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The solubility product of a sparingly soluble salt $A _{2} X _{3}$ is $1.1 \times 10^{-23}$. If specific conductance of the solution is $3 \times 10^{-5} \,S\, m ^{-1}$, the limiting molar conductivity of the solution is $x \times 10^{-3} \,S\, m ^{2}\, mol ^{-1}$. The value of $x$ is ...
$30$
$54$
$3$
$90$
Solution
$A _{2} X _{3( s )} \rightleftharpoons 2 A _{( aq )}^{+3}+3 X _{( aq )}^{-2}$
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. $3$
