Question

The Molecular weight of an oxide of an element is \[44\] . The equivalent weight of the element is \[14\] . The atomic weight of the element is:
A.\[14\]
B.\[28\]
C.\[42\]
D.\[56\]

Answer 1

Hint: The mass of a molecule is its molecular mass. The mass of one equivalent, that is, the mass of a given substance that will combine with or displace a fixed quantity of another substance is known as equivalent weight.

Complete answer:
The mass of an atom is its atomic mass. It's a weighted average of all the element's isotopes, with each isotope's mass multiplied by its abundance. Atomic weight is also known as atomic mass, but "mass" is a more accurate term.
Consider an element with the atomic mass of \[M\] g/mol and valency of \[x\] .
As a result, the chemical formula for the element oxide is \[{M_2}{O_x}\]
And the compound's molecular weight \[ = 2M + 16x\] \[\left( {16} \right.\] is the atomic mass of oxygen, \[\left. O \right)\]
Given that the oxide has a molecular mass of \[44\] g/mol.
That is, \[2M + 16x = 44\]
After simplifying the equation, we got
\[M + 8x = 22............................\left( 1 \right)\]
Given that the equivalent weight of the element \[ = 14\]
Here we are using the formula,
\[\text{Equivalent Weight} = \dfrac{\text{Atomic Mass}}{{Valency}}\]
After substituting the value, we got
\[14 = \dfrac{M}{x}\]
That is, \[M = 14x........................\left( 2 \right)\]
From equation \[\left( 1 \right)\] and \[\left( 2 \right)\]
\[14x + 8x = 22\]
\[ \Rightarrow 22x = 22\]
\[ \Rightarrow x = 1\]
Then we are substituting the value of \[x\] in the equation \[\left( 2 \right)\] to find the atomic mass of the element.
That is, \[M = 14 \times 1\]
\[ \Rightarrow M = 14\] g/mol
Hence we found that the atomic mass of the element is \[14\] g/mol.
Nitrogen is the element with the atomic mass \[14\] g/mol.
So, the correct answer is Option A.

Note:
Remember the formula \[\text{Equivalent Weight} = \dfrac{\text{Atomic Mass}}{{Valency}}\] . Owing to binding energy mass loss, the atomic mass of atoms, ions, and atomic nuclei is significantly less than the sum of the masses of their constituent protons, neutrons, and electrons.