1. Why do we use molality instead of molarity in the context of colligative properties?
2. Calculate the molality of a solution prepared by dissolving 125.0 g of glucose (C6H12O6) in 500 g of water.
3. Calculate the molality of a solution prepared by dissolving 78.0 grams of butanone (C4H8O) in 800mL of acetic acid. The density of acetic acid is 1.049 g/mL.
4. You dissolve 93.24 g of an unknown in 1000 g of water and obtain ΔTf = 2.34°C. What is themolecular weight of the unknown if Kf = 1.86 °C ·kg/mol?
Freezing point depression 求指教
2013-11-09 9:01 am
回答 (1)
2013-11-10 6:27 am
✔ 最佳答案
1.唔知點解，你知道答案之後最好講俾我知，但 freezing point lowering 和 boiling point elevation 都是用 molality 去計的。2. 125 g of C6H12O6 = 125/(12*6+1*12+16*6) = 0.694 mole
molality = 0.694/(500/1000) = 1.39 mol / kg
3. 78 g of C4H8O = 78/(12*4+1*8+16) = 1.083 mole
mass of acetic acid = 800*1.049 = 839.2 g
molality = 1.083/(839.2/1000) = 1.29 mol / kg
4. ΔTF = KF · b · i
Assume the unknown will not dissociate in water, then i = 1
2.34 = (1.86)(b) ........... b is the concentration in molality
b = 1.258 mole / kg
93.24/M = 1.258 ........... M is the molecular weight
M = 93.24/1.258 = 74 g / mole
2013-11-09 22:31:30 補充：
I found this from http://answers.yahoo.com/question/index?qid=20070224221646AACQ3Hx
It makes sense.
ok..molarity uses moles/liter of solution......colligative properties have to do with how the concentration affects the solvent. So,
2013-11-09 22:32:03 補充：
to determine how the concentration has an affect on the solvent we must know HOW MUCH solvent.....hence molality is moles per KG solvent. Colligative properties are concerned with the solvent.
收錄日期: 2021-04-27 20:33:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131109000051KK00012