Calculate the boiling point of a 2.00 molal solution of magnesium perchlorate, mg(clo4)2, in water.

103.06°C.

Explanation:

ΔTb = i.Kb.m,

Where, i is the van 't Hoff factor.

Kb is the molal boiling point  elevation constant of water (Kb = 0.51°C/m).

m is the molality of the solution (m = 2.0 m).

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

Mg(ClO₄)₂ → Mg²⁺ + 2ClO₄⁻,

1 mol of Mg(ClO₄)₂ produces 3 mol of ions (1 mol Mg²⁺ and 2 mol ClO₄⁻).

∴ i = 3/1 = 3.

∴ ΔTb = i.Kb.m = (3)(0.51 °C/m)(2.0 m) = 3.06 °C.

∵  ΔTb = the boiling point in presence of solute (Mg(ClO₄)₂) - the boiling point of pure water.

The boiling point of pure water = 100.0°C.

∴ The boiling point in presence of solute (Mg(ClO₄)₂) = ΔTb + the boiling point of pure water = 3.06 °C + 100.0°C = 103.06°C.

the answer for sodium is 0.186

the answer for rubidium is 0.251