Question

What is the boiling point of a solution of 0.515 grams of acenaphthene (C12H10) in 15 grams chloroform (CHCl3) given pure chloroform has a b.p. of 61.7 C and a Kbp of +3.63 C.

Thanks!

Answer 1

mols C12H10 = grams/molar mass.
Solve for mols.

m = molality = mols/kg solvent.
Solve for m

delta T = Kb*m
Solve for delta T.

Add delta T to the normal boiling point to find the new boiling point.

Answer 2

Molar mass of C12H10 is 154.2078 g/mols

C12H10 = .003 mols

.2 molal

delta T = 123. 4 C

Answer 3

nope. I didn't work it through but that's way too large an amount.

If I divide 0.515/154 = about 0.00334 and that divided by 0.015 = about 0.223.
I don't know why you threw two perfectly numbers away (the 23 of the 0.223) but that's in the ball park. So your mistake must be in the next to last step.
delta T = Kb*m

Answer 4

Is this right?

Delta T= (3.63C)(.223)= .809 C

Answer 5

Yes. Now to find the new boiling point just add delta T to the normal boiling point given in the problem.

Answer 6

Thank you so much for your help! :)