If the solubility of salt in water is .36 g/mL, a solution containing 72 g salt in 200ml water can be described as:
A. diluted
B. saturated
C. unsaturated
D. supersaturated

If the solubility of salt in water is .36 g/mL, a solution containing 72 g salt in 200ml water can be described as:

A. diluted
B. saturated
C. unsaturated
D. supersaturated


If the solution were saturated at 0.36 g/mL, you could get this much salt into 200 mL:

0.36\,\frac {g}{mL} \times\, 200\, mL = ??\,g

If you have more salt than that, the solution would be supersaturated.
If you have exactly that amount, the solution would be saturated.
If you have less than that amount, the solution would be either diluted or unsaturated. I'm not sure where to draw the boundary between them.

If the solution were saturated at 0.36 g/mL, you could get this much salt into 200 mL:

0.36\,\frac {g}{mL} \times\, 200\, mL = ??\,g

If you have more salt than that, the solution would be supersaturated.
If you have exactly that amount, the solution would be saturated.
If you have less than that amount, the solution would be either diluted or unsaturated. I'm not sure where to draw the boundary between them.

What's the letter of the answer?

Oh come on! Take a guess!

You can see that 0.36 \times 200 = 72g. That's a saturated solution (in 200 mL). Right? So, you have a 200 mL solution with exactly 72 grams in it. What is it?
A - a diluted solution?
B - a saturated solution?
C - an unsaturated solution?
D - a supersaturated solution?

If the solution were saturated at 0.36 g/mL, you could get this much salt into 200 mL:

0.36\,\frac {g}{mL} \times\, 200\, mL = ??\,g

If you have more salt than that, the solution would be supersaturated.
If you have exactly that amount, the solution would be saturated.
If you have less than that amount, the solution would be either diluted or unsaturated. I'm not sure where to draw the boundary between them.

I got it, you mean it is saturated! I was about to send this when I got the alert in your second reply.

If the solution were saturated at 0.36 g/mL, you could get this much salt into 200 mL:

0.36\,\frac {g}{mL} \times\, 200\, mL = ??\,g

If you have more salt than that, the solution would be supersaturated.
If you have exactly that amount, the solution would be saturated.
If you have less than that amount, the solution would be either diluted or unsaturated. I'm not sure where to draw the boundary between them.

Please give me examples of supersaturated, diluted and unsaturated...

Examples? Hmm. That's not really a problem, since they're all alike -- just the solubility is different. I think this explanation will be better than giving you problems, because the problems will be fairly trivial.

1. For any given solution, you have a substance to be dissolved (a solute), and a liquid to dissolve it in (solvent). At a given temperature, only a certain amount of that solute can be dissolved in the solvent. If you were to add more than that amount, you will most likely have solid on the bottom of the flask. The solution, in this case, is said to be "saturated". The amount of solute (A) that can be dissolved in the solvent (B) is called the solubility of A in B. Normally, you can't get more than the solubility of A in B into a solution.

2. If you have less of A dissolved in B than the solubility, you have an unsaturated solution.

3. If the solution is very dilute, you have a dilute solution. A dilute solution is also an unsaturated solution -- a very unsaturated solution. However, an unsaturated solution may not be a dilute solution. (think about it).

4. Finally, there's the supersaturated solution. This is the case where there's more of A dissolved in B than the solubility says there should be! How is it possible? This is weird. It's usually created by putting more solute (A) in a solvent (B) than the solubility limit. You then heat the mixture up until everything dissolves. You cool it (usually you have to cool it slowly, without shaking). There will be a temperature at which the solution will be saturated. Below that solution, you might expect (I would) crystals of the solute to start forming. Once in a while, they won't do so spontaneously. As you cool, you form a supersaturated solution.

There are a bunch of cool properties of supersaturated solutions. If you drop something into the solution (for example, a crystal of A), some of the solute will spontaneously come out of solution and form crystals. The crystals that form will fall to the bottom of the flask and the solution will become saturated.

Supersaturation can be difficult to observe in a laboratory. Nonetheless, supersaturation is a real problem in the chemical industry. When you put something into the solution that causes crystals to form, it is said that you're putting nucleating agents in the mixture. Nucleii (plural of nucleus) are usually crystals of the solute, but it's possible (in some cases) to use crystals of other materials that are similar, in crystal structure, to the crystals of the solute. Even dust can sometimes nucleate solutions.

A nucleus is the location where crystals start growing from. Crystals seem to require some template for them to lay down on. Nucleus is the same word as the atomic nucleus but it's something entirely different. By the way, don't say NU-CUE-LUR or NU-CUE-LUS like politicians and other such individuals; say NU-CLEE-UR and NU-CLEE-US. You can occasionally fool people into thinking you're educated if you do this. :p :rolleyes:

This contains a video of crystallization from a supersaturated solution (sodium acetate forms supersaturated solutions fairly easily. Most substances won't).

How to Demonstrate a Supersaturated Solution (http://www.instructables.com/id/Seeding-A-Crystal/)

Here's another reference:

Supersaturated Solutions Demonstration Sheet (http://chemed.chem.purdue.edu/demos/demosheets/15.2.html)

Here's a reference on solubility, itself (probably more than you want to know) and on supersaturation:

Wikipedia - Solubility (http://en.wikipedia.org/wiki/Solubility)

Wikipedia - Supersaturation (http://en.wikipedia.org/wiki/Supersaturation)

OK. That's the lecture.