Data:
Solute: 100.0 grams of ethylene glycol
Solvent 900.0 grams of
water,
Question: what is the freezing temperature of the solution formed?
Solution
Follow
these steps.
Calculate the molar mass of
ethylene glycol: 62.07 g/mol
ethylene glycol = C2H6O2
=> molar mass = 2 * 12.011 g/mol + 6 *1.008 g/mol + 2 * 15.999 g/mol = 62.068 g/mol = 62.07 g/mol
Calculate the number of moles of ethylene
glycol in the solution
number of moles = mass in grams / molar mass = 100.0 g / 62.07 g/mol = 1.611 mol
Calculate the molality of ethylene glycol
m = moles of solute / kg of solvent = 1.611 mol / 0.9000 kg = 1.79 m
Calculate the freezing point depression using the Kf from the Chemistry B
Information Sheet and the molality that you calculated
ΔTf = kf * m
kf = 1.86°C / m
ΔTf = 1.86 °C / m * 1.79 m = 3.329 °C
Calculate the
freezing point of the solution.
Tf = 0°C - 3.329°C = - 3.329 °C.
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?
Given that the typical mid-winter temperature we experience in Michigan is way below - 3.329 °C, this concentration of ethylene glycol is not high enough to use in car radation in the winter, because the water would freeze at - 3.329°C.
