91Ó°ÊÓ

2Cl^−(aq) + Hg(NO₃)₂(aq)⟶2±·°¿₃^−(aq) + HgCl₂(s)

Hg(NO₃)₂(aq) = 8.25×10^−4 M

Volume of Hg(NO₃)₂ = 1.46 mL = 0.01146 L

Sample volume = 0.25 mL = 2.5 x 10 ^-4 L

Molarity(M) = mol/ L

So, no of mol of Hg(NO₃)₂ = M x L

= 8.25×10^−4 M x 0.00146 L

=1.2045 x 10^-6 Hg(NO₃)₂

\(\begin{aligned}{}\frac{{1.2045{\rm{ }}x{\rm{ }}{{10}^{ - 6}}Hg{{\left( {N{O_3}} \right)}_2} \times 2\,mol\,of\,C{l^ - }}}{{1\,mol\,Hg{{\left( {N{O_3}} \right)}_2}}}\\ = 2.409 \times {10^{ - 6}}\,mol\,of\,C{l^ - }\end{aligned}\)

Required concentration of Cl

\(\begin{aligned}{} = \frac{{2.409 \times {{10}^{ - 6}}\,mol\,of\,C{l^ - }}}{{2.5 \times {{10}^{ - 4}}L}}\\ = 9.6 \times {10^{ - 3}}M\end{aligned}\)