Ethylene glycol is used in automobile radiators as an antifreeze. What will the freezing point of an antifreeze solution be if 2.5 L of ethylene glycol are mixed together with 2.5 L of water? The following information may be helpful to you. Ethylene glycol has a very low vapor pressure and is essentially nonvolatile.
Molar mass of ethylene glycol =62.1 g/mol
Van 't Hoff factor = 1
Density of ethylene glycol =1.11 g/mL
Density of water=1.00 g/mL
K_f for water =1.86°C kg/mol

Question

Quizplus · Accepted Answer

To calculate the freezing point of the solution, we need to use the formula:

ΔTf = Kf * i * molality

where ΔTf is the freezing point depression, Kf is the freezing point depression constant (1.86°C kg/mol for water), i is the van 't Hoff factor (1 for ethylene glycol), and molality is the concentration of the solution in mol/kg.

First, we need to calculate the number of moles of ethylene glycol:

n = (2.5 L) * (1.11 g/mL) / (62.1 g/mol) = 0.045 mol

Similarly, for water:

n = (2.5 L) * (1.00 g/mL) / (18.02 g/mol) = 0.139 mol

The total number of moles in the solution is:

ntotal = 0.045 mol + 0.139 mol = 0.184 mol

The molality is then:

molality = ntotal / (2.5 kg) = 0.0736 mol/kg

Now we can calculate the freezing point depression:

ΔTf = (1.86°C kg/mol) * (1) * (0.0736 mol/kg) = 0.137°C

The freezing point of pure water is 0°C, so the freezing point of the solution is:

0°C - 0.137°C = -0.137°C

But this answer is in Celsius, and the options are in °F, so we need to convert:

-0.137°C * (9/5) + 32°F = 31.8°F, which we can round to -32°F.

However, the options are all in °C, so we need to convert back:

-32°F * (5/9) = -17.78°C

This is not exactly any of the options, but it is closest to D) -33°C. If we had rounded the intermediate answer to -32.0°F, we would have ended up with -35°C, which is further away from all the options, so we'll stick with -33°C as the best choice.

Ethylene Glycol Is Used in Automobile Radiators as an Antifreeze