Question

List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m potassium bromide (KBr), 0.030 m phenol (C6H5OH). Rank solutions from highest to lowest freezing point. To rank items as equivalent, overlap them.

Answer 1

Answer:

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol

Explanation:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f$ = Depression in freezing point

i = van'T Hoff fcator

$K_f$ = Molal depression constant of solvent

m = molality of the solution

Higher the value of depression in freezing point at  lower will be freezing temperature the solution.

1. 0.040 m glycerin

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 ( organic molecule)

m = 0.040 m

$\Delta T_{f,1}=1\times\times 1.86^oC/m\times 0.040 m$

$\Delta T_{f,1}=0.0744^oC$

2. 0.020 m potassium bromide

Molal depression constant of water = $K_f=1.86^oC/m$

i = 2 (ionic)

m = 0.020 m

$\Delta T_{f,2}=2\times\times 1.86^oC/m\times 0.020 m$

$\Delta T_{f,2}=0.0744^oC$

3. 0.030 m phenol

Molal depression constant of water = $K_f=1.86^oC/m$

i = 1 (organic)

m = 0.030 m

$\Delta T_{f,3}=1\times\times 1.86^oC/m\times 0.030 m$

$\Delta T_{f,3}=0.0558^oC$

$0.0744^oC=0.0744^oC > 0.0558^oC$

$\Delta T_{f,1}=\Delta T_{f,2}>\Delta T_{f,3}$

Solutions from highest to lowest freezing point:

0.040 m glycerin = 0.020 m potassium bromide > 0.030 m phenol