Question

1. If 100.0 grams of ethylene glycol are dissolved in 900.0 grams of water, what is the freezing temperature of the solution formed? Follow these steps: Show all of your calculations.


Calculate the molar mass of ethylene glycol:
62.07 g/mol

Calculate the number of moles of ethylene glycol in the solution:


Calculate the molality of ethylene glycol:


Calculate the freezing point depression using the Kf from the Chemistry B Information Sheet and the molality that you calculated:


Calculate the freezing point of the solution.


2. Think about the result. Think about the typical mid-winter temperature we experience in Michigan. Is this concentration of ethylene glycol high enough to use in a car radiator in the winter? Why or why not?





please hurry

Answer 1

A solution will solidfy (freeze) at a lower temperature than the pure solvent. This is the colligative property called freezing point depression.

The more solute dissolved, the greater the effect. An equation has been developed for this behavior. It is:

ΔT = i Kf m

The temperature change from the pure to the solution is equal to two constants times the molality of the solution. The constant Kf is actually derived from several other constants and its derivation is covered in textbooks of introductory thermodynamics. Its technical name is the cryoscopic constant. The Greek prefix cryo- means "cold" or "freezing." In a more generic way, it is called the "molal freezing point depression constant."