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  • The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochroic acid are 280, 860 and 426 S cm2 mol-1 respectively. The molar conductivity at infinite dilution of barium sulphate is _______

The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochroic acid are 280, 860 and 426 S cm2 mol-1 respectively. The molar conductivity at infinite dilution of barium sulphate is _______ S cm2 mol-1. (Roundede off to the nearest integer)
 

Answers (1)

best_answer

Given,

\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{BaCl}_{2}\right)=280

\begin{aligned} &\lambda_{\mathrm{M}}^{\infty}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)=860 \\ &\lambda_{\mathrm{M}}^{\infty}(\mathrm{HCl})=426 \end{aligned}

Find \lambda_{M}^{\infty}\left(\mathrm{BaSO}_{4}\right)=? ?

We know ,

From Kohlrausch's law

\Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{BaSO}_{4}\right)=\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{Ba}^{2+}\right)+\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{SO}_{4}^{2-}\right)

\Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{BaSO}_{4}\right)=\Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{BaCl}_{2}\right)+ \Lambda_{\mathrm{m}}^{\infty}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) -2 \Lambda_{\mathrm{m}}^{\infty}(\mathrm{HCl})

\lambda_{M}^{\infty}\left(\mathrm{BaSO}_{4}\right) = 280 + 860 -2 (426)

\lambda_{M}^{\infty}\left(\mathrm{BaSO}_{4}\right) =288 \ \mathrm{Scm}^{2} \mathrm{~mol}^{-1}

Ans = 288

Posted by

Kuldeep Maurya

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