The solubility product of a sparingly soluble salt A2X3 is 1.1×10−23. If specific conductance of the solution is 3×10−5Sm−1, the limiting molar conductivity of the solution is x×10−3Sm2mol−1. The value of x is ________.

A2X3(s)⇌2A(aq)+3+3X(qq)−2solubility sM 2s 3s(2s)2(3s)3=1.1×10−23108s5=1.1×10−23s≃10−5M=10−5molL=0.01molm3 Now ∧m≃∧m∞=km=ks⇒∧m∞=3×10−50.01=3×10−3S−m2/mol

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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 $3 \times 10^{- 5} {Sm}^{- 1}$ , the limiting molar conductivity of the solution is $x \times 10^{- 3} {Sm}^{2} {mol}^{- 1}$ . The value of x is ________.

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Detailed Solution

$A_{2} X_{3 (s)} ⇌ 2 A_{(aq)}^{+ 3} + 3 X_{(qq)}^{- 2}$
solubility sM 2s 3s
$\begin{array}{l} (2 s)^{2} (3 s)^{3} = 1 .1 \times 10^{- 23} \\ 108 s^{5} = 1 .1 \times 10^{- 23} \\ s ≃ 10^{- 5} M = 10^{- 5} \frac{mol}{L} = 0 .01 \frac{mol}{m^{3}} \\ Now \land_{m} ≃ \land_{m}^{\infty} = \frac{k}{m} = \frac{k}{s} \\ \Rightarrow \land_{m}^{\infty} = \frac{3 \times 10^{- 5}}{0 .01} = 3 \times 10^{- 3} S - m^{2} / mol \end{array}$