you have 100 mL of a saturated solution of Pb(NO3)2 at 0 degrees celsius. If the solution is evaporated to dryness how many grams of Pb(NO3)2 do you expect to remain in the crucible?

Im not sure how to do this because i don't have a number to work with besides the 100ml. Help please

I found the solubility at a number of temperatures but not at zero. You might have better luck on google than I. Type in solubility of lead nitrate.

To determine how many grams of Pb(NO3)2 would remain in the crucible after evaporating the solution to dryness, you need additional information, particularly the solubility of Pb(NO3)2 at 0 degrees Celsius.

The solubility of a substance refers to the maximum amount that can dissolve in a given amount of solvent at a specific temperature. Once you know the solubility of Pb(NO3)2 at 0 degrees Celsius, you can calculate how much of the compound would remain if 100 mL of the saturated solution is evaporated to dryness.

You can find the solubility of Pb(NO3)2 from reference books or reliable online sources. Let's assume, for example, that the solubility of Pb(NO3)2 is 200 g/L (grams per liter) at 0 degrees Celsius.

To calculate the amount of Pb(NO3)2 remaining in the crucible after evaporating the solution to dryness, follow these steps:

1. Convert 100 mL to liters by dividing by 1000:
100 mL ÷ 1000 = 0.1 L

2. Calculate the maximum amount of Pb(NO3)2 that can dissolve in 0.1 L of the saturated solution using the solubility:
0.1 L × 200 g/L = 20 g

Therefore, if the solubility of Pb(NO3)2 is 200 g/L at 0 degrees Celsius, you would expect approximately 20 grams of Pb(NO3)2 to remain in the crucible after evaporating the solution to dryness.