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# freezing point + density problem for antifreeze

Submitted by APchem on Mon, 11/17/2008 - 23:46

What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 10.9 L of water to produce an antifreeze solution with a freezing point of -38.5°C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)

Does anyone know how to solve this problem?

the freezing point depression is given by the equation:
?T=m X Kf

where m is the molality of the solution
and Kf is the freezing point depression constant (for water, this constant is 1.86*C/m)

Figure out what you know:

-water freezes at 0C and you need the solution to freeze at -38.5C
-10.9L of water ~ 10.9kg of water
-molality is defined as (moles of solute/ kg of solvent)= (moles of solute/10.9kg H2O)

-plug all your known values into the above equation to figure out your unknown

0C-38.5C=(mole of solute/10.9 kg H2O) X 1.86*C/m

It thus appears that your unknown here is the moles of solute. Solve the equation for your moles of solute:

it should come out to be ~ 1.90 moles of ethylene glycol

now calculate the mass of ethylene glycol: molar mass of ethylene glycol is ~62.0g/mol

(62.0g/mol) X (1.90 moles) =117.8g of ethylene glycol

Now find the volume by relating the mass you just found to the density:

density = mass/volume

1.11g/cm^3=117.8g/V

V=106 mL