An unknown inorganic salt S is treated with aqueous NaOH and...

When salt $\mathrm{S}$ is treated with aqueous $\mathrm{NaOH}$ and warmed, a colourless gas $\mathrm{G}$ is evolved which gives a brown precipitate with Nessler's reagent. This confirms that gas $\mathrm{G}$ is ammonia ($\mathrm{NH}_3$) and the salt $\mathrm{S}$ contains the ammonium cation ($\mathrm{NH}_4^+$).

When the aqueous solution of salt $\mathrm{S}$ is treated with dilute $\mathrm{HNO}_3$ and $\mathrm{AgNO}_3$, a curdy white precipitate is formed. This indicates the presence of chloride ions ($\mathrm{Cl}^-$), as silver chloride ($\mathrm{AgCl}$) is a curdy white precipitate.

The precipitate $\mathrm{AgCl}$ is completely soluble in aqueous ammonia (aqueous solution of gas $\mathrm{G}$) due to the formation of the soluble complex $[\mathrm{Ag}(\mathrm{NH}_3)_2]\mathrm{Cl}$.

Therefore, the unknown salt $\mathrm{S}$ is ammonium chloride ($\mathrm{NH}_4\mathrm{Cl}$).

Answer: $\mathrm{NH}_4\mathrm{Cl}$