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? hurry

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.

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