What is the boiling point of a solution of 24.6 g of camphor in 98.5 g of benzene?
The normal boiling point of benzene is 80.1 °C.
The boiling point elevation constant for benzene is #"2.53 °C·kg·mol"^"-1"#
The normal boiling point of benzene is 80.1 °C.
The boiling point elevation constant for benzene is
2 Answers
The boiling point of your solution will be
So, you're dealing with a boiling point elevation problem in which an added solute, in your case camphor, will increase the boiling point of a pure solvent, benzene.
The equation for boiling point elevation is
Since camphor is a covalent compound, it will not dissociate when placed in the solvent, which implies a van't Hoff factor equal to 1.
Since molality is defined as moles of solute per kilogram of solvent, you need to determine how many moles of camphor you have in 24.6 g
Now plug your data into the equation for boiling point elevation
Rounded to three sig figs, the answer will be
Okay, I will try to answer this but I may be off due to rounding off numbers. I am also a little concerned about the
Molality = moles of solute/kilograms of solvent.
Next, I used the formula for the change in the temperature of the boiling point , which is
This would equal
If you add the change to the original boiling point of 80.1 °C, you get the new boiling point of 84.3 °C.
I am not sure that all of the math is correct, but you have a couple formulas that might help you.