Question

Number of moles of electrons in 4.2g of a nitride ion?

Answer 1

Mass N3- ion = 14 g/mol. So, 

4.2 g X (1 mol/14 g) = 0.3 mol N3- 

Number of N3- ions = 0.3 mol X 6.02 X 10^23 ions/mol = 1.8 X 10^23 ions 

Each N3- ion has 8 valence electrons, so total number of valence electrons = 1.8 X 10^23 X 8 = 1.4 X 10^24 valence electrons

Answer 2

Answer : The moles of electrons 4.2 g of a nitride ion is, 3 moles

Solution : Given,

Mass of $N^{3-}$ = 4.2 g

Molar mass of $N^{3-}$ = 14 g/mole

First we have to calculate the moles of $N^{3-}$ ion.

$\text{ Moles of }N^{3-}=\frac{\text{ Given mass of }N^{3-}}{\text{ Molar mass of }N^{3-}}=\frac{4.2g}{14g/mole}=0.3moles$

Now we have to calculate the number of electrons on $N^{3-}$ ion.

The number of electrons in $N^{3-}$ = 7 + 3 = 10 electrons

Now we have to calculate the number of electrons in $N^{3-}$ ion,

As, 1 mole of $N^{3-}$ ion contains $10\times (6.022\times 10^{23})$ number of electrons

So, 0.3 mole of $N^{3-}$ ion contains $0.3\times 10\times (6.022\times 10^{23})=3\times (6.022\times 10^{23})$ number of electrons

Now we have to calculate the moles of electrons in $N^{3-}$ ion,

$\text{Moles of electrons of}N^{3-}=\frac{\text{Number of electrons of}N^{3-}}{\text{Avogadro's number}}$

$\text{Moles of electrons of}N^{3-}=\frac{3\times (6.022\times 10^{23})}{6.022\times 10^{23}}=3moles$

Therefore, the moles of electrons 4.2 g of a nitride ion is, 3 moles