Question

Calculate the moles, molecules, and total number of atoms in $1.1 \times {10^{ - 4}}$ kg of carbon dioxide ( $C{O_2}$ )
(Given $C = 12$ , $O = 16$ , $N = 14$ , $H = 1,He = 4$ )

Answer 1

Hint: The mole is represented by Avogadro’s number, which is \[6.022 \times {10^{23}}\] atoms or molecules per mol.
-To convert from moles to molecules, multiply the molar amount by Avogadro’s number.
-To convert from molecules to atoms, multiply the number of molecules by the atomicity of the atom.

Complete answer:
Given ,
Mass of \[C{O_2} = 1.1 \times {10^{ - 4}}Kg\]
Molar mass of \[C{O_2} = 12 + 32\]
\[ = 44g/mol\]
$ = 44 \times {10^{ - 3}}kg/mol$
(i) Number of moles (n) \[ = \dfrac{given\,\,mass\,\,of\,\,C{O_2}}{molar\,\,mass\,\,of\,\,C{O_2}}\] ​​
\[ = 1.1 \times {10^{ - 4}}Kg/44 \times {10^{ - 4}}Kg\]
\[n = 2.5 \times {10^{ - 3}}mol\]
(ii) Number of molecules of ​ $C{O_2}\, = $ n $ \times $ Avogadro's number
\[ = 2.5 \times {10^{ - 3}} \times 6.022 \times {10^{23}}\]
\[ = 1.5 \times {10^{21}}\] molecules
(iii) Number of atoms present $ = $ no. of molecules $ \times $ atomicity
\[ = 1.5 \times {10^{21}} \times 3\]
\[ = 4.5 \times {10^{21}}\] atoms
\[C{O_2}\] is triatomic, hence multiplied by $3$ .

Note:
Avogadro’s number is typically dimensionless, but when it defines the mole, it can be expressed as \[6.022 \times {10^{23}}\] elementary entities/mol. This form shows the role of Avogadro’s number as a conversion factor between the number of entities and the number of moles. Therefore, given the relationship 1 mol $ = $ \[6.022 \times {10^{23}}\] atoms, converting between moles and atoms of a substance becomes a simple dimensional analysis problem.