JEE MAIN - Chemistry (2022 - 28th June Morning Shift - No. 19)
The solubility product of a sparingly soluble salt A2X3 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 m2 mol$$-$$1. The value of x is ___________.
Answer
3
Explanation
$A_{2} X_{3} \rightleftharpoons \underset{2S}{2 \mathrm{~A}}+\underset{3S}{3 \mathrm{X}}$
$$ \begin{aligned} &\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^{2}(3 s)^{3}=1.1 \times 10^{-23} \\\\ &\mathrm{~S} \approx 10^{-5} \end{aligned} $$
For sparingly soluble salts
$$ \begin{aligned} \wedge_{m} &=\wedge_{m}^{0} \\\\ \wedge_{m} &=\frac{\mathrm{k}}{\mathrm{S} \times 10^{3}} \\\\ &=\frac{3 \times 10^{-5}}{10^{-5}} \times 10^{-3} \\\\ &=3 \times 10^{-3} ~ \mathrm{Sm}^{2} \mathrm{~mol}^{-1} \end{aligned} $$
$$ \begin{aligned} &\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^{2}(3 s)^{3}=1.1 \times 10^{-23} \\\\ &\mathrm{~S} \approx 10^{-5} \end{aligned} $$
For sparingly soluble salts
$$ \begin{aligned} \wedge_{m} &=\wedge_{m}^{0} \\\\ \wedge_{m} &=\frac{\mathrm{k}}{\mathrm{S} \times 10^{3}} \\\\ &=\frac{3 \times 10^{-5}}{10^{-5}} \times 10^{-3} \\\\ &=3 \times 10^{-3} ~ \mathrm{Sm}^{2} \mathrm{~mol}^{-1} \end{aligned} $$
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