Chem 1/2 question! Need help :)

[imaginayres]

Hey I was just wondering how would I approach this question?

At 80°C a saturated solution of NaNO3 had a total mass of 70g. This solution was then cooled to 40°C. What mass of salt will precipitate from the solution?

Thanks in advance!

[zsteve]

You'll need a graph of the solubility vs. temperature for the compound in aqueous solution in order to solve this problem.

Was there any additional info for this problem?

[imaginayres]

I have one with me right now! Although, I don't know what to do in terms of finding out what I need to find out with those given values from the question. There was no more given information to this question unfortunately!

[zsteve]

I have one with me right now! Although, I don't know what to do in terms of finding out what I need to find out with those given values from the question. There was no more given information to this question unfortunately!

Hmm. In my opinion, there should have been given that information with the question, but just for example:
Say I've got a saturated solution of NaNO3 with concentration 100g/L at 80 C. If I cooled it to 40 C, at which the solubility is 60 g/L, then I know that for each litre, 100 g/L - 60 g/L = 40 g/L must precipitate out.

Another way of looking at it is:
At 80 C, in 1 L, we have 100g of dissolved NaNO3
Now let us cool it to 40 C
Now, in 1 L, we have 60 g of dissolved NaNO3. But we originally had 100g NaNO3, so the remainder must have precipitated, 100-60 = 40g NaNO3 precipitates.

You'll have to refer to additional data for this question, as the relationship of solubility to temperature is generally non-linear.

[imaginayres]

Haha alright, thanks for trying to help me out ^_^