The problem asks to identify the colourless salt(s) H based on a two-step reaction sequence.

Step 1: Analyze the first reaction

The problem states: "A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime."

1. Reactants: Salt H + excess NaOH (boiling).
2. Observation: Production of a non-flammable gas.
3. Inference: The reaction of a salt with a strong base (NaOH) upon heating to produce a gas is characteristic of ammonium salts ($NH_4^+$). The gas produced is ammonia ($NH_3$). $NH_4^+(aq) + OH^-(aq) \xrightarrow{\Delta} NH_3(g) + H_2O(l)$ Ammonia gas is considered non-flammable in this context. The gas evolution stops when all the $NH_4^+$ ions are consumed.
4. Check options: All the given options are colourless ammonium salts: A: $NH_4NO_3$ B: $NH_4NO_2$ C: $NH_4Cl$ D: $(NH_4)_2SO_4$ Therefore, all four salts will react with NaOH to produce $NH_3$ gas. This first step confirms H is an ammonium salt but does not distinguish between the options.

Step 2: Analyze the second reaction

The problem states: "Upon addition of Zn dust to the same solution, the gas evolution restarts."

1. Reactants: The resulting solution from Step 1 + Zn dust.
2. Composition of the solution: The solution contains the sodium salt formed from the anion of H and excess NaOH.
   • From A: $NaNO_3$ and excess NaOH.
   • From B: $NaNO_2$ and excess NaOH.
   • From C: $NaCl$ and excess NaOH.
   • From D: $Na_2SO_4$ and excess NaOH.
3. Reaction with Zn dust: Zinc is a reducing agent, especially in an alkaline medium. We need to check if Zn can react with the anion in the solution to produce a gas.

Step 3: Evaluate each option

• A: $NH_4NO_3$ After the first reaction, the solution contains $NaNO_3$ and excess NaOH. Zinc dust reduces nitrate ions ($NO_3^-$) in an alkaline medium to ammonia gas ($NH_3$). $4Zn + NO_3^- + 7OH^- \rightarrow 4ZnO_2^{2-} + NH_3(g) + 2H_2O$ Gas evolution ($NH_3$) restarts. This matches the description. Thus, A is a correct option.

• B: $NH_4NO_2$ After the first reaction, the solution contains $NaNO_2$ and excess NaOH. Zinc dust also reduces nitrite ions ($NO_2^-$) in an alkaline medium to ammonia gas ($NH_3$). $3Zn + NO_2^- + 5OH^- \rightarrow 3ZnO_2^{2-} + NH_3(g) + H_2O$ Gas evolution ($NH_3$) restarts. This also matches the description. Thus, B is a correct option.

• C: $NH_4Cl$ After the first reaction, the solution contains $NaCl$ and excess NaOH. The chloride ion ($Cl^-$) is not reduced by zinc under these conditions. However, zinc is an amphoteric metal and reacts with excess NaOH to produce hydrogen gas ($H_2$). $Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2(g)$ While gas evolution does restart, the gas produced is $H_2$, which is flammable. This contradicts the initial description of the gas as non-flammable. This reaction is a general property of Zn in alkali and does not depend on the specific anion ($Cl^-$). The question is designed to test the specific reactivity of the anion. Therefore, C is likely incorrect.

• D: $(NH_4)_2SO_4$ After the first reaction, the solution contains $Na_2SO_4$ and excess NaOH. The sulfate ion ($SO_4^{2-}$) is not reduced by zinc under these conditions to produce a gas. Similar to option C, zinc will react with excess NaOH to produce flammable hydrogen gas ($H_2$). For the same reason as C, this option is unlikely to be correct.

Conclusion

The key distinction lies in the reaction of the anion with zinc dust in the alkaline solution. Only nitrate ($NO_3^-$) and nitrite ($NO_2^-$) ions are reduced by zinc to produce ammonia gas, the same non-flammable gas evolved in the first step. Therefore, the salts must be ammonium nitrate and ammonium nitrite.