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 If the equivalent conductivity of  \mathrm{H}_2 \mathrm{SO}_4  at infinite is  381 \Omega^{-1} \mathrm{~cm}^2 \mathrm{eq}^{-1}  If  49 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4  per litre is present in solution and specific resistance is 15.58 \Omega  then what will be degree of dissociation

Option: 1

12%


Option: 2

3%


Option: 3

16%


Option: 4

8%


Answers (1)

best_answer

Equivalence of    \mathrm{H}_2 \mathrm{SO}_4=\frac{49}{49}=1 \mathrm{~N}

\mathrm{\text { Specific conductance } =\frac{1}{\text { specific resistance }} }
                                                \mathrm{ =\frac{1}{18.4} }
                                      \mathrm{ \lambda_{\text {eq }} =\frac{1000 \times \mathrm{K}}{\mathrm{N}} }
                                    \mathrm{\lambda \text {eq } =\frac{1000 \times 1}{15.58}}

                                    \mathrm{ \lambda_{\text {eq }}=64.1 }
\mathrm{ \text { Degree of dissociation }(\alpha)=\frac{\lambda_{\text {eq }}^0}{\lambda^{\infty }_{\text {eq }}} }
                                         \mathrm{ \alpha=\frac{64}{381} \\ }
                                     \mathrm{ \alpha \%=16 \%}.
 

Posted by

Devendra Khairwa

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