Question

You have 100ml of a 1.00M solution of Sodium Nitrate at 10 degrees C. Use tables T and G in your reference tables to determine if this is a saturated, unsaturated, or supersaturated solution.
I need help with this ASAP Thank you so much!!!

Answer 1

Answer:

• The solution is unsaturated.

Explanation:

1. First, detemine the amount of sodium nitrate dissolved in the 100 ml of your solution.

• # of moles = molarity × volume in liters

• # of moles = 1.00M × 100ml × 1 liter/1,000ml = 0.100mol

• mass in grams = # of moles × molar mass

• molar mass of sodium nitrate (NaNO₃) = 84.9947 g/mol

• mass in grams = 0.100 mol × 84.9947 g/mol = 8.49947 g ≈ 8.50g

That is 8.50g of solute in 100 ml of solution.

2. What is the solubility of sodium nitrate at 10 ºC?

See the graph attached. There, I indicated the solubility of sodium nitrate at 10ºC: 80 g of solue per 100 g of water.

In the previous part, it was determined that the solution has 8.50 g of solute in 100 ml of solution.

To compare, you should have the same basis; this is, either 100 ml of solution or 100 g of water for both data.

To transform the ml of solution to grams of solution, and then into grams of water, you would need the density of the solution, which you do not have.

Nevertheless, you can work with some estimations.

100 g of water is equal to 100 g of water, then, the solubility of 80g of solute per 100 g of water is the same as 80g of solute / 100 ml of water.

The volume of a solution is mainly the solvent, thus, only as an approximation, 80g of solute / 100 ml of water is close to 80 g of solute / 100 ml of solution. At least, this permits you to ensure that the concentraion of the 1.00M solution of sodium nitrate, 8.50 g / 100 ml o water, is way below the solubility of the solution.

Then, the solution can dissolve more solute, meaning that it is unsaturated.