Freezing Point Depression and Ethylene glycol

In summary, ethylene glycol has a freezing point of 1.86 degrees Celsius/m. To find the freezing point of radiator fluid that is 50% antifreeze by mass, divide the freezing point of water by the molality of ethylene glycol, which is 0.805 mol/100 g.
  • #1
sam.
16
0

Homework Statement



Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C_2H_6O_2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze.

What is the freezing point of radiator fluid that is 50% antifreeze by mass? K_f for water is 1.86 degrees Celsius/m.

Homework Equations



[tex]\Delta[/tex]T_f = K_f * m

m (molality) = # of moles of solute/mass of solution (kg)

The Attempt at a Solution



Okay, so I assumed 100 g of solution. And the molar mass of ethylene glycol is 62.08 g/mol. So you find the moles of ethylene glycol by multiplying 50 g by the molar mass which gives you 0.805 mol. Then you assume 100 g of water (which is 0.1 kg). And you solve for molality (m) and then you just sub it into the first equation? Is this correct?
 
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  • #2
sam. said:

Homework Statement



Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C_2H_6O_2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze.

What is the freezing point of radiator fluid that is 50% antifreeze by mass? K_f for water is 1.86 degrees Celsius/m.

Homework Equations



[tex]\Delta[/tex]T_f = K_f * m

m (molality) = # of moles of solute/mass of solution (kg)

The Attempt at a Solution



Okay, so I assumed 100 g of solution. And the molar mass of ethylene glycol is 62.08 g/mol. So you find the moles of ethylene glycol by multiplying 50 g by the molar mass which gives you 0.805 mol. Then you assume 100 g of water (which is 0.1 kg). And you solve for molality (m) and then you just sub it into the first equation? Is this correct?

Not quite. Remember this:

m (molality) = # of moles of solute/mass of solution (kg)

do not "...assume 100 g of water (which is 0.1 kg)."
 
  • #3
m (molality) = moles of solute/ mass of SOLVENT (KG) "NOTTTTTTTTTT SOLUTION"
 
  • #4
You not only necroposted in the thread that is over two years old, but you also missed the fact that chemisttree already addressed the problem.

--
 
  • #5
Boy! Those were the good old days. eh?
 

What is freezing point depression and how does it relate to ethylene glycol?

Freezing point depression is a phenomenon where the freezing point of a solution is lowered compared to the pure solvent. Ethylene glycol, a common antifreeze, is known to cause this effect due to its ability to disrupt the crystal formation of water molecules.

What is the role of ethylene glycol in antifreeze?

Ethylene glycol is the main active ingredient in antifreeze and is responsible for lowering the freezing point of the solution. It also has a high boiling point, making it useful for preventing engines from overheating.

How does ethylene glycol prevent freezing in cars?

When added to water, ethylene glycol forms a solution that has a lower freezing point than water alone. This allows the solution to remain liquid at lower temperatures, preventing the engine from freezing and potentially causing damage.

What are the potential dangers of ethylene glycol?

Ethylene glycol is highly toxic to humans and animals if ingested. It can also be harmful to the environment if not disposed of properly. It is important to handle and dispose of ethylene glycol-containing products with caution.

How does the concentration of ethylene glycol affect its freezing point depression?

The concentration of ethylene glycol in a solution directly affects the amount of freezing point depression that occurs. The higher the concentration, the greater the decrease in freezing point. However, too high of a concentration can also decrease the effectiveness of the antifreeze. A typical antifreeze mixture contains around 50% ethylene glycol.

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