Question

How many moles are there in $9.033 \times {10^{24}}$ atoms of Helium $\left( {He} \right)$ $?$

Answer 1

One mole of a substance is equal to $6.022 \times {10^{23}}$ units of that substance (such as atoms, molecules, or ions). The number $6.022 \times {10^{23}}$ is known as Avogadro's number or Avogadro's constant. i.e., One mole $=$ $6.022 \times {10^{23}}$ atoms, molecules, protons, etc.
To convert from moles to atoms, multiply the molar amount by Avogadro's number. To convert from atoms to moles, divide the atom amount (number of atoms of a substance) by Avogadro's number i.e., Number of moles $= \dfrac{{number\;of\;atoms}}{{6.022 \times {{10}^{23}}}}$ .

Complete answer:
Helium $\left( {He} \right)$ is a chemical element with the atomic number $2$ . It is a colourless, odourless, tasteless, non-toxic, inert, monatomic gas, the first element in the noble gas group in the periodic table. Its boiling point is the lowest among all the elements.
Number of atoms in $He$ $= 9.033 \times {10^{24}}$ atoms
Number of moles in $6.022 \times {10^{23}}$ atoms $= 1$ mole.
Since we know that, Number of moles is the ratio of the number of atoms to the Avogadro number.
Hence, number of moles in $9.033 \times {10^{24}}$ atoms of Helium $\left( {He} \right)$ $= \dfrac{{number\;of\;atoms\;inHe}}{{6.022 \times {{10}^{23}}}}$
$= \dfrac{{9.033 \times {{10}^{24}}}}{{6.022 \times {{10}^{23}}}}$
$= \dfrac{{9.033 \times {{10}^{24 - 23}}}}{{6.022}}$
$= \dfrac{{9.033 \times 10}}{{6.022}}$
$= \dfrac{{90.33}}{{6.022}}$
$= \dfrac{{90330}}{{6022}} = 15$ .
Hence the number of moles in $9.033 \times {10^{24}}$ atoms of Helium $\left( {He} \right)$ is $15$ .

Note:
Note that Avogadro number is used to measure the products obtained from the chemical reaction. The unit is denoted by $mol$ .
The formula for the number of moles from grams is expressed as $Number\;of\;Moles = \dfrac{{Mass\;of\;substance}}{{Mass\;of\;one\;mole}}$
Each element has a different molar mass depending on the weight of $6.022 \times {10^{23}}$ of its atoms. Example atoms in $12$ grams are the same as $12$ grams of carbon. Also note that, Helium is the second lightest and second most abundant element in the periodic table (Hydrogen is the lightest and most abundant element).