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Chemistry 13 Online
OpenStudy (anonymous):

Help calculating this? The boiling point of an aqueous solution is 102.17 °C. What is the freezing point?

OpenStudy (anonymous):

Also, how do I even begin this problem without knowing the element(s)?

OpenStudy (aaronq):

you don't need to know the element, the freezing point depression formula works on the basis of colligative properties. use this formula: \[DeltaT=iK _{f}m\] find m, molality, using Kb then plug into the same formula using the constant Kf

OpenStudy (anonymous):

Ka * Kb = Kw, correct?

OpenStudy (aaronq):

yeah but this is not related to acid and bases. Kf is the freezing point depression constant, and Kb is the boiling point elevations constant

OpenStudy (anonymous):

I honestly have no idea how to even start this. Can you tell me what to do first?

OpenStudy (aaronq):

read my first reply

OpenStudy (anonymous):

Well, I guess I don't know how to tell the delta T. Is it 100 degrees less than the boiling point?

OpenStudy (aaronq):

if the boiling point is 102.17 °C, then dT is 2.17 °C

OpenStudy (anonymous):

So, 2.17=Kb(9649)(i)?

OpenStudy (aaronq):

you're solving for m first, then for dT using the other constant

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