The conductivity of a saturated solution of BaSO4 is 3.06×10−6 ohm−1 cm−1 and its equivalent conductance is 1.53 ohm−1 cm2 equiv−1. The Ksp for BaSO4 will be

1.53=1000×3.06×10−6NormalityNormality = 2×10−3NMolarity = 2×10−32=10−3MKsp=10−6M

The conductivity of a saturated solution of BaSO4 is 3.06×10−6 ohm−1 cm−1 and its equivalent conductance is 1.53 ohm−1 cm2 equiv−1. The Ksp for BaSO4 will be

1.53=1000×3.06×10−6NormalityNormality = 2×10−3NMolarity = 2×10−32=10−3MKsp=10−6M

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The conductivity of a saturated solution of ${BaSO}_{4}$ is $3 .06 \times 10^{- 6} {ohm}^{- 1} {cm}^{- 1}$ and its equivalent conductance is $1 .53 {ohm}^{- 1} {cm}^{2} {equiv}^{- 1}$ . The $K_{sp}$ for ${BaSO}_{4}$ will be

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$4 \times 10^{- 12}$

$2.5 \times 10^{- 13}$

$25 \times 10^{- 9}$

$10^{- 6}$

answer is D.

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$\begin{array}{l} 1 .53 = \frac{1000 \times 3 .06 \times 10^{- 6}}{Normality} \\ Normality = 2 \times 10^{- 3} N \\ Molarity = \frac{2 \times 10^{- 3}}{2} = 10^{- 3} M \\ K_{sp} = 10^{- 6} M \end{array}$

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The conductivity of a saturated solution of BaSO4 is 3.06×10−6 ohm−1 cm−1 and its equivalent conductance is 1.53 ohm−1 cm2 equiv−1. The Ksp for BaSO4 will be

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The conductivity of a saturated solution of ${BaSO}_{4}$ is $3 .06 \times 10^{- 6} {ohm}^{- 1} {cm}^{- 1}$ and its equivalent conductance is $1 .53 {ohm}^{- 1} {cm}^{2} {equiv}^{- 1}$ . The $K_{sp}$ for ${BaSO}_{4}$ will be