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Why does the pure solvent shows a horizontal curve when solidfication occurs, but the curves for the solution slopes downward slightly?

2007-11-27 12:45:29 · 3 answers · asked by Rushi P 1 in Science & Mathematics Chemistry

3 answers

This is due to what is known as freezing-point depression.

A pure solvent, water is simple, will freeze at a specific temperature and in our case 0 degrees C.

If you place a solute such as sugar or salt or another liquid that is miscible in water, the freezing point of your solvent will lower slightly depending on the solute and it's concentration in the solvent.

Because solids are highly ordered, the molecules must be arranged in a specific geometry. When placing a solute in your solvent, the molecules of your solvent must rearrange themselves in order to aquire a highly ordered structure(become a solid).

In this case for example, water has specific geometry where a single water molecule will be surrounded by 6 other water molecules. If a solute is placed then perhaps now a single water molecule will now need 5 or 4 or 7 other molecules to become highly orderd.

If you want to look at something very interesting look up the following terms on google.com

ice I, ice II, ice III, ice IV....There are different forms of ice all varying with temperature and pressure. It's really cool.

2007-11-27 12:53:57 · answer #1 · answered by obscurusvita 4 · 3 0

a pure substance will freeze at a precise temperature.

A solution will freeze depending on the concentration of solute (colligative properties) As the solution freezes, the concentration changes, so the temperature of solidification will slope.

2007-11-27 20:53:23 · answer #2 · answered by reb1240 7 · 1 1

Try this. See the answers this person got for this question. I think it will work. It is a very similar question to what you asked. Isn't it? I think so. It is similar to what you asked. 'K.

2007-11-27 21:04:04 · answer #3 · answered by Aubreigh aka The Female T-Pain 4 · 0 3

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