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!! =)
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
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.
To calculate the amount of FeCl2 that should be dissolved in 1 kg of water to raise its boiling point to 101.56 degrees Celsius, we need to use the formula for boiling point elevation.
Boiling point elevation (∆Tb) is calculated using the equation:
∆Tb = Kb * m * i
Where:
∆Tb = boiling point elevation (in degrees Celsius)
Kb = boiling point elevation constant, which is given as 0.52 degrees Celsius/m for water
m = molality of the solution (moles of solute per kilogram of solvent)
i = van't Hoff factor or the number of particles into which the solute dissociates when it dissolves
In this case, we are given that the solute (FeCl2) is completely dissociated, so the van't Hoff factor will be 3 (2 Cl- ions and 1 Fe2+ ion).
First, let's calculate the boiling point elevation (∆Tb):
∆Tb = 101.56 - 100.00 = 1.56 degrees Celsius
Next, we need to calculate the molality (m) of the solution.
Molality (m) is calculated using the equation:
m = moles of solute / mass of solvent (in kg)
We are given the mass of the solvent as 1 kg, so we only need to calculate the moles of solute.
The molar mass of FeCl2 is:
Fe = 55.845 g/mol
Cl = 35.453 g/mol (x2 because of two chloride ions in FeCl2)
Molar mass of FeCl2 = 55.845 + 35.453 x 2 = 126.751 g/mol
Using the molar mass, we can calculate the moles of FeCl2:
moles of FeCl2 = mass of FeCl2 / molar mass of FeCl2
Since we don't know the mass of FeCl2, we need to solve for it using the molecular weight of FeCl2 and the molality of the solution:
mass of FeCl2 = molar mass of FeCl2 * molality * mass of solvent (in kg)
mass of FeCl2 = 126.751 g/mol * m * 1 kg
We already know the molality (∆Tb) and the identity of the solvent (water), so we can substitute these values into the equation:
mass of FeCl2 = 126.751 g/mol * (∆Tb / Kb) * 1 kg
Finally, substitute the value of ∆Tb = 1.56 degrees Celsius and Kb = 0.52 degrees Celsius/m into the equation and solve for mass of FeCl2:
mass of FeCl2 = 126.751 g/mol * (1.56 / 0.52) * 1 kg
mass of FeCl2 = 4.727 g
Therefore, approximately 4.7 grams of FeCl2 should be dissolved in 1 kg of water to raise its boiling point to 101.56 degrees Celsius.