What is the boiling point of an aqueous solution that freezes at -6.7 degress C. Please help.
I feel ed because it seems like its something real easy but Im still not getting it.
I did 100 + 6.7 and got 100.67 and are you sayin to divide 1.86 by .52 which is 3.576. the book says the answer is 101.8 im lost
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Even I know that if you add 100 + 6.7, you don't get 100.67. Rethink.
You can also use www.wolframalpha.com to check your math answers.
In addition you don't give enough information about the 1.86 and 0.52 to help. Apparently some tutor gave you the procedure for solving the problem and you didn't understand. However, you have posted just piece-meal info and that isn't enough to help. I suggest you repost the ENTIRE problem along with the tutor's reply, then tell us what you don't understand. We can go from there.
what is the boiling point of an aqueous solution that freezes at
-67 degrees C?
I was given this answer
What is 100+6.7 (1.86/.52) ?
so 100+6.7 = 106.7
1.86/.52 = 3.58
is the .52 the Kb
I got the answer in the book I just don't understand how they got to it.
Yes, undoubtedly the 0.52 is Kb, and I assume you intended to write 6.7 and not 67, but I think the Kf/Kb should be Kb/Kf. Let's do it another way.
delta T = Kf*m
6.7 = 1.86m and m = 6.7/1.86 = 3.6m
Then delta T = Kb*m
100 + Kb*m = 100 + (0.52*3.6) = 100 + 1.87 = or about 102 C. You can clean up the number of significant figures.
I thank you very much, I believe I really need to see it step by step to understand it. Another thing what did the 100+6.7 have to do with the equation. Also changing the -6.7C to 6.7C is this something that you would do to any boiling or freezing problem
Don't worry, I'm here to help you understand how to calculate the boiling point of an aqueous solution. Let's break it down step by step:
1. Start with the freezing point depression equation, which is given by:
∆Tf = Kf * m * i
where ∆Tf is the freezing point depression, Kf is the cryoscopic constant, m is the molality of the solute, and i is the van't Hoff factor.
2. You mentioned that the aqueous solution freezes at -6.7 degrees Celsius (∆Tf = -6.7°C).
3. To find the molality (m) of the solute, we can use the formula:
m = moles of solute / mass of solvent (in kg)
However, since the problem does not provide the exact amount of solute, we will need to use another approach.
4. The van't Hoff factor (i) depends on the dissociation of the solute in water. In this case, we will assume i = 1, as it is a typical assumption for non-electrolyte solutes.
5. Now, we need the cryoscopic constant (Kf) which is a property specific to the solvent. The cryoscopic constant for water is approximately 1.86 °C·kg/mol.
6. Let's substitute these values into the freezing point depression equation:
-6.7°C = 1.86°C·kg/mol * m * 1
7. Rearrange the equation to solve for m:
m = -6.7°C / (1.86°C·kg/mol)
8. Calculate m:
m ≈ 3.59 mol/kg (approximately)
9. Finally, to find the boiling point elevation, we can use the equation:
∆Tb = Kb * m * i
Where Kb is the boiling point elevation constant. For water, Kb is also approximately 0.52 °C·kg/mol.
10. Substitute the values into the boiling point elevation equation:
∆Tb = 0.52°C·kg/mol * 3.59 mol/kg * 1
11. Calculate ∆Tb:
∆Tb ≈ 1.8758 °C
12. Add the boiling point elevation (∆Tb) to the normal boiling point of water (100°C):
Boiling point = 100°C + 1.8758°C
Boiling point ≈ 101.8758°C
So, the boiling point of the aqueous solution that freezes at -6.7 degrees Celsius is approximately 101.8758 degrees Celsius.