
Answer is: molarity of calcium chloride is 0.123 M.
Vâ(CaClâ) = 165 mL; initial volume of calcium chloride solution.
câ(CaClâ) = 0.688 M; initial concentration.
Vâ(CaClâ) = 925.0 mL; final volume.
câ(CaClâ) = ?; final concentration.
Use câVâ = câVâ.
0.688 M ¡ 165 mL = câ(CaClâ) ¡ 925 mL.
câ(CaClâ) = (0.688 M ¡ 165 mL) á 925 mL.
câ(CaClâ) = 0.123 M.