What is the expected freezing point of a 3.50 m aqueous solution of BaCl2 (Kfp=1.86C/m)? What is the boiling point (Kbp=0.512C/m? & which is affected to the greater extent, the boiling point or the freezing point?
Ah, the wonderful world of colligative properties! Let's break it down, shall we?
For the freezing point depression, we can use the formula:
ΔTf = Kfp * m
where:
- ΔTf is the freezing point depression
- Kfp is the freezing point constant (1.86C/m for water)
- m is the molality of the solution
In this case, we have a 3.50 m aqueous solution of BaCl2. Plugging those values in, we get:
ΔTf = 1.86C/m * 3.50 m
Simplifying that, we find that the freezing point depression is 6.51°C.
As for the boiling point elevation, we use a similar formula:
ΔTb = Kbp * m
where:
- ΔTb is the boiling point elevation
- Kbp is the boiling point constant (0.512C/m for water)
- m is the molality of the solution
Since we have the same m value of 3.50, plugging in the values:
ΔTb = 0.512C/m * 3.50 m
Simplifying, we find that the boiling point elevation is 1.79°C.
Now, which is affected to a greater extent? Based on the magnitude of the values, the freezing point is affected to a greater extent. The freezing point depression is about 3.6 times greater than the boiling point elevation.
Hope that answers your question!
To calculate the expected freezing and boiling points of a solution, we can use the formulas:
Freezing Point Depression (ΔTf) = Kfp * m
Boiling Point Elevation (ΔTb) = Kbp * m
where Kfp is the freezing point constant and Kbp is the boiling point constant. "m" represents the molality of the solution, which is the number of moles of solute per kilogram of solvent.
Given:
molarity (M) of BaCl2 solution = 3.50 m
Kfp (freezing point constant) = 1.86 °C/m
Kbp (boiling point constant) = 0.512 °C/m
To find the molality:
Molality (m) = moles of solute / mass of solvent (in kg)
Since we are given the molarity, we need to convert it into moles of solute using the formula:
moles of solute = Molarity * volume (in liters)
1. Calculate moles of BaCl2:
moles of BaCl2 = 3.50 m * volume (L)
2. Convert moles of BaCl2 to mass of BaCl2:
mass of BaCl2 = moles of BaCl2 * molar mass of BaCl2
3. Calculate the mass of water (solvent):
mass of water = volume (L) * density of water (g/mL) * 1000 (to convert mL to grams)
4. Calculate molality (m):
molality (m) = moles of solute / mass of solvent (in kg)
5. Calculate the freezing point depression (ΔTf):
ΔTf = Kfp * m
6. Calculate the boiling point elevation (ΔTb):
ΔTb = Kbp * m
7. Determine which is affected to the greater extent:
Compare the values of ΔTf and ΔTb and see which one is larger.
By following these steps, you can find the expected freezing and boiling points of the BaCl2 solution and identify which is affected to a greater extent, the boiling point or the freezing point.
To find the expected freezing point and boiling point of a solution, you can use the equation:
ΔT = i * Kf * m
Where:
- ΔT is the change in temperature (freezing or boiling)
- i is the van't Hoff factor (the number of particles into which a compound dissociates in solution)
- Kf is the molal freezing point constant (Kfp) or molal boiling point constant (Kbp)
- m is the molality of the solution
First, let's calculate the freezing point depression:
ΔTf = i * Kfp * m
The van't Hoff factor (i) is determined by the number of ions produced when the compound dissociates in water. For BaCl2, it dissociates into three ions: Ba2+ and two Cl− ions. Therefore, i = 3.
Next, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is BaCl2, which has a molar mass of 208.23 g/mol.
m = (moles of solute) / (mass of solvent in kg)
Given that the solution is 3.50 m, it means there are 3.50 moles of BaCl2 dissolved in 1 kg of water (solvent). Therefore, m = 3.50 mol/kg.
Now we can substitute the known values into the equation to calculate ΔTf:
ΔTf = 3 * 1.86 ℃/m * 3.50 mol/kg
By performing the calculation, we find that the freezing point depression (ΔTf) is -19.71 ℃.
To find the boiling point elevation, we apply the same logic:
ΔTb = i * Kbp * m
Since BaCl2 dissociates into three ions, the van't Hoff factor (i) is also 3.
The molality (m) is the same as before: 3.50 mol/kg.
Substituting all the known values into the equation, we find:
ΔTb = 3 * 0.512 ℃/m * 3.50 mol/kg
By performing the calculation, we find that the boiling point elevation (ΔTb) is 5.38 ℃.
Now, to answer your final question, we compare the freezing point depression (ΔTf) of -19.71 ℃ to the boiling point elevation (ΔTb) of 5.38 ℃. We can see that the freezing point is affected to a greater extent than the boiling point.
Just look at the constants, freezing point depression is bigger.
FP=orig - 1.86*3.50*3
the three comes from the three ions per molecule.