I expect if I look really hard I could find exactly where you went off track but it is easier to show you how to do the problem.
Plan: Determine molarity CaCl2 and molarity NaCl, then multiply M CaCl2 by 2 (to find M Cl) and add to molarity NaCl.
mols CaCl2 = 10/111 = 0.09
M = moles/L = 0.09/0.1 = 0.9 M CaCl2
M Cl in CaCl2 is 2 x 0.9 = 1.8
moles NaCl = 10/58.44 = 0.171
M = moles/L = 0.171/0.1 = 1.71 M NaCl.
then 1.8 + 1.71 = 3.51 M to three s.f.
If 10.0g of CaCl2 and 10.0g of NaCl are dissolved in 100.0mL of solution, what is the concentration of chloride ions? (the molar mass of CaCl2 is 111.0g/mol; the molar mass of NaCl is 58.44g/mol.
I found individual molar mass of:
Ca-40.08
Cl-35.45
Na-22.99
So I divided 10g CaCl2/111.0g CaCl2 + 10g NaCl/58.44g NaCl = .262205764. I than divided by .100L = 2.61205764.
I than divided 70.9g/111.0g and 35.45g/58.44g, added them together and came up with 1.245305065.
The final answer is 3.51M.
I cannot figure out where this answer came from. Help please.
1 answer
2 answers