Question

The number of moles of electrons present in 42 g of N3- ion are

Answer 1

no.of moles of N=W/M
=>n=42/14
=>n=3
now,
total no. of N3- electrons=10
hence,
no. of moles of electrons=n*avogadro's number*10
=>n'=3*6.022*10^23*10
=>n'=18.066*10^24
hope u got me

Answer 2

Answer: Electrons present in 42 g of azide ion are $1.32\times 10^{25}$.

Explanation:

Mass of azide ions given = 42 g

Moles of azide ion $N_{3}^-=\frac{\text{mass of azide ion}}{\text{Molar mass of azide ion}}=\frac{42 g}{42 g/mol}= 1mol$

1 mole = $=N_A=6.022\times 10^{23} molecules$

Number azide ions molecule in 1 mole = $6.022\times 10^{23} molecules$

Numbers of electron in 1 molecule azide ion = 22 electrons.

So. number of electrons in $6.022\times 10^{23}$ molecules of azide ions:

$=22\times 6.022\times 10^{23}=1.32\times 10^{25} electrons$

Electrons[/tex] are present in 1 mole of azide ion are $1.32\times 10^{25}$.