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.
-.52 is the Kf. 1.86 is the Kb You need to understand the signs.
first, find the molality.
-6.7=-.52*m
m=6.7/.52
Now the bp
New bp=100+kb*m=100*1.86*(6.7/.52)
Thanks I under stank Kf but in my book it says that Kf is 0.512
To determine the boiling point of an aqueous solution, you can use the equation:
ΔTb = Kb * m
where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant, and m is the molality of the solute in the solution.
In the given question, you are given the freezing point of the solution (-67 degrees C). The boiling point elevation will be the difference between the boiling point of the solution and the boiling point of the pure solvent (in this case, water).
To find the boiling point elevation, you need to convert the freezing point from degrees Celsius to Kelvin. Remember that the Kelvin scale is the same as the Celsius scale, but shifted by 273.15. So, -67 degrees Celsius is 206.15 Kelvin.
Now, you need to find the molality of the solute in the solution. Unfortunately, this information is not given in the question, so we cannot proceed with answering it accurately. However, if you have the molality (moles of solute per kilogram of solvent) value, you can calculate the boiling point elevation using the equation mentioned above.
Regarding the answer you were given (100+6.7 (1.86/.52)), it seems unrelated to the question about the boiling point of an aqueous solution. The calculation appears to involve addition, multiplication, and division, but without any context or explanation, it is difficult to determine its purpose or relevance.