Asked by Aimee

If you are given .060m. sugar (glucose-C6H12O6), .030m. AlCl3 and .025m. NaOH.

How do you figur out lowest boiling point and lowest vapor pressure of the given solutions?

Answers

Answered by DrBob222
Lowest boiling point.
delta T = i*Kb*m
Since Kb is a constant, you only need to do i*molality.
i = # dissolved particles. m is molality.
delta T will be the lowest for the smallest product of i*m
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