Question

The no. of electrons in $2{\text{gm}}$ ion of nitrate ion ($N{O_3}^ - $ ) is:

Answer 1

Hint:This question, we have to calculate the moles of $N{O_3}^ - $. Simply use the unitary method to find the answer to this question. \[6.022 \times {10^{23}}\]is Avogadro's number and represents the number of atoms/molecules present in one mole of the substance. The mole concept will be used to find the number of moles present from which we shall calculate the number of electrons in one ion of $N{O_3}^ - $ . Then, we can calculate the total number of electrons present.
 Formula used:
 ${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
The total number of electrons in a molecule of nitrate ion is = $7 + 3 \times 8 + 1 = 32$${e^ - }$
Molar mass of nitrate ion is = $62{\text{g}}$.
Total no. of moles in $2{\text{gm}}$ ion of nitrate ion = $\dfrac{2}{{62}}$.
As we know that 1 mole of any compound contains \[6.022 \times {10^{23}}\] atoms. Thus:
Total no. of molecules in $2{\text{gm}}$ ion of nitrate ion = $\dfrac{2}{{62}} \times 6.02 \times {10^{23}}$.
Total no. of electrons in $2{\text{gm}}$ ion of nitrate ion = $\dfrac{2}{{62}} \times 6.02 \times {10^{23}} \times 32{e^ - }$
So, the total number of electrons in 2g of nitrate ion is: $ = 6.2 \times {10^{23}}{e^ - }$

Note:
The concept that a mole of any substance contains the same number of particles was formed out of research which was conducted by Italian physicist Amedeo Avogadro. Avogadro constant can be defined as the number of molecules, atoms, or ions in one mole of a substance: $6.022 \times {10^{23}}$ per mol. It is derived from the number of atoms of the pure isotope $^{{\text{12}}}{\text{C}}$ in 12 grams of that substance and is the reciprocal of atomic mass in grams. The formula for mole concept can be summarized as:
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}$.