Answer:
molarity = 1.5925 mol/L
Step-by-step explanation:
To make an aqueous solution of CaI₂ (the 2 is due to the ionic charge of Calcium being Ca²⁺, and Iodide being I⁻), the concentration will be calculated by: concentration (c) (or molarity) = number of moles present (n) ÷ volume needed (V) (in litres)
since we don't have moles, we can calculate moles by:
number of moles (n) = mass present (m) (in grams) ÷ molar mass (M) (in grams per mole), which we can find using a standard IUPAC Periodic Table
∴ n(CaI₂) = m/M = 234/(40.08 + 2×126.9) = 0.796243 mol
Now we have the number of moles present, we can calculate concentration:
∴ c(CaI₂) = n/V = 0.796243/0.5L = 1.5925 mol/L