Problem: If each substance listed here costs the same amount per kilogram, which would be most cost-effective as a way to lower the freezing point of water? (Assume complete dissociation for all ionic compounds.)a. HOCH2CH2OHb. NaClc. KCld. MgCl2e. SrCl2

FREE Expert Solution

We’re going to use the equation for Freezing Point Depression.

Tf = iKfm

∆Tf = change in freezing point = Tf pure solvent –Tf solution
Kf = freezing point depression constant
i = van' t Hoff factor of the solute = no. of ions
m = molality


The higher the i, the stronger it is to lower the freezing point. 

The highest number of i per gram should be the most cost-effective. 


a. HOCH2CH2OH

Molar mass = 62.08 g/mol

ig = 11 mol HOCH2CH2OH ×1  mol HOCH2CH2OH 62.08 g = 0.016


b. NaCl

Molar mass =58.44

ig = 21 mol NaCl ×1  mol NaCl 58.44 g = 0.034


c. KCl

Molar mass =74.55

ig = 21 mol KCl ×1  mol KCl 74.55 g = 0.026


d. MgCl

Molar mass =95.21

ig = 31 mol MgCl2 ×1  mol MgCl2 95.21 g = 0.032

e. SrCl2

Molar mass =158.52

ig = 31 mol SrCl2 ×1  mol SrCl2 158.52 g = 0.019


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Problem Details

If each substance listed here costs the same amount per kilogram, which would be most cost-effective as a way to lower the freezing point of water? (Assume complete dissociation for all ionic compounds.)
a. HOCH2CH2OH
b. NaCl
c. KCl
d. MgCl2
e. SrCl2

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