How many g of CaCl2 are required to prepare 0.250 L of a 0.200 M solution?

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

Then mols = g/molar mass. you know molar mass and mols, solve for grams.

To find out how many grams of CaCl2 are required to prepare a 0.200 M solution in 0.250 L, we need to use the formula:

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

First, let's rearrange the formula to solve for moles of solute:

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

Next, insert the known values:

moles of solute = 0.200 M x 0.250 L

Now, we have the number of moles of CaCl2 needed to prepare the solution. To determine the grams of CaCl2 required, we need to use the molar mass.

The molar mass of CaCl2 can be calculated by adding up the atomic masses of calcium (Ca) and two chlorine (Cl) atoms:

Ca: 40.08 g/mol
Cl: 35.45 g/mol

Molar mass of CaCl2 = (1 x Ca) + (2 x Cl) = (1 x 40.08) + (2 x 35.45) = 110.98 g/mol

Now, we can calculate the grams of CaCl2 required:

grams of CaCl2 = moles of solute x molar mass of CaCl2

Substituting the known values:

grams of CaCl2 = (0.200 M x 0.250 L) x 110.98 g/mol

Finally, calculate the result:

grams of CaCl2 = 0.050 g

Therefore, 0.050 grams of CaCl2 are required to prepare 0.250 L of a 0.200 M solution.