what is the boiling point of a solution containing 233 grams

what is the boiling point of a solution containing 2.33 grams of C8H10N4O2 , dissolved in 15 grams of benzene?

the boiling point of pure benzene is 80.1 C . and the boiling point elevation constant ,Kbp, is 2.53C/m

Solution

ANSWER

the 2.02 is how much the bp increased. you need to add that to 80.1 to get the new bp.. like this...

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dT = Kb x m x i

dT = change in BP = BP solution - BP pure
Kb = ebullioscopic constant = 2.53 C/m
m = molality = moles solute / kg solvent
i = van\'t hoff factor = # ions the solute dissociates into.

we need to find m and i first....

moles caffeine = 2.33 g x (1 mole / 194.2 g) = 1.200 moles
molality = 1.200 moles / 0.015 kg = 0.800 m

i = 1.. caffeine is a non polar molecule and benzene is a non polar solvent. so caffeine will not dissociate in benzene and i = 1 for non dissociating molecules...

dT = Kb x m x i

BP solution - BP pure = Kb x m x i

BP solution = Kb x m x i + BP pure

BP solution = 2.53 C/m x 0.800 m x 1 + 80.1 C = 2.02 C + 80.1 C = 82.1 C