Answer:
(4) 2.0 M CaCl₂(aq).
Explanation:
ΔTb = i.Kb.m,
where, ΔTb is the elevation in boiling point.
i is the van 't Hoff factor.
Kb is the molal elevation constant of water.
m is the molality of the solution.
(1) 1.0 M KCl(aq):
i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.
suppose molarity = molality, m = 1.0 m,
∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).
(2) 2.0 M KCl(aq):
i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.
suppose molarity = molality, m = 2.0 m,
∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).
(3) 1.0 M CaCl₂(aq):
i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.
suppose molarity = molality, m = 1.0 m,
∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).
(4) 2.0 M CaCl₂(aq):
i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.
suppose molarity = molality, m = 2.0 m,
∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).