Question

An oxide of nitrogen has a molecular weight 92. Find the total number of electrons in one-gram mole of that oxide (N is Avogadro number)
(A) $ 4.6 $ N
(B) 46N
(C) 23N
(D) $ 2.3 $ N

Answer 1

Hint: In the above question, the molecular weight of the compound is given and we have to find out the number of electrons present in $ {\text{1g}} $ mole of that oxide. Since, the number of electrons depends upon the atomic number. So, first we have to find the relation between atomic number and molecular weight using the formula given. Then we can find the number of electrons in $ {\text{1g}} $ mole.
Formula Used : In the above question, we will check if $ \dfrac{{\text{A}}}{{\text{Z}}} $ ratio is equal or not where:
 $ {\text{A = atomic mass of the substance}} $
 $ {\text{Z = atomic number of the substance}} $.
Complete Step by step solution
For oxygen:
Z= atomic number =8
A=atomic mass = 16
 $ \dfrac{{\text{A}}}{{\text{Z}}}{\text{ = }}\dfrac{{{\text{16}}}}{{\text{8}}}{\text{ = 2}} $
For nitrogen:
Z= atomic number =7
A=atomic mass = 14
 $ \dfrac{{\text{A}}}{{\text{Z}}}{\text{ = }}\dfrac{{{\text{14}}}}{7}{\text{ = 2}} $
So, both nitrogen and oxygen atomic mass is double of their atomic number. Atomic number is equal to the number of electrons present.
Hence, an oxide of nitrogen having molecular weight 92 will have number of electrons present as $ \dfrac{{92}}{2} = 46 $
 $ {\text{1g}} $ of the compound = 46 electrons
And we know that 1 g mole of the compound contains $ {{\text{N}}_{\text{A}}} $ g of the compound
1 g mole of the compound contains $ {{\text{N}}_{\text{A}}} \times {\text{46 electrons}} $
1 g mole of the compound contains $ {\text{46}}{{\text{N}}_{\text{A}}}{\text{ electrons}} $
$ \therefore $ Option B is the correct answer.

Note
-Avogadro’s number is kind of generalized number, hence, we can write:
 $ {\text{1 mole}} $ = $ 6.022 \times {10^{23}} $ atoms, or molecules, or protons, or electrons etc.
-In short, it is the number of particles in a mole.
-Hence, in order to solve these types of questions, we have to first decide which quantity (electron, proton, atom) is equivalent to Avogadro’s number.