posted by Jaques on .
What would be the freezing and boiling points of a solution prepared by
dissolving 1.00g of benzoic acid (C6H5CO2H) in 10.0g of benzene?
mols acid = grams/molar mass
m acid = mols/kg solvent
delta T = Kf*m
Solve for T and add to normal freezing point of benzene to find the new f.p.
Do the same for b.p.
mols acid is same
m acid is same
delta T = Kb*m
Add delta T to normal b.p. to find new b.p.