for a question it is given that there is 14.6 g of calcium chloride in 246 g of water. Kf is 1.86 C/m
we need to find the freezing point of this solution.
when i do the calculations,
i get the freezing point to be -0.99 (without using the vant hoff factor i)
or i get -2.98 (using the van't hoff factor i for CaCl3 = 3)
which one is the correct answer and why? when do i use the van't hoff factor and when do i not?
please explain, thank you
The -2.98C using 3 for i is the correct value to use. In truth you ALWAYS use the van't Hoff factor of i.
The equation is delta T = i*Kf*m
The van't Hoff factor is always 1 for almost all organic compounds that do not ionize. Since they don't ionize there is 1 particle/molecule so a 1 always goes in for i. Since it's a 1 we usually don't include it.
For ALL ionic compounds, the ionization determines the value of i.
For units that ionize to give just 1 particle/molecule i = 1. But I can't think of any ionic compounds that do this.
NaCl, KBr, LiF, etc ionize to give 2; therefore, i = 2.
Na2SO4, K2SO4 etc give i = 3
K3PO4 etc give i = 4.
I think this will be what you are looking for.
To find the freezing point of a solution, you need to consider the concept of freezing point depression. The freezing point of a pure solvent is higher than that of a solution containing a solute. The amount by which the freezing point decreases depends on the concentration of the solute and the number of particles it dissociates into when it dissolves.
In this case, you have calcium chloride (CaCl2). However, you mentioned the van't Hoff factor of 3, which suggests that you are considering CaCl3 (calcium chloride trihydrate). Let's assume we are dealing with CaCl2 for now.
To calculate the freezing point depression, you can use the equation:
ΔT = Kf * m * i,
where:
- ΔT is the change in freezing point,
- Kf is the molal freezing point depression constant,
- m is the molality of the solution (moles of solute per kg of solvent), and
- i is the van't Hoff factor (the number of particles a solute dissociates into when it dissolves).
Given:
- Mass of calcium chloride (CaCl2): 14.6 g
- Mass of water: 246 g
- Kf = 1.86 °C/m
- Van't Hoff factor (i) for CaCl2: 2
First, calculate the molality (m):
Moles of CaCl2 = mass / molar mass of CaCl2
Moles of CaCl2 = 14.6 g / (40.08 g/mol + 2 * 35.45 g/mol) = 0.2 mol
Mass of water (in kg) = 246 g / 1000 = 0.246 kg
Molality (m) = moles of solute / mass of solvent in kg
Molality (m) = 0.2 mol / 0.246 kg = 0.813 mol/kg
Now, we can calculate the freezing point depression (ΔT):
ΔT = Kf * m * i
Substituting the given values:
ΔT = 1.86 °C/m * 0.813 mol/kg * 2 = 3.02 °C
To find the freezing point of the solution, subtract the calculated ΔT from the freezing point of the pure solvent (water):
Freezing point = 0 °C - 3.02 °C = -3.02 °C
Therefore, the correct answer is -3.02 °C, assuming we are dealing with CaCl2.
Now, let's address your initial calculations:
- Without using the van't Hoff factor: -0.99 °C
- With the van't Hoff factor (i = 3): -2.98 °C
The correct value is -3.02 °C, so both of your solutions are close. However, the use of the van't Hoff factor depends on whether the solute dissociates into multiple particles when it dissolves. In this case, CaCl2 dissociates into two particles (Ca2+ and 2Cl-), so the van't Hoff factor should be 2, not 3.
It is essential to be careful with the van't Hoff factor and make sure you apply it correctly based on the nature of the solute.