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!! =)
6 answers
2.32
Could you show how you solved this problem?
Use Raoult's law.
You will find an explanation at
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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
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
thank you very much!
why sulfuric acid need atmospheric temperature?
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.
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.
4 answers