what is the freezing point of an aqueous solution that boils at 102.1 C?
Kf=1.86
Kb=0.512
102.1-100.0=2.1
2.1/0.512=4.10
4.10*1.86=7.62 C
7.62 C
Is this correct?
Well, I hate to burst your bubble, but there seems to be a mix-up here. The information you provided is about boiling point elevation, not freezing point depression. Boiling point elevation and freezing point depression are related phenomena, but they involve different calculations. So, unfortunately, the answer you calculated is not correct for the freezing point of the aqueous solution.
If you have the molality of the solution and the known freezing point depression constant (Kf), you can use the formula ΔT = Kf * m to calculate the freezing point depression. However, without the specific molality or any other information about the solute, it's not possible to accurately determine the freezing point of the solution.
But hey, it's not the end of the world! We all have our off-days with calculations. Just remember, when it comes to freezing points, it's chill to ask for help!
No, your calculation is not correct. To find the freezing point of an aqueous solution, you need to use the equation:
ΔTf = Kf * m
Where:
ΔTf = change in freezing point
Kf = freezing point depression constant (given as 1.86)
m = molality of the solution (moles of solute per kilograms of solvent)
Given that the solution boils at 102.1 °C, the boiling point elevation can be calculated using the equation:
ΔTb = Kb * m
Where:
ΔTb = change in boiling point
Kb = boiling point elevation constant (given as 0.512)
m = molality of the solution (moles of solute per kilograms of solvent)
Since you already have the boiling point change (2.1 °C) and the boiling point elevation constant (0.512), you can calculate the molality of the solution (m):
m = ΔTb / Kb
= 2.1 / 0.512
≈ 4.10 mol/kg
Now, using this molality value, you can calculate the freezing point depression (ΔTf):
ΔTf = Kf * m
= 1.86 * 4.10
≈ 7.626 °C
Therefore, the freezing point of the aqueous solution is approximately 7.626 °C.
Yes, your calculation is correct. To determine the freezing point of an aqueous solution, you need to use the equation:
△T = Kf * m
where △T is the change in freezing point, Kf is the molal freezing point depression constant, and m is the molality of the solution.
In this case, you are given the boiling point of the solution, which is 102.1°C, and the boiling point elevation constant (Kb) is not relevant to finding the freezing point. Therefore, we can ignore it in this calculation.
To find △T, subtract the normal freezing point (0°C) from the boiling point of the solution:
△T = 102.1 - 0 = 102.1°C
Next, divide the change in temperature (△T) by the molal freezing point depression constant (Kf):
m = △T / Kf = 102.1 / 1.86 = 54.89 mol/kg
Finally, multiply the molality (m) by the molal freezing point depression constant (Kf) to find the freezing point depression:
△T = Kf * m = 1.86 * 54.89 = 102.09°C
Therefore, the freezing point of the aqueous solution is approximately 7.62°C (100°C - 7.62°C ≈ 92.38°C).