A white crystalline substance dissolves in water. On passing ${\mathrm{H}}_2\mathrm{S}$ in this solution, a black precipitate is obtained. The black precipitate dissolves in this solution, a black precipitate is obtained. The black precipitate dissolves completely in hot ${\mathrm{HNO}}_3$ • On adding a few drops of concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ , a white precipitate is obtained. This precipitate is that of :

On adding the concentrated sulphuric acid the white precipitate of lead sulphate is obtained.

The following reaction is obtained as the result,

$\begin{array}{l}{\mathrm{Pb}}^{2+}+{\mathrm{H}}_2\mathrm{S}\stackrel{\text{ acidic }}{⟶}\underset{\text{ (Black ppt) }}{\mathrm{PbS}}\downarrow +{\mathrm{H}}_2\mathrm{O}\uparrow \\ 3\mathrm{PbS}+8{\mathrm{HNO}}_3\to 3\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NO}+3\mathrm{S}+4{\mathrm{H}}_2\mathrm{O}\\ 3\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to \underset{(\text{ white pot) }}{ }{\mathrm{PbSO}}_4\downarrow +2{\mathrm{HNO}}_3\left(\mathrm{aq}\right)\\ \text{ (white ppt) }\\ {\mathrm{PbSO}}_4+2{\mathrm{CH}}_3{\mathrm{COONH}}_4\to {\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\mathrm{Pb}+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\end{array}$

The obtained white precipitatedoes not dissolve in ammonium acetate. The precipitates of BaS and SrS are not black. Hence on adding the concentrated sulphuric acid the white precipitate of lead sulphate is obtained.

Hence the correct option is C.