In the test of sulphate radical,the white precipitate of sulphate is soluble in

Choose the correct matching:-

Column-$I$	Column-$II$
$(P) NaBr + \text{conc. } H_2SO_4 \rightarrow$	$(A) \text{Colourless gas}$
$(Q) Na_2S + \text{dil. } HCl \rightarrow$	$(B) \text{Brown colour}$
$(R) NaNO_2 + \text{dil. } HCl \rightarrow$	$(C) \text{Rotten smell}$
$(S) NaNO_3 + \text{conc. } H_2SO_4 \rightarrow$	$(D) \text{Paramagnetic}$

On addition of conc. $H_2SO_4$ to a chloride salt,colourless fumes are evolved,but in the case of an iodide salt,violet fumes are observed. This is because:

Aqueous solution of a salt $(A)$ forms a dense white precipitate with $BaCl_{2}$ solution. The precipitate dissolves in dilute $HCl$ to produce a gas $(B)$ which decolourises acidified $KMnO_{4}$ solution. $A$ and $B$ respectively are

$S^{2-}$ and $SO_3^{2-}$ can be distinguished by using

$A$ salt gives violet vapours when treated with concentrated $H_2SO_4$,it contains