4.5 g of aluminium (atomic mass 27 amu) is | vedclass

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(C) According to Faraday's law of electrolysis,the number of equivalents of substances deposited or produced by the same quantity of charge is equal.E

Vedclass · Accepted Answer

$5.6$

Vedclass · Answer

(C) According to Faraday's law of electrolysis,the number of equivalents of substances deposited or produced by the same quantity of charge is equal.Equivalents of $Al = 	ext{Equivalents of } H_2$$	ext{Equivalents of } Al = \frac{	ext{mass}}{	ext{equivalent mass}} = \frac{4.5}{27/3} = \frac{4.5}{9} = 0.5$Since the number of equivalents of $H_2$ is also $0.5$,we use the relation: $	ext{Equivalents of } H_2 = 	ext{moles of } H_2 	imes n	ext{-factor}$.For the reaction $2H^+ + 2e^- 	o H_2$,the $n$-factor is $2$.$0.5 = n_{H_2} 	imes 2 \implies n_{H_2} = 0.25 	ext{ moles}$.At $STP$,the volume of $1 	ext{ mole}$ of gas is $22.4 \ L$.$V_{H_2} = 0.25 	imes 22.4 = 5.6 \ L$.

$4.5 \ g$ of aluminium (atomic mass $27 \ amu$) is deposited at the cathode from an $Al^{3+}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at $STP$ from $H^{+}$ ions in the solution by the same quantity of electric charge will be .............. $L$.

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