Asked by el

How many grams of FeCl2 should be dissolved in 1 kg of water to make its boiling point equal to 101.56 degrees centrigrade at atmospheric pressure of 1 atm? (Kb=0.52 degrees Celsius/m; solute is completely dissociated).

Thanks!! =)
Answered by jinny
2.32
Answered by el
Could you show how you solved this problem?
Answered by drwls
Use Raoult's law.
You will find an explanation at
(Broken Link Removed)

The vant Hoff factor for FeCl2 will be 3, I believe. It dissociates into three ions.

You are trying to elevate the boiling point by 3 Kb.

3 = molality * (vant Hoff factor)

Molality = 1.0
Answered by el
thank you very much!
Answered by Anonymous
why sulfuric acid need atmospheric temperature?
Answered by Anonymous
the answer is 127g ( rounded)
First, you would find the change in temperature: 101. 56- 100 =1.56 *C
The equation that will be used for this is : ∆Tb = i Kb m. We want to find m .
So it will be 1.56 *C/ (3 *.52*C/m) = 1 m, then 1 m * 1 Kg of H20 = 1 mol.
The molar mass of FeCl2 is roughly 127 g, therefore the answer is 127 g.
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