# 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

∆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

b. NaCl

Molar mass =58.44

c. KCl

Molar mass =74.55

d. MgCl

Molar mass =95.21

e. SrCl2

Molar mass =158.52

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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