The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric ac

Question

The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280,860 , and $426 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. The molar conductivity at infinite dilution of barium sulphate is ______ $S \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. (Round off to the Nearest Integer).\

Professor PC Thomas Classes · Answer

(288) :\ $\Lambda_{m_{\mathrm{BaCl}_{2}}}^{\circ}=280 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$\ $\Lambda_{m_{\mathrm{H}_{2} \mathrm{SO}_{4}}}^{\circ}=860 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$\ $\Lambda_{m_{\mathrm{HCl}}}^{\circ}=426 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$\ $\Lambda_{m_{\mathrm{BaSO}_{4}}}^{\circ}=$ ?\ Add equations (i) \& (ii) and subtract twice of equation (iii),\ $\Lambda_{\mathrm{BaSO}_{4}}^{\circ}=\Lambda_{\mathrm{Ba}^{2+}}^{\circ}+\Lambda_{\mathrm{SO}_{4}^{2-}}^{\circ}$\ $=\Lambda_{m_{\mathrm{BaCl}_{2}}}^{\circ}+\Lambda_{m_{\mathrm{H}_{2} \mathrm{SO}_{4}}^{\circ}}^{\circ}-2 \Lambda_{\mathrm{HCl}}^{\circ}$\ $=280+860-(426 	imes 2)=288 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$