I need to determine how many grams of Cesium Iodide following solutes would be needed to make 5.00 102 mL of a 0.300 M solution.

How many mols do you need? That's M x L = ?

Then mols = grams/molar mass. Solve for grams.

it said that was wrong

NO ITS RIGHT

To determine the grams of Cesium Iodide (CsI) needed to make the solution, you need to use the formula:

moles = Molarity × Volume

First, let's convert the given volume from milliliters (mL) to liters (L):

5.00 × 10^2 mL = 5.00 × 10^2 / 1000 L = 0.500 L

Now, we can calculate the moles of CsI using the formula:

moles = 0.300 M × 0.500 L = 0.150 mol

To determine the grams of CsI needed, you need to know its molar mass. The molar mass of CsI can be calculated by summing up the atomic masses of cesium (Cs) and iodine (I) from the periodic table:

Cs: 1 atom × atomic mass of Cs = 1 × 132.91 g/mol = 132.91 g/mol
I: 1 atom × atomic mass of I = 1 × 126.90 g/mol = 126.90 g/mol

Molar mass of CsI = 132.91 g/mol + 126.90 g/mol = 259.81 g/mol

Finally, we can calculate the grams of CsI needed using the formula:

grams = moles × molar mass = 0.150 mol × 259.81 g/mol ≈ 38.97 grams

Therefore, approximately 38.97 grams of Cesium Iodide would be needed to make a 0.300 M solution with a volume of 5.00 × 10^2 mL.