200 l of hard water contains 44.4 g of Cacl2. determine the calcium content in 1 liter of such water

Given:
V1 (H2O) = 200 l
m1 (CaCl2) = 44.4 g
V2 (H2O) = 1 L

To find:
m2 (Ca) -?

Decision:
1) 200 l of H2O – 44.4 g of CaCl2;
1 l H2O – (x) g CaCl2;
x = 1 * 44.4 / 200 = 0.222 g;
m2 (CaCl2) = x = 0.222 g;
2) n2 (CaCl2) = m2 (CaCl2) / M (CaCl2) = 0.222 / 111 = 0.002 mol;
3) n2 (Ca) = n2 (CaCl2) = 0.002 mol;
4) m2 (Ca) = n2 (Ca) * M (Ca) = 0.002 * 40 = 0.08 g.

Answer: Ca mass is 0.08 g.