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what mass of ethylene glycol C2H6O2 the main component of anitfreeze, must be added to 10.0L water to produce a solution for use in a car's radiator that freezes at -23.3°C. Assume the density for water is exactly 1 g/m.L
delta T = Kf*m
You know delta T and Kf (1.86 for water), calculate molality.
molality = mols/kg.
You know molality and kg, solve for moles.
mols = grams/molar mass
You know mols and molar mass, solve for grams. Post your work if you get stuck.
I feel like I have made a mistake because the mass seems too small. I don't know where I went wrong :S
Molality = ÄTf / (Kf * i)
= 23.3°C / (1.86°C/m * 3)
= 4.18m (mol C2H6O2 / kg water)
The amount to dissolve is:
n_C2H6O2 = molality * m_water
= 4.18 mol/kg * 0.1kg
= 0.0418 mol
m_C2H6O2 = n_C2H6O2 * M_C2H6O2
= 0.0418mol * 62g/mol
Let's see. Why are you using i. i is the van't Hoff factor which is 1 for ethylene glycol.
delta T = Kf*m
23.3 = 1.86*m
m=23.3/1.86 = 12.527 molal.
m = mols/kg solvent
12.527 = mols/10 kg = 125.27 moles.
grams = mols x molar mass = 125.27 x 62.068 = 7,775.18 g ethylene glycol. You need to redo the problem and straighten the significant figures out. I usually carry one more than I'm allowed, then round at the end. I think 23.3 is the most s.f. and 1.86 both have 3 places, so the answer would round to 7.78 x 10^3 grams . Check my work.
I did it again this time I got 7774.51g which is much better. Thanks alot!
I don't understand why is it 10kg solvent?