What is the molality of a solution of water and KCl if the boiling point of the solution is 103.07°C? (K subscript b equals 0.512 degrees C/m solution; KCl is an ionic compound right parenthesis

Question

A: .300
B: .600
C: 3.00
D: 6.00

I'm really confused as to how to solve this, and can't make an educated guess, so any advice or help would be appreciated.

Answer 1

Educated deduction would be wise.

Tb-100C=Kb*Kions*molality

Now, Kb you have
Tb you have
Kions: 2 because in KCl it breaks up into two ion particales, K+, and Cl-
molality you are solving for.

Some texts use for Kions other symbols, check your text.

Answer 2

As Bob P points out, Kion may be called the van't Hoff factor, i and for KCl it is 2.

Answer 3

1. b
2. a
3. b
4. c

100%

Answer 4

hlsml is correct! 100%