The mass of CaCl2 required to prepare 100.mL of 0.100MCl−(aq) ions is 0.555 g.
Given that;
[tex]CaCl2(s) ------> Ca^2+(aq) + 2Cl^-(aq)[/tex]
Number of moles of Cl- = concentration × Volume of solution
concentration Cl- = 0.100M
Volume = 100.mL or 0.1 L
Number of moles of Cl- = 0.100M × 0.1 L
Number of moles of Cl- = 0.01 moles of Cl-
If 1 mole of CaCl2 contains 2 moles of Cl-
x moles of CaCl2 contains 0.01 moles of Cl-
x = 1 × 0.01/2
x = 0.005 moles of CaCl2
Mass = number of moles of CaCl2 × molar mass of CaCl2
Molar mass of CaCl2 = 111g/mol
Mass = 0.005 moles of CaCl2 × 111g/mol
Mass = 0.555 g of CaCl2
Learn more: https://brainly.com/question/9743981
