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

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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

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\[iK _{b = }DeltaT\] 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.
Ethylene glycol is a non electrolyte so it won't produce ions in solution so the van hoff factor will just be 1
\[175g H _{2}O x (1kg/1000g) = 0.175 kg of H2O \]

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\[38.6g x (1mol/62.07g) = (38.6/62.07) = moles of ethylene glycol \]
\[Molality = \frac{ moles of solute }{ kilograms of solution } = \frac{ .62mol }{ 0.175kg } = 3.5 mol/kg\]
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.

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