If N_A is Avogadro’s number,then the number o | vedclass

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(A) The molar mass of nitrogen $(N)$ is $14 \ g/mol$. Since the mass of electrons is negligible,the molar mass of nitride ions $(N^{3 -})$ is also $14

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(A) The molar mass of nitrogen $(N)$ is $14 \ g/mol$. Since the mass of electrons is negligible,the molar mass of nitride ions $(N^{3 -})$ is also $14 \ g/mol$.One mole of $N^{3 -}$ ions contains $1 \ mol$ of $N$ atoms. The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$,so it has $5$ valence electrons. The $N^{3 -}$ ion gains $3$ electrons,resulting in $5 + 3 = 8$ valence electrons per ion.Therefore,$1 \ mol$ $(14 \ g)$ of $N^{3 -}$ ions contains $8 \ N_A$ valence electrons.For $4.2 \ g$ of $N^{3 -}$ ions,the number of valence electrons is:$\frac{8 \ N_A 	imes 4.2 \ g}{14 \ g} = 2.4 \ N_A$.

If $N_A$ is Avogadro’s number,then the number of valence electrons in $4.2 \ g$ of nitride ions $(N^{3 -})$ is ................... $N_A$.

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