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  •  If the molar conductivity at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 252,926 and <img alt="336~ \mathrm{~Scm}^2 \mathrm{~mol}^{-1}" src="https://entranc

 If the molar conductivity at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 252,926 and 336~ \mathrm{~Scm}^2 \mathrm{~mol}^{-1}  respectively. What will be the molar conductivity at infinite dilution of barium sulphate 

Option: 1

506 \mathrm{scm}^2 \mathrm{~mol}^{-1}


Option: 2

402 \mathrm{scm}^2 \mathrm{~mol}^{-1}


Option: 3

112 \mathrm{scm}^2 \mathrm{~mol}^{-1}


Option: 4

142 \mathrm{scm}^2 \mathrm{~mol}^{-1}


Answers (1)

best_answer

From Kohlrausch's law:
\Lambda_m^{\infty}\left(\mathrm{BaSO}_4\right)=\lambda_m^{\infty}\left(\mathrm{Ba}^{2+}\right)+\lambda_m^{\infty}\left(\mathrm{SO}_4^{2-}\right) \\
\lambda ^{\infty}_m\left(\mathrm{BaSO}_4\right)=\lambda_m^{\infty}\left(\mathrm{BaCl}_2\right)+\lambda_m^{\infty}\left(\mathrm{H}_2 \mathrm{SO}_4\right)-2 \lambda_m^{\infty}(\mathrm{HCL})
\mathrm{ \lambda_ m^{\infty} { (BaSO_4) }=252+926-2 \times 336 }
\mathrm{ \lambda _m^{\infty}(\mathrm{BaSO_4})=506 \mathrm{Scm}^2 \mathrm{~mol}^{-1} }

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rishi.raj

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