What is the boiling point of a 0.75-molal solution of a non-volatile non-electrolyte solute in water? Kb for water = 0.52 degrees C/ molal.
Answer

99.61 °C

100.0 °C

100.39 °C

101.44 °C

Question

What is the boiling point of a 0.75-molal solution of a non-volatile non-electrolyte solute in water? Kb for water = 0.52 degrees C/ molal.
Answer
		
99.61 °C
		
100.0 °C
		
100.39 °C
		
101.44 °C

anonymous · Answer

@aaronq

aaronq · Answer

Find the temperature change with:  $\Delta T=i*m*K_b$ 

m=molality
$i$=1 (non-electrolyte)

anonymous · Answer

ok...so...
T = i x .75 x .52
but what is i?

aaronq · Answer

$i$ is the van't hoff constant, it's equal to the number of particles the solute dissociates into.  In this case, it is one because you are told that the solute is a non-electrolyte.

anonymous · Answer

oh...
so...
T = 1 x .75 x .52 ?

anonymous · Answer

.39?

aaronq · Answer

yep, note that you're finding the change in temperature $\Delta T$, not the temperature.

anonymous · Answer

alright, but .39 isn't an option...I'm guessing you add 100 to get 100.39, which is an option?

aaronq · Answer

yep, because 100 is the BP, and you found the change

anonymous · Answer

ahhhh...makes sense. Thank you! :D

aaronq · Answer

no problem!