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Whch of the following will have the highest freezing point? a) 0.1M KCl
b) 0.1M glucose c) 0.1M BaCl2 d) 0.1M AlCl3

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Raoult's law originally was meant to describe the properties of non-electrolytes soltions. When it comes to the solutions of electrolytes, such as salts, you need to introduce the van't Hoff factor, sometimes called isotonic coefficient.
To deal with it, you need to know the folmula:
i=1+α(n-1),
where
i - van't Hoff factor
α - dissociation degree
n - number of ions formed from 1 ionic formula of electrolyte
In the case of strong electrolytes, we just assume α=1, so i=n.
Thus, you just need to calculate the number of ions which are formed from 1 ionic formula (for instance, it's 5 for Al2(SO4)3, as it goes into 5 ions - 2 aluminums and 3 sulfates), and multiply the cryoscopic constant and molality by it.
So in your case the biggest number of ions comes from d)

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  • $\begingroup$ What if concentration and van't hoff factor is same? Can i consider molar mass then? $\endgroup$ – Jui Sep 15 '17 at 8:36
  • $\begingroup$ No, colligative properties are about the number of particles, not their nature by definition. Like, 0.01 M solutions of KCl and NaBr would have same freezing points, as both dissociate giving 2 ions (if we assume their dissociation is 100% complete) $\endgroup$ – MEL Science Sep 15 '17 at 8:38
  • $\begingroup$ But, depression in freezing point is a colligative property, not freezing point. Am I right? $\endgroup$ – Jui Sep 15 '17 at 8:41
  • $\begingroup$ You are right. If depressions for both compounds are the same, aqueous solutions of these compounds would have the same FP. $\endgroup$ – MEL Science Sep 15 '17 at 9:02
  • $\begingroup$ Let's say the depression for both KCl and NaBr of some concentration is 0.5 degrees. Then the freezing point of both solutions is (FP of water)-(depression)=-0.5 degrees Celcius $\endgroup$ – MEL Science Sep 15 '17 at 9:03

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