The solubility product of a sparingly soluble salt A2X3 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$-$3 S m2 mol$-$1. The value of x is ___________.
Answer (integer)
3
Solution
$A_{2} X_{3} \rightleftharpoons \underset{2S}{2 \mathrm{~A}}+\underset{3S}{3 \mathrm{X}}$
<br/><br/>
$$
\begin{aligned}
&\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^{2}(3 s)^{3}=1.1 \times 10^{-23} \\\\
&\mathrm{~S} \approx 10^{-5}
\end{aligned}
$$
<br/><br/>
For sparingly soluble salts
<br/><br/>
$$
\begin{aligned}
\wedge_{m} &=\wedge_{m}^{0} \\\\
\wedge_{m} &=\frac{\mathrm{k}}{\mathrm{S} \times 10^{3}} \\\\
&=\frac{3 \times 10^{-5}}{10^{-5}} \times 10^{-3} \\\\
&=3 \times 10^{-3} ~ \mathrm{Sm}^{2} \mathrm{~mol}^{-1}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
This question is part of PrepWiser's free JEE Main question bank. 67 more solved questions on Equilibrium are available — start with the harder ones if your accuracy is >70%.