What is the predictided change in the boiling point of water when 1.5 g of barium chloride (bacl2) is dissolved in 1.5 kg of water

0.0074 °C.

Explanation:

ΔTb = i.Kb.m,

where, ΔTb is the elevation in boiling water.

i is van 't Hoff factor, The 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. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

i for BaCl₂ = 3/1 = 3.

Kb is the molal elevation constant of water (Kb = 0.512 °C/m).

m is the molality of the solution.

m = (no. of moles of BaCl₂)/(mass of water (kg)) = (mass/molar mass of BaCl₂)/(mass of water (kg)) = (1.5 g/208.23 g/mol)/(1.5 kg) = 0.0048 m.

∴ ΔTb = i.Kb.m = (3)(0.512 °C/m)(0.0048 m) = 0.0074 °C.

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)