Question

What is the charge required to deposit $40.5$ gram of $Al$ (atomic mass $= 27\dfrac{g}{{mol}}$) from the fused $A{l_2}{(S{O_4})_3}$ ?

Answer 1

In the given question,first of all we need to see how much charge is generated when aluminium ions become aluminium metal and then the ratio of moles of electrons to the moles of aluminium metal would remain constant. Now calculate moles of aluminium in one gram of aluminium and equate it to the mole ratio of charge and aluminium metal and then calculate the charge.

Complete answer:
In the above given question, we are asked how much charge is required from aluminium ion to deposit one gram of aluminium metal, now the aluminium ion is present fused $A{l_2}{(S{O_4})_3}$, which means these ions gets diffused very easily and are now are in their ion forms.
Now first we need to analyse the reaction for the conversion of aluminium ion to aluminium metal.
$A{l^{3 + }} + 3{e^ - } \to Al$
Now from the above given equation we can say that one mole of aluminium ion takes in three moles of electrons to form one mole of aluminium metal.
Now we are given that we need to deposit $40.5$ gram of aluminium metal, now first let’s calculate the moles of aluminium deposited when $40.5$ gram of aluminium is there
$moles = \dfrac{{given\,mass}}{{molecular\,mass}}$
$\Rightarrow moles = \dfrac{{40.5}}{{27}}$

\dfrac{1}{{3 \times moles\,of\,Al}} \\\ \dfrac{1}{{40.5}} \\\ $$.Now from the reaction $A{l^{3 + }} + 3{e^ - } \to Al$ , we can say that the moles of aluminium metal formed are thrice of the moles of electrons consumed, therefore if $\dfrac{{40.5}}{{27}}$ moles of aluminium are formed then thrice of $\dfrac{{40.5}}{{27}}$ moles of charge are consumed, that is $3 \times moles\,of\,Al = 3 \times \dfrac{{40.5}}{{27}} = \dfrac{{40.5}}{9}$ Now we need to calculate the charge of $\dfrac{{40.5}}{9}$ moles of electrons. Now each electron has a charge of $1.6 \times {10^{ - 19}}$ coulomb and there are $6.022 \times {10^{23}}$ units of electrons therefore $\dfrac{{40.5}}{9}$ moles of electron will have the charge equivalent to $charge = \dfrac{{40.5}}{9} \times 6.022 \times {10^{23}} \times 1.6 \times {10^{ - 19}}$ $ \Rightarrow charge = 4.34 \times {10^5}C$ **Therefore, the charge required to deposit $40.5$ gram of $Al$ metal from aluminium ion is $4.34 \times {10^5}C$ coulomb.** **Note:** Aluminium is a silvery white and a lightweight metal which is used for various purposes like it is used in power lines, high rise buildings, window frames, consumer electronics, household and industrial appliances, aircraft components, ships, etc.