A Levels Chemistry (9701)•9701/13/M/J/19
Explanation
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
Topic: Analytical techniques
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