Question

What is the normal boiling point in ∘ C ^\circ {\rm C} of a solution prepared by dissolving 1.50 g {\rm g} of aspirin (acetylsalicylic acid, C 9 H 8 O 4 {\rm C}_{9}{\rm H}_{8}{\rm O}_{4}) in 75.00 g {\rm g} of chloroform (CHCl 3 ) ({\rm CHCl}_{3})? The normal boiling point of chloroform is 61.7 ∘ C ^\circ {\rm C}, and K b {\rm K}_{{\rm b}} for chloroform is 3.63 ( ∘ C⋅kg)/mol \rm (^\circ C \cdot kg)/mol.

Answer 1

If I can sort through this maze of hieroglyphics
mols aspirin = grams/molar mass
Substitute and solve for mols.

m = mol/kg solvent
Substitute and solve for m

delta T = Kf*m
Substitute and solve for delta T.

Then add delta T to normal boiling point to find the new boiling point.