Question
I'm really confused on this question so if someone could show the steps and how to get the answer, that would be great. I understand that you would use delta T=Kbm, but don't know how to use the equation to find the answer. Thanks!!
Calculate the boiling point of a solution containing 7.58 grams of C10H8 in 120.0 grams of benzene. The boiling point of benzene is 80.1 degrees Celcius and the Kb = 2.52 degrees celcius/molal.
Calculate the boiling point of a solution containing 7.58 grams of C10H8 in 120.0 grams of benzene. The boiling point of benzene is 80.1 degrees Celcius and the Kb = 2.52 degrees celcius/molal.
Answers
DrBob222
Note the correct spelling of celsius.
mols C10H8 = grams/molar mass = 7.58/molar mass C10H8 = ?
Then molality = m = mols/kg solvent
m = mols from above/0.120 = ?
Then delta T = Kb*m. You know Kb and m, solve for delta T, then add that to the normal boiling point of benzene in the problem to find the new boiling point.
mols C10H8 = grams/molar mass = 7.58/molar mass C10H8 = ?
Then molality = m = mols/kg solvent
m = mols from above/0.120 = ?
Then delta T = Kb*m. You know Kb and m, solve for delta T, then add that to the normal boiling point of benzene in the problem to find the new boiling point.
Serena
I know what I did wrong. Some reason I divided the moles of C10H8 by .0120 instead of .120, so my answer was 11 times the answer it should have been. Thank you very much!