I believe you are over thinking the problem.
delta T for ethylene glycol is
dT = Kf*m. You can calculate m, then dT.
Substitute that dT and Kf and solve for m of CH3OH. Then m = mols/kg; you know kg and m, solve for mols. I think that is 5 mols CH3OH.
The reason for the problem is to recognize that KNO3 is not a non-electrolyte; it dissolves into two particles and you must use the version of the formula that is
dT = i*Kf*m where i is the van't Hoff factor which is 1(for non-electrolytes like ethylene glycol and methyl alcohol (CH3OH) but 2 for KNO3. It would be 3 for Na2SO4 and 2 for NaCl, etc.
So set up dT and i and Kf and solve for m and mols. I think that answer will be 2.5 mols needed although I didn't put a pen and paper to it and it's possible I goofed in my memory banks.
How many moles of each of the following are needed to give the same freezing point lowering as 5.0mol of the nonelectrolyte ethylene glycol in 1 kg of water?CH3−OH, a non electrolyte, KNO3, a strong electrolyte.
Im confused on what they are asking, and how in the world would you set this up
1 answer
1 answer