Question

Find number of electrons present in 34 g of $N{H_3}$ (g).
A. $2{N_A}$
B. ${N_A}$
C. $20{N_A}$
D. $10{N_A}$

Answer 1

Firstly, we will calculate the number of electrons in one mole of $N{H_3}$. For a neutral atom, the number of electrons is equal to atomic number. So, one mole of N contains 7 electrons and 3 moles of H contains 3 electrons, in this way we will calculate the number of moles of $N{H_3}$. And hence, to calculate the electrons present in 34 g of $N{H_3}$ (g), we will multiply number electrons with Avogadro’s number.

Complete step by step answer:
Given in the question is,
Weight of $N{H_3}$ (g) = 34 g
Molar mass of $N{H_3}\left( g \right) = 14 + \left( {3 \times 1} \right) = 14 + 3 = 17$
Number of moles (n) = given mass / molar mass
Therefore, Number of moles of $N{H_3}\left( g \right) = \dfrac{{34}}{{17}} = 2$
Number of electrons in one molecule of $N{H_3}$ (g) = 7 +3 = 10
Hence, number of electrons in 2 moles of $N{H_3}\left( g \right) = 2 \times 10 \times 6.022 \times {10^{23}}\; = 20{N_A}$
Since, Avogadro’s number $= 6.022 \times {10^{23}} = {N_A}$
So, $20{N_A}$ of electrons present in 34 g of $N{H_3}$ (g).

Therefore, the correct answer is option (C).

Note: The compound $N{H_3}$ is Ammonia and its other name is Azande. It is a colourless alkaline compound and is not highly flammable. It is formed during the decomposition of organic materials in the body when a large number of reactions happen. Ammonia has alkaline properties and is corrosive. $N{H_3}$ is the most important compound which is used for the growth of the plant. Ammonia gas easily dissolves in water.