Question

The charge on one gram ion of $A{l^{3 + }}$ ion is: $\left[ {e = 1.6 \times {{10}^{ - 19}}} \right]$
A: $\dfrac{1}{{27}}{N_A} \times e{\text{ coulomb}}$
B: $\dfrac{1}{{27}}{N_A} \times e{\text{ coulomb}}$
C: $\dfrac{1}{9}{N_A} \times e{\text{ coulomb}}$
D: $3 \times {N_A} \times e{\text{ coulomb}}$

Answer 1

Hint:
One gram atom of a substance is equal to the one mole of substance. One mole of a substance contains Avogadro's number of particles $\left( {{N_A}} \right)$. Charge due to one electron is $e{\text{ coulomb}}$ and charge due to one mole of electrons is ${N_A} \times e{\text{ coulomb}}$.

Complete step by step answer:
In this question we have to find charge on one gram of aluminium ion $A{l^{3 + }}$. One gram atom of a substance is equal to the one mole of substance. One mole of a substance contains Avogadro's number of particles $\left( {{N_A}} \right)$. Charge due to one electron is $e{\text{ coulomb}}$ and charge due to one mole of electrons is ${N_A} \times e{\text{ coulomb}}$.
In the given aluminium ion, charge is ${3^ + }$, therefore one mole of given aluminium ion charge will be equivalent to three mole of electrons.
One mole of electrons$ = \left( {{N_A} \times e{\text{ coulomb}}} \right){\text{charge}}$
Three mole of electrons$ = 3 \times \left( {{N_A} \times e{\text{ coulomb}}} \right){\text{charge}}$
In this question we have to find the charge on one gram of aluminum ions. Mass of one mole of aluminium ions is $27g$ and the charge that we have calculated is on one mole of aluminium ions. Therefore,
Charge on $27g$ of aluminium ion$ = 3 \times \left( {{N_A} \times e{\text{ coulomb}}} \right)$
Charge on $1g$ of aluminium ion$ = \dfrac{3}{{27}} \times \left( {{N_A} \times e{\text{ coulomb}}} \right)$
Solving this we get,
Charge on $1g$ of aluminium ion$ = \dfrac{1}{9} \times \left( {{N_A} \times e{\text{ coulomb}}} \right)$
Therefore charge on one gram of aluminium ion is $\dfrac{1}{9}{N_A} \times e{\text{ coulomb}}$.
So, the correct answer is option C.



Note:
In this question we have to find the charge on one gram of aluminium ion this is why we divided the charge (that was on one mole of the ions) with molecular mass. If we were asked to find the charge on one gram atom of substance then it would be equal to the charge present on one mole of aluminium ions.