Okay so the question is: what would be the molarity of a saturated aqueous solution of NaI prepared at 25 C, assuming that the added NaI does not add significantly to the volume of the water?
My question is, I got confused on what formula I use to for the question, because the formula for Molarity is (M=#of moles/# of L), but it doesn't give me that.
You must have a table or a graph or something like that which will tell you the number of grams of NaI in a saturated solution. From the data you should get something like grams/mL or grams/L. Whatever you get change that to grams/L then convert grams to mols.
mols = grams NaI/molar mass NaI
To calculate the molarity of a saturated aqueous solution of NaI prepared at 25°C, you will need to use the solubility data for NaI at the given temperature.
Here's a step-by-step explanation of how to solve the problem:
1. Find the solubility of NaI at 25°C: Solubility is usually given in terms of grams per 100 grams of water or grams per liter of water. Let's say the solubility of NaI at 25°C is given as 184 grams per 100 grams of water.
2. Convert the solubility to moles per liter: To calculate the moles of NaI in one liter of the solution, you need to convert the grams of NaI to moles. Use the molar mass of NaI, which is the sum of the atomic masses of sodium (Na) and iodine (I). The molar mass of NaI is approximately 149.89 grams per mole.
Moles of NaI = (Grams of NaI / Molar mass of NaI)
Moles of NaI = (184 g / 149.89 g/mol)
Moles of NaI = 1.2302 moles
3. Calculate the molarity: Now that you know the number of moles of NaI in one liter of solution, you can calculate the molarity.
Molarity (M) = Moles of solute (NaI) / Volume of solution (in liters)
The volume of the solution is assumed to be negligibly affected by the addition of NaI, so it can be considered equal to 1 liter.
Molarity (M) = 1.2302 moles / 1 liter
Molarity (M) = 1.2302 M
Therefore, the molarity of the saturated aqueous solution of NaI prepared at 25°C would be approximately 1.2302 M.