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(C) The molar mass of $N^{3-}$ is approximately $14 \, g/mol$ (as electrons have negligible mass).Number of moles of $N^{3-} = \frac{	ext{mass}}{	ex

Vedclass · Accepted Answer

$1.44 	imes 10^{24}$

Vedclass · Answer

(C) The molar mass of $N^{3-}$ is approximately $14 \, g/mol$ (as electrons have negligible mass).Number of moles of $N^{3-} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{4.2 \, g}{14 \, g/mol} = 0.3 \, mol$.One nitrogen atom $(N)$ has $5$ valence electrons. An $N^{3-}$ ion has $5 + 3 = 8$ valence electrons.Total number of valence electrons = (Number of moles) $	imes$ (Avogadro's number) $	imes$ (Valence electrons per ion).Total valence electrons = $0.3 	imes 6.022 	imes 10^{23} 	imes 8$.Total valence electrons = $1.8066 	imes 10^{23} 	imes 8 = 1.44528 	imes 10^{24} \approx 1.44 	imes 10^{24}$.

Calculate the number of valence electrons in $4.2 \, g$ of $N^{3-}$ ions.

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