Asked by Layla

What is the boiling point of a solution of benzene contain 0.2 moles of a non volatile solute in 125 grams of benzene given that pure benzene has a b.p. of 80.1 Celcius and a Kbp of +2.53 C/m.

Thanks!
Answered by DrBob222
m = molality = mols/kg solvent. You know mols and kg solvent, solve for m
Then substitute m into the below equation.

delta T = Kb*m
Solve for delta T and add to the normal boiling point to find the new boiling point.
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