Question

Colligative Properties

Mixtures and Solutions
Acéllus Learning System
What is the boiling point (in °C) of a 0.743 m
aqueous solution of KCI?
Enter your rounded answer with
3 decimal places.
Kb for water = 0.512 °C/m
Enter

Answer 1

100.761°C

Step-by-step explanation:

ΔTb≡Tb−Tb*=iKbm,

where:

ΔTb is the change in boiling point in ∘C, from that of the pure solvent, Tb*, to that of the solution, Tb.

i is the van't Hoff factor, i.e. the effective number of solute particles in solution.

Kb=0.512∘C/m is the boiling point elevation constant of water.

m is the molality of the solution... mol solute/kg solvent. Is the solute volatile or nonvolatile?

Assuming 100% dissociation...

KCl(aq)→K+(aq)+Cl−(aq)

and 1+1=2≈i, so...

Tb=Tb*+iKbm

=100°C+2×0.512°C/m×0.743 m

=100.761°C