# Reducing temperature on dissolved Pb(NO3)2

I have pretty much zero experience in chemistry whatsoever but I wonder about how much crystal will I get from fully dissolved element when I reduce water temperature.

Variables

• at $\mathrm{40^\circ C,\ 171}\pu{g}$ of $\ce{Pb(NO3)2}$ is fully dissolved in water

• at $\mathrm{18^\circ C,\ }\ce{Pb(NO3)2}$ - How much is dissolved now? Everything that was not dissolved turned to crystals?

I know that it probably has something to do with this graph right here but I can't turn my mind to right direction. Can someone explain this problem and how can I find solution?

• You need to know how much water is being used, not just the amount of the salt. Commented Dec 13, 2013 at 13:25
• @NicolauSakerNeto Well since it's fully dissolved I calculated that it is dissolved in 244,28ml of water.
– sed
Commented Dec 13, 2013 at 13:28
• Being fully dissolved and being fully dissolved in the least amount of water are two different things. If you know the solution is saturated (i.e. it cannot dissolve any more of the salt), then you are correct. Commented Dec 13, 2013 at 13:32
• @NicolauSakerNeto So what are next moves if we know that it is dissolved with the last amount of water possible?
– sed
Commented Dec 13, 2013 at 13:41

Now, when cooling down to 18°C the solubility drops to ~55g / 100g water. This means that in our solution we still have $$228~\text{g}~\ce{H2O} \times \frac{55~\text{g}~\ce{Pb(NO3)2}}{100~\text{g}~\ce{H2O}}=125.4~\text{g}~\ce{Pb(NO3)2}\; .$$