The conductivity of a saturated solution of BaSO 4 is 3.06 × 1 0 − 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

Solublity = Λ eq conductivity × 1000 = 1.53 3.06 × 1 0 − 6 × 1000 = 2 × 1 0 − 3 K sp = S 2 = 4 × 1 0 − 6

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ChemistryElectrochemistryConductance/conductivity/^mHard2 minPYQ_2009

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The conductivity of a saturated solution of $BaSO_{4}$ is $3.06 \times 1 0^{- 6} ohm^{- 1} cm^{- 1}$ 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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Solution:

Solublity $= \frac{conductivity \times 1000}{Λ _{eq}}$ $= \frac{3.06 \times 1 0 ^{- 6} \times 1000}{1.53} = 2 \times 1 0^{- 3} K_{sp} = S^{2} = 4 \times 1 0^{- 6}$

Stream:BITSATSubject:ChemistryTopic:ElectrochemistrySubtopic:Conductance/conductivity/^m

⏱ 2mℹ️ Source: PYQ_2009

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