solubility $= sM\quad 2 s \quad 3 s$
$(2 s )^{2}(3 s )^{3}=1.1 \times 10^{-23}$
$108 s ^{5}=1.1 \times 10^{-23}$
$s \simeq 10^{-5} M =10^{-5} \,\frac{ mol }{ L }=0.01 \,\frac{ mol }{ m ^{3}}$
Now $\wedge_{ m } \simeq \wedge_{ m }^{\infty}=\frac{ k }{ m }=\frac{ k }{ s }$
$\Rightarrow \wedge_{ m }^{\infty}=\frac{3 \times 10^{-5}}{0.01}=3 \times 10^{-3} \,S - m ^{2} / mol$
Ans. $3$
$\begin{array}{*{20}{c}}
{C{H_3}C{H_2}OH{\text{ }} + {\text{ }}{H_3}{O^ + }\, \to \,C{H_3}C{H_2} - {O^ + } - H\, + {H_2}O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H\,\,\,\,\,\,}
\end{array}$
$(i)\, CO_2 + H_2O $ $\rightleftharpoons$ $ H_2CO_3 (ii) NH_3+ H_2O $ $\rightleftharpoons$ $ NH_4OH (iii) HCl + H_2O$ $\rightleftharpoons $ $ Cl^-+ H_3O^+$
$K_{sp}$ એ $AgCl = 1.2\times 10^{-10} \,K_{sp}$ એ $AgI = 1.7 \times 10^{-16}$
$K_{sp}$ એ $AgSCN = 7.1 \times 10^{-7} \,K_{sp}$ એ $AgBr = 3.5 \times 10^{-13}$