If you have a 29.6 mass percent KCl in benzene,what is the new boiling point?
29.6% (w/w) is 29.6 g KCl/100 g solution.
That is 29.6g KCl in (29.6g KCl + 70.4 g H2O.
mols KCl = grams/molar mass = ?
kg H2O = 0.0704
molality of the solution is mols/kg solvent
Then delta T = i*Kb*m
i = 2 for KCl. Solve for delta T and add to 100 C to find new boiling point.