Question

The charge on 1-gram ions of $A{l^{ + 3}}$ is: ( ${N_A} = $ Avogadro number, e=charge on one electron)
A. $\dfrac{1}{{27}} \times {N_A} \times e$ coulomb
B. $\dfrac{1}{3} \times {N_A} \times e$ coulomb
C. $\dfrac{1}{9} \times {N_A} \times e$ coulomb
D. $3 \times {N_A} \times e$ coulomb

Answer 1

Hint: Here we find the charge on $A{l^{ + 3}}$ by given charge on ions. Charge on one molecule of electron is found by the Avogadro number and calculate the charge on one mole of $A{l^{ + 3}}$ ion. The gram of atomic weight of an element is called one-gram ion which is equal to the atomic weight in gram is called as gram atom.

Complete step by step answer:
Aluminum is an aluminum cation which contains a charge of +3. It is an aluminum cation, a monatomic traction and a monatomic aluminum.
We find the charge of one gram of ion of $A{l^{ + 3}}$ , as we know that 1 gm of atom is equal to 1 mole of atom. So, charge on one mole of electrons is equal to charge on one electron and Avogadro number.
1 mole of electron
 $ = {N_A} \times $ charge on electron
 $
   = 6.023 \times {10^{22}} \times 1.6 \times {10^{ - 19}}C \\
   = 9.6368 \times {10^4} \\
 $
Therefore, charge on one mole of
 $A{l^{ + 3}} = 3 \times 96368$
= 289104 C (or $3{N_A}{e^ - }$ )
Since, 27 g of $A{l^{ + 3}}$ ion having charge of 289104 C.
Thus, 1 g of having $A{l^{ + 3}}$ ion having charge \[ = \dfrac{{289104}}{{27}}\] C.
= 10707.5 C. (or $\dfrac{{{N_a}{e^ - }}}{9}$ )

Hence, option (C) is the correct answer.

Note: The 1 g of having $A{l^{ + 3}}$ ion having charge which is defined as the charge of aluminum. Here we remember that 1 mole of electrons is equal to Avogadro number and charge on electron. The charge of an aluminum ion is 13 protons. Aluminum is a soft metal in the boron group on the periodic table of elements.