Q47 / 197 · NEET 2009 (Mains) · Chemistry

$\mathrm{Al}_2 \mathrm{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^4$ amperes of current is passed through molten $\mathrm{Al}_2 \mathrm{O}_3$ for 6 hours, what mass of aluminium is produced ? (Assume 100\% current efficiency, at. mass of $\mathrm{Al}=27 \mathrm{~g} \mathrm{~mol}^{-1}$ ) -

  • A
    $1.3 \times 10^4 \mathrm{~g}$
  • B
    $9.0 \times 10^3 \mathrm{~g}$
  • Correct answer
    C
    $8.1 \times 10^4 \mathrm{~g}$
  • D
    $2.4 \times 10^5 \mathrm{~g}$

Solution

$$ \begin{aligned} \mathrm{W} & =\frac{\mathrm{E}}{96500} \times \mathrm{I} \times \mathrm{t} \\ \mathrm{W} & =\frac{9}{96500} \times 4.0 \times 10^4 \times 6 \times 3600 \\ & =8.1 \times 10^4 \mathrm{~g} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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