalpy of formation + (Enthalpy of sublimation of metal + Enthalpy of dissociation of non-metal) + (Electron affinity of non-metal)

Lattice energy = -411 kJ/mol + (107 kJ/mol + 242 kJ/mol) + (-349 kJ/mol) = -607 kJ/mol

Therefore, the lattice energy of sodium chloride using the Born-Haber cycle is -607 kJ/mol.

3. Calculating pH pH = -log [H+] [H+] = Concentration of HCl = 0.01 M pH = -log (0.01) = 2

Therefore, the pH of a 0.01 M solution of hydrochloric acid is 2.

4. Determining Ksp Ksp = [Ca2+][CO32-] = (Solubility of CaCO3)^2 Ksp = (0.0089 g/L / 100 g/mol)^2 = 8.01 × 10^-14

Therefore, the solubility product constant (Ksp) of calcium carbonate in water at 25 °C is 8.01 × 10^-14.