Q23 / 160 · AP EAMCET 2012 · Chemistry

The number of moles of electrons required to deposit $36 \mathrm{~g}$ of $\mathrm{Al}$ from an aqueous solution of $\mathrm{Al}\left(\mathrm{NO}_3\right)_3$ is (At. wt. of $\mathrm{Al}=27$ )

  • Correct answer
    A
    $4$
  • B
    $2$
  • C
    $3$
  • D
    $1$

Solution

$\mathrm{Al}^{3+}+\underset{3 \mathrm{~mol}}{3 e^{-}} \longrightarrow \underset{1 \mathrm{~mol}=27 \mathrm{~g}}{\mathrm{Al}}$ $\because 27 \mathrm{~g}$ of $\mathrm{Al}$ is deposited by 3 moles of electrons $\therefore 36 \mathrm{~g} \mathrm{Al}$ will be deposited by electrons $=\frac{3}{27} \times 36=4 \mathrm{~mol}$

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

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