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jruiz3
Junior Member


Joined: 18 Sep 2006
Posts: 8

PostPosted: Mon Jan 22, 2007 1:47 pm    Post subject: Someone please help - thank you so much Reply with quote

You add 1.00kg of the antifreeze ethylene glycol (C2H6O2) to your car radiator, which contains 4450g of water. The constants are: Kb=0.512*C and Kf=1.86*C. What are the boiling and freezing points (*C) of the solution?
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gypsum
Probably not a bot. Junior Member.


Joined: 23 Nov 2006
Posts: 3

PostPosted: Fri Jan 26, 2007 3:36 pm    Post subject: hey Reply with quote

actually, this is a colligative properties...boiling point elevation and freezing point depression....you just calculate first the number of moles of ethylene glycol(62.01 g/mole) in a given amount of water, then calculate the number of moles of ethylene glycol in a 1 L of water or the molality, m...

#mole EG = (1000 g * 1mole/62.01 g) = 16.126 moles

m = 16.126 mole EG/4.45 Kg water = 3.624 m

delta Tf = (1.86 C/m)(3.624 m) = 6.740 degrees celcius

however, the water freezes at 0 degrees celcius, therefore, the freezing point of EG is - 6.74 degress celcius

delta Tb = (0.512)(3.624) = 1.856 C

but, water will boil at 100 c, then Tb for EG is 101.856 degress celcius.

just check it out...
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