In inorganic salt analysis, for the indicatory test of \(Cl^-\), conc. \(H_2SO_4\) is used. This is because:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → conc. H2SO4 has lower volatility than HCl
A less volatile acid displaces a more volatile acid from its salt: when the mixture is warmed, the more volatile acid escapes as a gas. In the indicatory test, a chloride warmed with concentrated sulphuric acid gives colourless, pungent HCl gas, which forms white fumes of \( \ce{NH4Cl} \) with a glass rod dipped in ammonia:
\( \displaystyle \ce{NaCl + H2SO4} \xrightarrow{\Delta} \ce{NaHSO4 + HCl ^} \)
\( \displaystyle \ce{HCl + NH3 ->} \underset{\text{white fumes}}{\ce{NH4Cl}} \)
So the test works because \( \ce{H2SO4} \) has lower volatility than HCl.
Option (1) conc. H2SO4 is a strong oxidizing agent and Cl2 is produced. — Incorrect. The gas given off is HCl, not \( \ce{Cl2} \). Chlorine is produced only when an oxidising agent such as \( \ce{MnO2} \) is added:
\( \displaystyle \ce{2NaCl + MnO2 + 3H2SO4 -> 2NaHSO4 + MnSO4 + 2H2O + Cl2 ^} \)
Option (2) conc. H2SO4 is a strong reducing agent and HCl is produced. — Incorrect. Concentrated \( \ce{H2SO4} \) is an oxidising acid, not a reducing agent. HCl forms here by simple displacement, with no change in oxidation state.
Option (3) conc. H2SO4 is a dehydrating agent and Cl2 is produced. — Incorrect. Its dehydrating property (as in charring sugar) plays no part in this test, and \( \ce{Cl2} \) is not formed.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Sulphuric Acid – “… The acid forms two series of salts: normal sulphates (such as sodium sulphate and copper sulphate) and acid sulphates (e.g., sodium hydrogen sulphate). Sulphuric acid, because of its low volatility can be used to manufacture more volatile acids from their corresponding salts.”