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# Sodium crystallizes in the bcc unit cell. Calculate the approximate number of unit cells in $9.2g$ of sodium.

Last updated date: 18th Jul 2024
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Hint: Number of atoms in a bcc unit is two. We can calculate the number of moles of sodium and multiply it by the number of atoms in a mole of an atom and when we further divide it by two, we get the number of unit cells.

It is given in the question that the sodium atom crystallises in a body centred cubic structure and if $9.2g$ of sodium is given then the number of unit cells present in that amount of sodium.
Now we are given that the sodium is in a body centred cubic structure, now what happens in such a structure is that the atoms are at the edges of the cube unit cell and also at the centre of the unit cell. Now the effective number of atoms in a unit cell of a body centered cubic structure is two.
We are given $9.2g$ of sodium and we already know that the atomic mass of sodium is $23u$.
Now to calculate moles of any element we need to divide the given mass by the molecular or the atomic mass of the element.
$moles = \dfrac{{given\,mass}}{{molecular\,weight}}$
$\Rightarrow moles = \dfrac{{9.2}}{{2.3}}$
$\Rightarrow moles = 0.4$
On dividing the given mass by molecular mass, we get $0.4$ moles of sodium.
Now one mole of an atom has $6.022 \times {10^{23}}$ number of atoms.
Number of atoms in $0.4$ moles of sodium is
$moles \times 6.022 \times {10^{23}}$
$= 0.4 \times 6.022 \times {10^{23}}$
$= 2.408 \times {10^{23}}$
Now $0.4$ moles of sodium have $2.408 \times {10^{23}}$ number of particles or atoms.
As we already know that in bcc unit cell of sodium, they have two atoms each, then if there are $2.408 \times {10^{23}}$ of atoms, then the number of unit cells would be the division of the number of particles by two or number of atoms in a unit cell
$\dfrac{{2.408 \times {{10}^{23}}}}{2} \\ = 1.204 \times {10^{23}} \\$
Now we know that there are $1.204 \times {10^{23}}$ unit cells in $9.2g$ of sodium.

Note:
The number of effective atoms in a face centred cubic is four because of the Presence of atoms on each face of the unit cell therefore increasing the number of effective atoms. While in simple cubic structure the number of effective atoms in a unit cell is one.