Chem hw..help please

Question

A solution is prepared by dissolving of CaCl2 in 300g of water. The density of the resulting solution is 1.05g/mL. The concentration of CaCl2 in this solution is ____molar.

Answer: 1.48

why??

Answer 1

55.5 g CaCl2 x 1 mole CaCl2/111 g = 0.50 moles

(300 g + 55.5)(1 ml/1.05 g) = 355.5/1.05 = 338.6 ml

0.50 moles/338.6 ml x 1000 ml/L = 1.48 mol/L = 1.48 M

Answer 2

**denotes previously known/common info

Things I know:

CaCl2 + H2O --> solution

start with 300 g H2O (**300 mL H2O)

**Density of water = 1.00 g/mL

Density of solution = 1.05 g/mL

Things I want to know:

The concentration of CaCl2 in the solution is ? M

(Xg CaCl2 + 300g H2O)/300 mL = 1.05 g/mL

1.05g/mL*300mL = 315 g

X g CaCl2 + 300 g H2O = 315 g solution

Therefore, X is 15. (15 + 300 = 315)

MW CaCl2 = (40 + (2 x 35)) = 110 g/mol CaCl2

15 g CaCl2 x (1 mol CaCl2 / 110 g CaCl2) = 0.1363 mol CaCl2

0.1363 mol/0.300 L = 0.4545 M CaCl2 ?????

idk what I did wrong, if anyone sees this, help lol