A saturated solution of sodium nitrate at 70oC is cooled to 40oC.  How many grams of precipitate are formed?

Do you have solubility tables? And don't you need to know how much solution you have?

This is all the info I have DrBob222

I don' have a solubility table. It doesn't say how much solution either,huh, is this something that can be answered?

I don't think so. Doesn't it make sense to you that if we had a liter and cooled it that we would get x grams NaNO3. If we had 2L we would get 2x grams NaNO3 and if we had 10 L we would get 10x grams NaNO3. Right?

yes

To calculate the amount of precipitate formed, we need to determine if the cooling temperature of the solution is below or above its saturation point at 40°C.

First, we need to find the solubility of sodium nitrate at both 70°C and 40°C. The solubility of a substance typically increases with temperature, so we can assume that the solubility at 70°C will be higher than at 40°C.

To find the solubility values, we can search for a solubility table or use a chemical database. Let's assume that the solubility of sodium nitrate at 70°C is 250 grams per 100 mL, and at 40°C is 150 grams per 100 mL.

Now, we can determine the amount of sodium nitrate that can be dissolved at both temperatures:

At 70°C:
- Solubility = 250 g/100 mL

At 40°C:
- Solubility = 150 g/100 mL

Let's assume we have 200 mL of the saturated solution of sodium nitrate at 70°C. To find the amount of sodium nitrate dissolved, we can use the solubility value:

At 70°C:
- Dissolved sodium nitrate = 250 g/100 mL * 200 mL
- Dissolved sodium nitrate = 500 g

Now, we need to determine how much sodium nitrate would remain if we cool the solution to 40°C:

At 40°C:
- Solubility = 150 g/100 mL
- Remaining sodium nitrate = 500 g - (150 g/100 mL * 200 mL)
- Remaining sodium nitrate = 500 g - 300 g
- Remaining sodium nitrate = 200 g

Therefore, when the saturated solution of sodium nitrate at 70°C is cooled to 40°C, 200 grams of precipitate will be formed.