Question

Calculate the number of electrons in \[100\] grams of \[C{O_2}\] .
Can a body have a charge of (a) \[0.32 \times {10^{ - 18}}C\] (b) \[0.64 \times {10^{ - 20}}C\] (c) \[4.8 \times {10^{ - 21}}C\] ?

Answer 1

Hint: For the first part, first calculate the number of moles of \[C{O_2}\] by the equation \[n = \dfrac{{Mass}}{{Molar{\text{ }}mass}}\] . Then the number of molecules of \[C{O_2}\] will be equal to \[6.022 \times {10^{23}} \times n\] and the total electrons will be \[6.022 \times {10^{23}} \times n \times 22\] . For the second part, the charge on a body is only possible when the charge is an integral multiple of \[1.6 \times {10^{ - 19}}C\] .

Complete answer:
Let’s first calculate the number of electrons in \[100\] grams of \[C{O_2}\] .
We know the equation for moles for a given amount of substance is
\[n = \dfrac{{Mass}}{{Molar{\text{ }}mass}}\]
\[n = \] The number of moles of that substance
\[Mass = \] Weight of the given substance in mass
\[Molar{\text{ }}mass = \] The molar mass of the given substance
The molar mass of a \[C{O_2}\] molecule \[ = \] Molar mass of 1 \[C\] atom \[ + \] Molar mass of 2 \[O\] atoms
The molar mass of a \[C{O_2}\] molecule \[ = 12 + 2 \times \left( {16} \right) = 12 + 32 = 44\]
So, the number of moles of \[C{O_2}\] molecule is
\[n = \dfrac{{100}}{{44}} = 2.28\]
We know that 1 mole of \[C{O_2}\] molecules is equal to \[6.023 \times {10^{23}}\] molecules of \[C{O_2}\] . So, \[2.28\] moles of \[C{O_2}\] molecule contains
\[2.28 \times 6.023 \times {10^{23}} = 1.37 \times {10^{24}}\] molecules of \[C{O_2}\]
Now, the total number of electrons in a \[C{O_2}\] molecule \[ = \] Number of electrons in \[C\] atom \[ + \] Twice the number of electrons in \[O\] atom
Total number of electrons in a \[C{O_2}\] molecule \[ = 6 + 2 \times \left( 8 \right) = 22\] electrons
So, the number of electrons in \[1.37 \times {10^{24}}\] molecules of \[C{O_2}\] is
\[1.37 \times {10^{24}} \times 22 = 3.02 \times {10^{25}}\] electrons

For the second part, we have to explain whether a body has a charge of
(a) \[0.32 \times {10^{ - 18}}C\] (b) \[0.64 \times {10^{ - 20}}C\] (c) \[4.8 \times {10^{ - 21}}C\]

When we say a body is charged, it means that electrons from one body have been transferred from one object to the other. When a body donates one electron to the other body, the donor body gets a charge of \[ + 1.6 \times {10^{ - 19}}C\] and the receiving body gets a charge of \[ - 1.6 \times {10^{ - 19}}C\] . So, the charge on the body will always be an integral multiple of \[ \pm 1.6 \times {10^{ - 19}}\] .
a.)Since, \[\dfrac{{0.32 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}} = \dfrac{{3.2 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 2\]
Since the charge is an integral multiple of the charge of an electron, so a body can have a charge of \[0.32 \times {10^{ - 18}}C\]

b.)Since, \[\dfrac{{0.64 \times {{10}^{ - 20}}}}{{1.6 \times {{10}^{ - 19}}}} = \dfrac{{0.64 \times {{10}^{ - 20}}}}{{16.0 \times {{10}^{ - 20}}}} = 0.665\]
Since the charge is not an integral multiple of the charge of an electron, so a body cannot have a charge of \[0.64 \times {10^{ - 20}}C\]

c.)Since, \[\dfrac{{4.8 \times {{10}^{ - 21}}}}{{1.6 \times {{10}^{ - 19}}}} = 0.03\]
Since the charge is not an integral multiple of the charge of an electron, so a body cannot have a charge of \[4.8 \times {10^{ - 21}}C\] .

Note:
In the second part of the solution, we discussed that charge on a body can only be an integral multiple of \[ \pm 1.6 \times {10^{ - 19}}\] , i.e. Charge \[ = \pm n \times 1.6 \times {10^{ - 19}}\] , where \[n = 1,2,3,...\] . This property is called the quantization of charge and is a fundamental property of all charged bodies.