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Q.

The conductivity of a saturated solution of BaSO₄ is $3.06 \times 10^{- 6} o h m^{- 1} c m^{- 1}$ and its equivalent conductance is 1.53 ohm^-1 cm² eq^-1 Then $K_{s p}$ for BaSO₄ will be _____

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a

$4 \times 10^{- 6}$

b

$2.5 \times 10^{- 13}$

c

$10^{- 6}$

d

$2.5 \times 10^{- 9}$

answer is D.

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Detailed Solution

$\begin{aligned} \land_{m}^{0} = \frac{k \times 1000}{s} \land_{m}^{0} = 2 \land_{e q}^{0} \\ s = \frac{k \times 1000}{\land_{m}^{0}} = \frac{3.06 \times 10^{- 6} \times 1000}{2 (1.53)} \\ S = 10^{- 3} \end{aligned}$
$K_{s p} = S^{2} = 10^{- 6} M^{2}$

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