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If at infinite dilution the equivalent  conductance of  \mathrm{Al}^{+3} and  \mathrm{\mathrm{SO}_4^{-2}}  ion are  \mathrm{175 \Omega ^{-1}\mathrm{cm}^{-1}eq^{-1}}  and  \mathrm{155 \Omega^{-1} \mathrm{cm}^2 \mathrm{eq}^{-1}} Calculate the equivalent & molar conductivity at infinite dilution of  \mathrm{\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3}.

Option: 1

\text { equivalent }=216 ; \text { molar }=316 \\


Option: 2

\text { equivalent }=816 ; \text { molar }=334 \\


Option: 3

\text { equivalent }=316 ; \text { molar }=816 \\


Option: 4

\text { equivalent }=428; \text { molar }=224\\


Answers (1)

best_answer

\lambda_{\text {eq }}^{\infty}\left[\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\right] =\frac{1}{3} \lambda_{\mathrm{Al}^{+3}}^{\infty}+\frac{1}{2} \lambda^{\infty} \mathrm{SO}_4^{-2} \\
\mathrm{=58.3+77.5 }

\mathrm{=135.8 \Omega^{-1} \mathrm{~cm}^2 \mathrm{eq}^{-1} \\ }

\mathrm{\approx 136 \Omega^{-1} \mathrm{~cm}^2 \mathrm{eq}^{-1} }

Molar Conductivity:- \mathrm{\lambda eq V.F }

\mathrm{= 136 \times 6 \\ }

\mathrm{= 816 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} }.
 

Posted by

Kuldeep Maurya

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