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 ...

  • A
    $30$
  • B
    $54$
  • C
    $3$
  • D
    $90$

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