At 291 K,a saturated solution of BaSO_4 was | vedclass

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(B) According to Kohlrausch's law:$\lambda_{m}^{\circ}(BaSO_4) = \lambda_{m}^{\circ}(Ba^{2+}) + \lambda_{m}^{\circ}(SO_4^{2-})$$\lambda_{m}^{\circ}(Ba

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$2.266 	imes 10^{-3} \ g \ L^{-1}$

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(B) According to Kohlrausch's law:$\lambda_{m}^{\circ}(BaSO_4) = \lambda_{m}^{\circ}(Ba^{2+}) + \lambda_{m}^{\circ}(SO_4^{2-})$$\lambda_{m}^{\circ}(BaSO_4) = 110 + 136.6 = 246.6 \ \Omega^{-1} \ cm^2 \ mol^{-1}$The conductivity of $BaSO_4$ is:$\kappa_{BaSO_4} = \kappa_{	ext{solution}} - \kappa_{	ext{water}}$$\kappa_{BaSO_4} = (3.648 	imes 10^{-6}) - (1.25 	imes 10^{-6}) = 2.398 	imes 10^{-6} \ S \ cm^{-1}$Solubility in $mol \ L^{-1}$ is given by:$	ext{Solubility} = \frac{\kappa 	imes 1000}{\lambda^{\circ}_{m}} = \frac{2.398 	imes 10^{-6} 	imes 1000}{246.6} \approx 9.724 	imes 10^{-6} \ mol \ L^{-1}$Molar mass of $BaSO_4 = 137 + 32 + (4 	imes 16) = 233 \ g \ mol^{-1}$Solubility in $g \ L^{-1} = 9.724 	imes 10^{-6} \ mol \ L^{-1} 	imes 233 \ g \ mol^{-1} \approx 2.266 	imes 10^{-3} \ g \ L^{-1}$

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