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The solubility product of a sparingly soluble salt $\mathrm{A}_{2} \mathrm{X}_{3} \text { is } 1.1 \times 10^{-23}$ . If specific conductance of he solution is $3 \times 10^{-5} \mathrm{Sm}^{-1}$ , the limiting molar conducivity of the solution is $x \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}$ . The value of x is _______.
Option: 1
3

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Let the solubility of $\mathrm{A_{2} X_{3} }$ be $\mathrm{s ~mol L^{-1} }$ .

We have,

$\mathrm{A_{2} X_{3} \longrightarrow 2 A^{3+}+3 X^{2-}}$

$\mathrm{2s \quad \; \; \; \; \; \; 3s}$

$\mathrm{k_{s p} \text { of } A_{2} X_{3}=(2s)^{2}(3s)^{3}}$

$\mathrm{1.1 \times 10^{-23}=108s^{5}}$

$\mathrm{s=1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}}$

$\mathrm{s=\frac{10^{-5}}{10^{-3}} \mathrm{~mol} / \mathrm{m}^{3} \quad\left(1 \mathrm{~L}=10^{-3} \mathrm{~m}^{3}\right)}$

$\mathrm{s=10^{-2} \mathrm{~mol} / \mathrm{m}^{3}}$

Now for a sparingly soluble salt,

$\mathrm{\Lambda _{m}^{\infty }=\frac{K}{C}~Sm^{2}mol^{-1}}\; \; \; \; \; \; \; \left [ \text{all terms in SI units}\right ]$

$\Lambda_{m}^{\infty}=\frac{3 \times 10^{-5}}{10^{-2}} \mathrm{~Sm}^{2} \mathrm{~mol}^{-1}\\$

$\Lambda_{m}^{\infty}=3 \times 10^{-3} \mathrm{~Sm}^{2} \mathrm{~mol}^{-1}$

$x=3$

Answer is 3

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Ritika Kankaria

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