# What is the number of nitrogen-based electrons in a 4.2*g mass of nitride anion?

## $A .$ 0.30xxN_A; $B .$ 30xxN_A; $C .$ 3.0xxN_A; $D .$ 4.2xxN_A; $E .$ $\text{cannot be determined...}$

May 2, 2017

Well, nitrogen has $Z = 7$............

#### Explanation:

And if $Z = 7$, and we have ${N}^{3 -}$, we have 10 electrons per atom, i.e. $10 \cdot {e}^{-} \times {N}_{A}$ electrons per mole of nitride anion.

But we have a mass of $4.2 \cdot g$, which corresponds to a molar quantity of $\frac{4.2 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1} = 0.30 \cdot m o l$.

And thus in a $0.30 \cdot m o l$ quantity of nitride anion, there are $10 \times 0.30 \cdot m o l \times {N}_{A}$ quantity of electrons........ So we choose $\text{option c}$.

Good question, I am stealing this.