Since molality is 1/5 and molar boiling point constant is equal to 1/molality here. Hence, correct answer is 5.

The number of free ions produced in this case is 3. Hence, van't Hoff factor (i) = 3.

Ca(NO₃)₂ → Ca⁺² + 2NO₃^-1

Van't Hoff factor (i) = 3

The change in the colligative properties depends on the number of particles of the non-volatile solute,

The 0.1 M solution of CaCl₂ will furnish 0.3 moles of ions (assuming α=1).

Since the particles of the non- volatile solute are maximum in CaCl₂ the change in the colligative properties will be maximum.

Therefore the solution of CaCl₂ will have the lowest vapour pressure.

0.1 mol of NaCl will furnish 0.2 moles (2 x 0.1) of ions. While urea and glucose non-ionisable. Therefore, i = 1, and both will gives 0.1 moles of ions.

In case of a buffer solution containing equimolar amount of ammonium hydroxide and ammonium chloride, the reserve basicity of the solution is due to the presence of ammonium hydroxide molecules.

Acidic buffer solution is produced on mixing the aqueous solutions of acid and acidic salt.

pH=-log[10^-6]=6

Use the expression: K_sp = [Ag⁺][Cl^-]

The oxidation state of sulphur in Na₂S₄O₆ is 5/2.

E_cell = E_Reduction (cathode) - E_Reduction (anode)

Or

E_cell = E_Oxidation (anode) - E_Oxidation (cathode)

A cell functions when Ecell is positive, i.e. when E_Reduction(cathode) > E_Reduction (anode) and E_Oxidation (anode) > E_Oxidation (cathode).

Since the order of oxidation potentials is: B > D > A > C.

Therefore, the cell will not function when A is at anode and B is at cathode.

In the reaction Cl₂ + H₂O₂ → 2HCl + O₂, the oxidation number of oxygen changes from -1 to 0.

Hence, H₂O₂ is oxidised and acts as a reducing agent.

The substances with a lower value of reduction potential are stronger reducing agents.

Hence, the increasing order of the strengths as reducing agents is:

Cl₂, Fe³⁺, H₂, Zn