Question

The number of atoms present in 92 amu of Na is:
A. 92 atoms
B. $\dfrac{{{N_A}}}{{92}} \times 23$
C. $\dfrac{{{N_A}}}{{23}} \times 92$
D. 4 atoms

Answer 1

An amu represents an atomic mass unit which is an international standard unit to weigh an element. All substances are made up of atoms and atoms are the smallest constituent particles of any substance and are responsible for all their properties. The atomic mass unit represents the mass of an atom. We shall multiply the moles of sodium present with Avogadro’s number so as to get the number of atoms.

Complete step by step answer:
First of all, we need to understand the atomic mass unit and molar mass. It represents the mass of a single atom. It is defined as one twelfth of the mass of one atom of carbon-12 isotope. We can also say that one amu is equal to the mass of one proton (mass of proton and neutron are almost similar).
We know that the mass of a mole of atoms in grams is equal to the mass of an atom in amu. Thus, if the molar mass of sodium is 23, then we can say that the atomic mass of sodium is 23 amu.
So, the number of moles in 92 amu of sodium are,
$n = \dfrac{{92}}{{23}}$
The number of atoms are given by ${\text{no}}{\text{. of atoms}}\left( n \right){\text{ }} \times {\text{ }}{{\text{N}}_A}$
Hence, we get, $N = \dfrac{{{N_A}}}{{23}} \times 92$

Thus, the correct answer is C.

Note:
One atomic mass unit is given by, ${\text{1a}}{\text{.m}}{\text{.u}}{\text{. = 1}}{{.66 \times 1}}{{\text{0}}^{{\text{ - 24}}}}{\text{g}}$.
Whereas, molar mass represents the mass of one mole of atoms or $6.023 \times {10^{23}}$atoms.
We are given 92 amu of sodium in the question. The molar mass of sodium is equal to 23 grams per mole. Thus, the mass of $6.023 \times {10^{23}}$ atoms of sodium is equal to 23 grams.
So we get the mass of one sodium atom as, $\dfrac{{23}}{{6.023 \times {{10}^{23}}}} = 3.82 \times {10^{ - 23}}g$
Converting it into amu, $1{\text{ Na atom}} = \dfrac{{3.82 \times {{10}^{ - 23}}}}{{{\text{1}}{{.66 \times 1}}{{\text{0}}^{{\text{ - 24}}}}}} \approx 23{\text{amu}}$.