how many grams of NaI are needed to prepare 400.0 mL of 0.0500 M NaI solution?

To prepare a 0.0500 M (molar) NaI solution, you first need to know the molar mass of NaI, which is the sum of the atomic masses of sodium (Na) and iodine (I). The atomic mass of Na is approximately 22.99 g/mol and that of I is approximately 126.90 g/mol. Therefore, the molar mass of NaI would be:

Molar mass of NaI = Atomic mass of Na + Atomic mass of I Molar mass of NaI ≈ 22.99 g/mol + 126.90 g/mol Molar mass of NaI ≈ 149.89 g/mol

Next, you can calculate the number of moles of NaI required using the molarity formula:

Molarity (M) = moles of solute / liters of solution

We need a 0.0500 M solution, and we are preparing 400.0 mL (or 0.4000 L) of it. To find the number of moles (n) of NaI needed:

0.0500 M = n / 0.4000 L n = 0.0500 M * 0.4000 L n = 0.0200 moles of NaI

Now, you can calculate the mass of NaI needed by using the molar mass:

Mass of NaI (g) = moles of NaI * molar mass of NaI Mass of NaI (g) = 0.0200 moles * 149.89 g/mol Mass of NaI (g) ≈ 3.00 g

Therefore, you would need approximately 3.00 grams of NaI to prepare 400.0 mL of a 0.0500 M NaI solution.