If nearly ${10^5}\,{\rm{C}}$ liberate ${\rm{1}}\,\,{\rm{g}}$ equivalent of aluminium, then the amount of aluminium (equivalent weight ${\rm{g}}$) deposited through electrolysis in ${\rm{20}}\,\,{\rm{min}}$ by a current of ${\rm{50}}\,\,{\rm{A}}$ will be

$\begin{array}{*{20}{l}} {{\rm{5}}{\rm{.4}}\,\,{\rm{g}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{0}}{\rm{.09}}\,\,{\rm{g}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{0}}{\rm{.6}}\,\,{\rm{g}}} \end{array}$

$\begin{array}{*{20}{l}} {{\rm{10}}{\rm{.8}}\,\,{\rm{g}}} \end{array}$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

1 Medium

1 of 1

Application +4 | -1

If nearly

{10^5}\,\mathrm{C}

liberate

\mathrm{1}\,\,\mathrm{g}

equivalent of aluminium, then the amount of aluminium (equivalent weight

\mathrm{g}

) deposited through electrolysis in

\mathrm{20}\,\,\mathrm{min}

by a current of

\mathrm{50}\,\,\mathrm{A}

will be

View answer and explanation

Correct option: C

\begin{array}{llllllllllllllllllll} {\mathrm{5}\mathrm{.4}\,\,\mathrm{g}} \end{array}

Option C (

5.4 \,\mathrm{g}

) is correct because it is the calculated mass of aluminum deposited using Faraday's laws of electrolysis.