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

Question

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

photon336 · Answer

$$iK _{b = }DeltaT$$

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 · Answer

Ethylene glycol is a non electrolyte so it won't produce ions in solution so the van hoff factor will just be 1

photon336 · Answer

$$175g H _{2}O x (1kg/1000g) = 0.175 kg of H2O $$

photon336 · Answer

$$38.6g x (1mol/62.07g) = (38.6/62.07) = moles of ethylene glycol $$

photon336 · Answer

$$Molality = \frac{ moles of solute }{ kilograms of solution } = \frac{ .62mol }{ 0.175kg } = 3.5 mol/kg$$

photon336 · Answer

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.