anonymous
  • anonymous
Ethylene glycol, C2H6O2, is a non-volatile substance unable to form ions in water. If 38.6 grams of ethylene glycol is dissolved in 175 grams of water, what is the freezing point of the solution? (Kf = 1.86°C/m) -6.61°C 1.82°C 6.61°C -1.82°C
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Photon336
  • Photon336
\[iK _{b = }DeltaT\] i = van' hoff factor, how many moles of ions produced in your solution. in this case it's just 1 because no ions have been produced in our solution. m = molality, moles of solute per kilogram of solvent. you must first convert grams of ethylene glycol to moles of ethylene glycol. then grams of water to kilograms . Your Kb = 1.86 C/m If i remember correctly your Delta T = gives you the change in how much the freezing point was lowered.
Photon336
  • Photon336
Ethylene glycol is a non electrolyte so it won't produce ions in solution so the van hoff factor will just be 1
Photon336
  • Photon336
\[175g H _{2}O x (1kg/1000g) = 0.175 kg of H2O \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Photon336
  • Photon336
\[38.6g x (1mol/62.07g) = (38.6/62.07) = moles of ethylene glycol \]
Photon336
  • Photon336
\[Molality = \frac{ moles of solute }{ kilograms of solution } = \frac{ .62mol }{ 0.175kg } = 3.5 mol/kg\]
Photon336
  • Photon336
You know Kb you know that i = 1 and you know molaity = 3.5 m and you know Kf, all you have to do is plug these into the formula and you get the Delta T which is your temperature change.

Looking for something else?

Not the answer you are looking for? Search for more explanations.