If 10.0g of CaCl2 and 10.0g of NaCl are dissolved in 100.0mL of solution, what is the concentration of chloride ions?

I found the molar mass of CaCl2 is 111.0g/mol; the molar mass of NaCl is 58.44g/mol.

I divided 10g CaCl2/111.0g CaCl2
I divided 10g NaCl/588.44g NaCl and added the figures together and divided by .100L. The answer came out to 2.612057642. After comparing to my answer key, the answer is 3.51M. Can someone tell me what I'm doing wrong?

I thought I worked this for you a couple of nights ago.

moles Cacl2 in 10 g CaCl2 = 10/111 = 0.090 moles CaCl2
moles Cl in that is 0.090 x 2 = 0.180 moles Cl.
moles NaCl in 10 g NaCl = 10/58.44 = 0.171 moles

total moles Cl = 0.171 + 0.180 = 0.351
M = moles/L = 0.351 moles Cl/0.100 L = 3.51 M.

Obtain a clear glass and fill it with water. Or, observe water as it pours from a faucet. List at least five physical properties of the water (five words that describe the water).

To find the concentration of chloride ions in the solution, you need to consider the number of moles of chloride ions present.

First, calculate the moles of CaCl2 and NaCl separately:
Moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2 = 10.0g / 111.0g/mol = 0.0901 mol
Moles of NaCl = mass of NaCl / molar mass of NaCl = 10.0g / 58.44g/mol = 0.171 mol

Now, you can determine the total number of moles of chloride ions by considering the ratio of chloride ions in each compound:
For CaCl2, there is 2 moles of chloride ions per 1 mole of CaCl2.
For NaCl, there is 1 mole of chloride ions per 1 mole of NaCl.

Total moles of chloride ions = (2 * moles of CaCl2) + (1 * moles of NaCl) = (2 * 0.0901 mol) + (1 * 0.171 mol) = 0.351 mol

Finally, calculate the concentration of chloride ions by dividing the moles by the volume of the solution:
Concentration of chloride ions = moles of chloride ions / volume of solution
= 0.351 mol / 0.100 L = 3.51 M

Therefore, the correct concentration of chloride ions in the solution is indeed 3.51 M.