Question

The charge on 1 gram of \[A{l^{3 + }}\] ions is (e= electronic charge):
A. $\dfrac{1}{{27}}{N_A}$ $e^-$ coulomb
B. $\dfrac{1}{3}{N_A}$ $e^-$ coulomb
C. \[\dfrac{1}{9}{N_A}\] $e^-$ coulomb
D. $3{N_A}$ $e^-$ coulomb

Answer 1

Hint: The mole (symbol: mol) is the unit of measurement for amount of substance in the International System of Units (SI). A mole of a substance or a mole of particles is defined as exactly $6.022 \times {10^{23}}$ particles, which may be atoms, molecules, ions, or electrons. Precisely, in short, for particles $1mol = 6.022 \times {10^{23}}$ ions/ atoms/ molecules, etc.

Complete step by step answer:
The mole is essentially a count of particles. Usually the particles counted are chemically identical entities, individually distinct. For example, a solution may contain a certain number of dissolved molecules that are more or less independent of each other. However, in a solid the constituent particles are fixed and bound in a lattice arrangement, yet they may be separable without losing their chemical identity. Thus the solid is composed of a certain number of moles of such particles.
As we know, 1 gram atom = $1mole$ of atom
The charge on one mole of electrons = ${N_A} \times $ charge on $1{e^ - }$ = ${N_A}e^-$ coulomb
Charge on one mole of $A{l^{3 + }}$ ion = $3 \times {N_A}e$ coulomb
$27g$ of $A{l^{3 + }}$ ion consists of a charge = $3 \times {N_A}e^-$ coulomb
$1g$ of $A{l^{3 + }}$ ion consists of a charge = $\dfrac{{3 \times {N_A}e}}{{27}} = \dfrac{1}{9}{N_A}e^-$ coulomb
Thus, the correct option is C \[\dfrac{1}{9}{N_A}\] $e^-$ coulomb.

Note:
In yet other cases, such as diamond, where the entire crystal is essentially a single molecule, the mole is still used to express the number of atoms bound together, rather than a count of multiple molecules. Thus, common chemical conventions apply to the definition of the constituent particles of a substance, in other cases exact definitions may be specified.