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# The number of electrons present in 1$c{m^3}$ of Mg (Density of${}_{12}^{24}Mg$ is 1.74 gm/$c{m^3}$) is :A.0$\times {10^{24}}$B.46$\times {10^{23}}$C.3$\times {10^{20}}$D.24$\times {10^{23}}$  Verified
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Hint:
The number of atoms in one mole of any substance is equal to Avogadro’s Number i.e. $6.023 \times {10^{23}}$ atoms per mole.

From the given data, we know that the atomic mass of Mg is 24 and the atomic number is 12. We also know that, no. of electrons in any given atom can be calculated by:
Number of electrons = Atomic Number of elements
From this relation, we can understand that there are 12 electrons in one atom of Mg.
Since it is given that the atomic mass of Mg is 24 gm, then we can say that one mole of Mg weighs 24gm.
In the given question, we are being asked the number of electrons present in 1$c{m^3}$ of Mg and the density of Mg has been given as $1.74gm/c{m^3}$
So, let’s break this problem into steps and solve:

Hence, as the problem demands data regarding $1c{m^3}$ of Mg, we consider the eight of the sample to be 1.74 gm.

${\text{24 gm of Mg }} \to {\text{ 1 mol}}$
$1.74gmofMg \to 'x'mol$
Hence, x = $\dfrac{{1.74 \times 1}}{{24}}$
x = 0.0725mol
As discussed earlier, we know that the number of electrons in an Mg atom is 12. And also, that the number of atoms in 1 mole of Mg is $6.023 \times {10^{23}}$ atoms.
Hence, from this, the number of electrons in one mole of Mg is $12 \times 6.023 \times {10^{23}}$ electrons.
Therefore, the number of electrons in $1c{m^3}$ of Mg can be calculated by the relation:

number of electrons in $1c{m^3}$ of Mg = $x \times 12 \times 6.023 \times {10^{23}}$
= $0.0725 \times 12 \times 6.023 \times {10^{23}}$
= $5.24 \times {10^{23}}$
Hence, The number of electrons present in 1$c{m^3}$ of Mg is $5.24 \times {10^{23}}$ electrons

Hence, Option D is the correct option.

Note:
Mole is the unit of measurement for the amount of substance in the SI units. The particles present in the substance can be molecules, ions, atoms, or electrons.