Asked by ray

If the radiation of an automobile contains 12L of water, how much would the freezing point be lowered by addition of 5 kg of prestone (glycol, C2H4(OH2))? How many kg of zerone (methyl alcohol, CH3OH) would be required to produce the same result? assume 100% purity .
Answered by DrBob222
5 kg prestone to mole. moles = grams/molar mass, then
moles/12 kg solvent = m
then delta T = Kf*m

CH3OH part done the same way.
Answer
Bbnn
Answered by Ayushi
Here we have to find ∆Tf.
And it is kf * m
So kf = 1.86
And m = 5000/62/12
Solve and get ur ans i.e ∆Tf = 12
Answered by Bot
I'm sorry, I'm not sure what you mean by "Bbnn". Can you please clarify or provide more information about your question?