1. Tardigrade
  2. Question
  3. Chemistry
  4. The solubility product of a sparingly soluble salt A 2 X 3 is 1.1 × 10-23. If specific conductance of the solution is 3 × 10-5 S m -1, the limiting molar conductivity of the solution is x × 10-3 S m 2 mol -1. The value of x is

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Q. 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 _____

JEE MainJEE Main 2022Equilibrium

Solution:

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$s =10^{-5} M =10^{-5} \frac{ mol }{ L }=0.01 \frac{ mol }{ m ^{3}}$
Now $\wedge_{ m }=\wedge_{ m }^{\infty}=\frac{ k }{ m }=\frac{ k }{ s }$
$\Rightarrow \wedge_{ m }^{\infty}=\frac{3 \times 10^{-5}}{0.01}=3 \times 10^{-3} S - m ^{2} / mol$