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$MnO_4^ - \, + \,x{e^ - }\,\xrightarrow[{(Alkaline\,\,medium)}]{}MnO_4^{2 - }$
$MnO_4^ - \, + \,y{e^ - }\,\xrightarrow{{(Acidic\,\,medium)}}M{n^{2 + }}$
$MnO_4^ - \, + \,z{e^ - }\,\xrightarrow{{(Neutral\,\,medium)}}Mn{O_2}$
$A.\,\,F -CH_2\,\,CH_2\,\,COOH$
$\begin{array}{*{20}{c}}
{B.\,\,Cl - CH - C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
$C.\,\,F -CH_2 -COOH$
$D.\,\,Br -CH_2-CH_2 -COOH$
Correct answer is
$\lambda_{\mathrm{m}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{x} \;\mathrm{S}\; \mathrm{cm}^{2} \mathrm{mol}^{-1}$
$\lambda_{\mathrm{m}\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{y} \;\mathrm{S\;cm}^{2} \mathrm{mol}^{-1}$
$\lambda_{\mathrm{m}(\mathrm{CH_3} \mathrm{COOK})}^{0}=\mathrm{z}\; \mathrm{S\;cm}^{2} \mathrm{mol}^{-1}$
$\lambda_{\mathrm{m}}^{0}\left(\mathrm{in}\; \mathrm{S} \;\mathrm{cm}^{2} \mathrm{mol}^{-1}\right)$ for $\mathrm{CH}_{3} \mathrm{COOH}$ will be