ryan85

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On one of the Kaplan G Chem subject tests, on question 27, it says

23 g of ethanol (C2H5OH) is dissolved in 250 g of
water. What is the approximate freezing point of this
solution? (kf of water = 1.86°C_kg/mol)

they say that the answer is -3.7 C.

I thought that ethanol would disocciate into C2H5O- and H+, but I assume that it doesn't from this solution.

Any one have any ideas as to how I could know if some thing disocciates in water or not?
 

aranjuez

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Just try to keep in mind what the general goal of this question is: freezing point. So it's best to find the variables that we know we're going to be using. Kf is provided. All we need now is m. That's just (1/2 mol)/(1/4 kg) = 2m. Multiply that by the provided Kf. And that's how much the freezing will depress by from 0.

-3.7


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ryan85

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i know that, but if the compound breaks up into two solutes, then you would have to multiply by 2 because the equation is K*molal*i.

so ethanol doesn't break up? is what i am asking.
 
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tom_servo_dds

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ryan85 said:
i know that, but if the compound breaks up into two solutes, then you would have to multiply by 2 because the equation is K*molal*i.

so ethanol doesn't break up? is what i am asking.
Don't think so. -OH is not a good leaving group by itself. Can be with the help of SOCl2 though.
 

GeauxTigers

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Here's some advice that may help. Think of freezing point depression as having three (3) variables and not (2). Here's what I mean:

T(f) = (i)(m)(K)

The Kaplan book doesn't mention (i) as a variable, but it's important.

(i) = number of dissociated compounds, for example NaCl --> Na+ and Cl-, therefore your (i) is 2. Also, organic compounds won't dissociate, which is what you have here with C2H5OH. They remain as a single compound and their (i) value are always 1.

I assume you understand m is molality and K is the constant.

Thus us you have T(f) = (1)(0.5/0.25)(1.86) = 3.7, and the answer is negative since we're dealing with freezing point depression.

Hope this helps :)
 
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