Question

Which one of the following will have the largest number of atoms?
A. 1g $Au$
B. 1g $Na$
C. 1g $Li$
D. 1g $C{{l}_{2}}$

Answer 1

Hint: An atom is the smallest unit of matter which forms a chemical element. Matter is anything which occupies space and has definite mass. Matter exists in three states which are known as states of matter; these are known as solid, liquid and gas.

Complete answer:
Every atom is composed of a nucleus and this nucleus is made up of protons and neutrons where protons are positively charged and neutrons are neutral in charge and electrons which are positively charged are revolving around the nucleus.
To find out which will have the largest number of atoms we have to calculate the number of atoms for every given element. Number of atoms can be calculated step by step:
1. First calculate the molar mass of the element.
2. Then calculate the number of moles.
3. Now convert moles of atoms using Avogadro’s number.
A. 1g$Au$: Given mass = 1g and molar mass or atomic mass of silver = 197
1g $Au$= $\dfrac{1}{197}$ mol atom of $Au$
\[\dfrac{1}{197}\times 6.022\times {{10}^{23}}\]Atoms of$Au$.
B. 1g $Na$= \[\dfrac{1}{23}\]mol atom of $Na$
\[\dfrac{1}{23}\times 6.022\times {{10}^{23}}\]Atoms of $Na$
C. 1g $Li$= \[\dfrac{1}{7}\]mol atom of $Li$
\[\dfrac{1}{7}\times 6.022\times {{10}^{23}}\] Atoms of $Li$
D. 1g $C{{l}_{2}}$= \[\dfrac{1}{71}\]mol atom of $C{{l}_{2}}$
\[\dfrac{1}{71}\times 6.022\times {{10}^{23}}\] Molecules of $C{{l}_{2}}$
\[\dfrac{2}{71}\times 6.022\times {{10}^{23}}\] Atoms of $C{{l}_{2}}$

Hence from this we can consider that 1g $Li$ has the largest number of atoms i.e. option C is the correct answer.

Note:
Avogadro’s number basically tells us about the number of particles present in one mole of the substance. These particles can be electrons, molecules or atoms. It can also be known by the name Avogadro’s constant which has a fixed value of \[6.022\times {{10}^{23}}\].