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A coin is made up of Al and weighs 0.75g. It has a square shape and its diagonal measures 17mm. It is electrically neutral and contains equal amounts of positive and negative charges. The magnitude of these charges is: (Atomic mass of Al = 26.98g)
(A) \[3.47 \times {10^4}C\]
(B) \[3.47 \times {10^2}C\]
(C) \[1.67 \times {10^{20}}C\]
(D) \[1.67 \times {10^{22}}C\]

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Last updated date: 22nd Feb 2024
Total views: 18.3k
Views today: 1.18k
IVSAT 2024
Answer
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Hint: Using Avogadro’s number, we can calculate the number of aluminium atoms in a single coin. The number of electrons in any element is equal to its atomic number. And the charge on a single electron(e) is 1.60 \[ \times {10^{ - 19}}C\].

Formula used: \[Number{\text{ }}of{\text{ }}atoms{\text{ }} = {\text{ }}\dfrac{{Avogadro{\text{ }}number \times given{\text{ }}mass}}{{Atomic{\text{ }}mass}}\] and
magnitude of charges = ZNe

Complete step-by-step solution:
Given that, aluminium is electrically neutral and has equal amount of positive and negative charges. As the atomic number of aluminium is 13, there are 13 protons and 13 electrons(Z) present in one aluminium atom.
We know that Avogadro’s number = 6.022\[ \times {10^{23}}\]atoms, given mass of one coin is 0.75g and atomic mass of aluminium is 26.98g, number of aluminium atoms in a coin can be calculated by the following formula-
\[Number{\text{ }}of{\text{ }}atoms{\text{ }} = {\text{ }}\dfrac{{Avogadro{\text{ }}number \times given{\text{ }}mass}}{{Atomic{\text{ }}mass}}\]
Putting the values in it, we get
\[N = \dfrac{{6.022 \times {{10}^{23}}}}{{26.98}} \times 0.75 = 1.67 \times {10^{22}}\]
\[\therefore \]number of aluminium atoms in one coin are \[1.67 \times {10^{22}}\].
Knowing the value of charge on an electron, we can calculate magnitude of positive and negative charges in one coin by using below formula-
magnitude of charges = ZNe
where Z is the number of electrons
N is number of aluminium atoms in a coin
And e is the charge on an electron
Putting the respective values in it, we get-
Magnitude of charges = \[13 \times 1.67 \times {10^{22}} \times 1.60 \times {10^{ - 19}}\]
= \[3.47 \times {10^4}C\]or 34.7kC

Hence, the correct option is (A).

Note: The calculated magnitude of these charges tells us that the high amount of charge is due to the presence of a large amount of positive and negative charges present in a neutral atom. ‘It has a square shape and its diagonal measures 17mm.’ This line in question is given just to distract you. It has nothing to do with the solution.