Question

When a solute B dissolve in solvent A , the molality of the resulting solution is 1, then what is the relation between ∆Tb and ∆ kb ?​

Answer 1

Answer:

Calculating Boiling Point Elevation and Freezing Point Depression

CHEM 30A

Boiling Point Elevation:

Tb = i Kb m

BP = BPnormal + Tb

Freezing Point Depression:

Tf = i Kf m

FP = FPnormal – Tf

Example: What are the BP and FP of a 7.187m NaCl solution?

For water: Kb = 0.512 ºC · kg H2O / mol particles

Kf = 1.86 ºC · kg H2O / mol particles

BPsoln = BPnormal + Tb = 100.00ºC + 7.35ºC = 107.35ºC

FPsoln = FPnormal - Tf = 0.00ºC – 26.7ºC = –26.7ºC

1.86 ºC · kg H2O

mol particles

2 mol particles

1 mol NaCl

7.187 mol NaCl

1 kg H2O

constant (different for each solvent)

molality (m) =

Change in BP

(add this number to the

normal BP)

Change in FP

(subtract this number

from the normal FP)

mol solute

kg solvent

mol particles

mol of solute van’t Hoff factor =

A. Romero 2008

Tb = i Kb m =

0.512 ºC · kg H2O

mol particles

= 7.35 ºC

2 mol particles

1 mol NaCl

7.187 mol NaCl

1 kg H2O

Tf = i Kf m = = 26.7 ºC

100ºC

to

0ºC

107.35ºC

to

–26.7ºC

the range of temperatures over which the solution

is a liquid is colligatively extended on both ends