Question

calculate the boiling point of 1 molar aq solution .(Density is 1.04 g/m) of KCl . Kb for water is 0.52 K Kg/mol . Mb of KCl = 74.5g/mol

Answer 1

Molar mass of KCl = 74.5g/mol
Moles of KCl = 1
Mass of solution = Volume × density = 1000 ​× 1.04 = 1004g
Mass of solvent = Mass of solution - Mass of solute = 1004 - 74.5g = 929.5g

Molality = MolesMass of solvent ×1000 = 1929.5×1000 = 1.07 m

For KCl, i = 2 since it produce 2 ions in the solution.

ΔTb = i × m × KbΔTb = 2×0.52×1.07∆Tb = 1.1128K

Boiling point of solution = To + ΔTb = 373 + 1.1128 = 374.1128K

Answer 2

Answer:

Molar mass of KCl = 74.5g/mol

Moles of KCl = 1

Mass of solution = Volume × density = 1000 ​× 1.04 = 1004g

Mass of solvent = Mass of solution - Mass of solute = 1004 - 74.5g = 929.5g

Molality = MolesMass of solvent ×1000 = 1929.5×1000 = 1.07 m

For KCl, i = 2 since it produce 2 ions in the solution.

ΔTb = i × m × KbΔTb = 2×0.52×1.07∆Tb = 1.1128K

Boiling point of solution = To + ΔTb = 373 + 1.1128 = 374.1128K