- A${H_2}O$ is the conjugate base of $HCl$ acid
- ✓$C{l^ - }$is the conjugate base of $HCl$ acid
- C$C{l^ - }$is the conjugate acid of ${H_2}O$ base
- D${H_3}{O^ + }$is the conjugate base of $HCl$
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Statement $I$: Aniline reacts with con. $\mathrm{H}_2 \mathrm{SO}_4$ followed by heating at $453-473 \mathrm{~K}$ gives $\mathrm{p}$ aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement $II$: In Friedel - Craft's alkylation and acylation reactions, aniline forms salt with the $\mathrm{AlCl}_3$ catalyst. Due to this, nitrogen of aniline aquires a positive charge and acts as deactivating group.
In the light of the above statements, choose the correct answer from the options given below :
Sucrose $\xrightarrow[{Cleavage\,\,(Hydrolysis)}]{{Gly\cos idic\,bond}}A + B\xrightarrow[{{\text{reagent}}}]{{{\text{Seliwanoff 's}}}}?$
$Pt \left| H _{2}( g )\right| H ^{+}( aq ) \| Cu ^{2+}(0.01 M ) \mid Cu ( s )$
is $0.576 \,V$ at $298\, K$. The $pH$ of the solution is $......\,.$ (Nearest integer)