Identification of calcium chloride via silver chloride solubility test

Steps:

• The white precipitate forms from AgNO3 reacting with the anion of the calcium compound in acidic medium, indicating a halide.
• Dilute HNO3 ensures no interference from basic anions like carbonate.
• The precipitate dissolves in dilute NH3, characteristic of AgCl (unlike AgBr or AgI).
• Thus, the anion is Cl⁻, so the compound is CaCl₂; M = 40.1 + (2 × 35.5) = 111.1.

Why C is correct:

• Matches the molar mass formula for CaCl₂, confirmed by the AgCl solubility in dilute ammonia per qualitative analysis.

Why the others are wrong:

• A: 54.0 does not match any calcium compound; too low for CaX₂.
• B: 75.6 approximates Ca(OH)₂ (74.1) but hydroxide yields no Ag⁺ precipitate.
• D: 99.9 approximates CaBr₂ (199.9) but AgBr is insoluble in dilute NH3.

Final answer: C