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
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.
1. b
2. a
3. b
4. c
100%
hlsml is correct! 100%
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.
As Bob P points out, Kion may be called the van't Hoff factor, i and for KCl it is 2.
To solve this problem, we can use the boiling point elevation formula:
ΔT = K_b * m
Where:
ΔT is the boiling point elevation,
K_b is the molal boiling point elevation constant (given as 0.512 degrees C/m for the solution),
and m is the molality of the solute (KCl) in the solution.
In this case, we are asked to find the molality (m) of the solution, given the boiling point elevation (ΔT) as 103.07°C.
Re-arranging the equation gives us:
m = ΔT / K_b
Plugging in the values, we get:
m = 103.07°C / 0.512 degrees C/m
m ≈ 201.41 mol/kg
However, the answer choices given are in units of molality (mol/kg), not mol/kg. To convert mol/kg to molality, we divide by 1000:
m = 201.41 mol/kg / 1000
m ≈ 0.20141 mol/kg
Since the question asks for the molality of the solution to the correct number of significant figures, the answer is approximately 0.201, which corresponds to option A: 0.300.
Therefore, the molality of the solution is 0.300 (option A).