$\begin{aligned} & \mathrm{A}_2 \mathrm{X}_{3(\mathrm{~s})} \rightleftharpoons 2 \mathrm{~A}_{(\mathrm{aq})}^{+3}+3 \mathrm{X}_{(\mathrm{aq})}^{-2} \\ & \text { solubility }=\mathrm{sM} 2 \mathrm{~s} 3 \mathrm{~s} \\ & (2 \mathrm{~s})^2(3 \mathrm{~s})^3=1.1 \times 10^{-23} \\ & 108 \mathrm{~s}^5=1.1 \times 10^{-23} \\ & \mathrm{~s} \simeq 10^{-5} \mathrm{M}=10^{-5} \frac{\mathrm{~mol}}{\mathrm{~L}}=0.01 \frac{\mathrm{~mol}}{\mathrm{~m}^3} \\ & \text { Now } \wedge_{\mathrm{m}} \simeq \wedge_{\mathrm{m}}^{\infty}=\frac{\mathrm{k}}{\mathrm{m}}=\frac{\mathrm{k}}{\mathrm{s}} \\ & \Rightarrow \wedge_{\mathrm{m}}^{\infty}=\frac{3 \times 10^{-5}}{0.01}=3 \times 10^{-3} \mathrm{~S}-\mathrm{m}^2 / \mathrm{mol}\end{aligned}$