What is the boiling point of a solution containing 2.33 g of caffeine, C8H10N4O2, dissolved in 15.0 g of benzene? The boiling point of pure benzene is 80.1 °C and the boiling point elevation constant, Kbp, is 2.53 °C/m.
1. 2.02 C
2. 78.1 C
3. 80.2 C
4. 82.1 C
5. 83.3 C
The answer is 4 (82.1 C), but I repeatedly got the 2.02 C. Any ideas?
the 2.02 is how much the bp increased. you need to add that to 80.1 to get the new bp.. like this...
*******************************
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
the 2.02 is how much the bp increased. you need to add that to 80.1 to get the new bp.. like this...
*******************************
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
