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4.5 g of aluminium (Atomic mass 27 amu) is deposited at cathode from $\mathrm{Al}^{3+}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $\mathrm{H}^{+}$ions in solution by the same quantity of electric charge will be
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5.6 L
From second law of Faraday
$\frac{m_{\mathrm{Al}}}{m_{\mathrm{H}}}=\frac{E_{\mathrm{Al}}}{E_{\mathrm{H}}}$
$\therefore \frac{4.5}{m_{\mathrm{H}}}=\frac{27 / 3}{1}$
or $m_{\mathrm{H}}=0.5 \mathrm{~g}$
$\because 2 \mathrm{~g} \mathrm{H}_2$ volume at $\mathrm{STP}=22.4 \mathrm{~L}$
$\therefore 0.5 \mathrm{~g} \mathrm{H}_2$ volume at STP
$=\frac{22.4 \times 0.5}{2} \mathrm{~L}=5.6 \mathrm{~L}$
$\frac{m_{\mathrm{Al}}}{m_{\mathrm{H}}}=\frac{E_{\mathrm{Al}}}{E_{\mathrm{H}}}$
$\therefore \frac{4.5}{m_{\mathrm{H}}}=\frac{27 / 3}{1}$
or $m_{\mathrm{H}}=0.5 \mathrm{~g}$
$\because 2 \mathrm{~g} \mathrm{H}_2$ volume at $\mathrm{STP}=22.4 \mathrm{~L}$
$\therefore 0.5 \mathrm{~g} \mathrm{H}_2$ volume at STP
$=\frac{22.4 \times 0.5}{2} \mathrm{~L}=5.6 \mathrm{~L}$
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