
Given, $Na$ crystallizes B.C.C. Find out the density of the crystal if edge length of the unit cell is $2\overset{\circ }{\mathop{\text{A}}}\,$ and consider Avogadro’s number $6\text{ x 1}{{\text{0}}^{23}}$.
Answer
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Hint: The density of the solid can be calculated by the formula $d=\dfrac{Z\text{ x M}}{{{N}_{A}}\text{ x }{{\text{a}}^{3}}}$, where d is the density, Z is the number of atoms, M is the molecular mass of the atom, a is the edge length, and ${{N}_{A}}$ is the Avogadro’s number.
Complete answer:
We can calculate the density of the solid by using the formula$d=\dfrac{Z\text{ x M}}{{{N}_{A}}\text{ x }{{\text{a}}^{3}}}$, where d is the density, Z is the number of atoms, M is the molecular mass of the atom, a is the edge length, and ${{N}_{A}}$ is the Avogadro’s number.
So in this question we have ${{N}_{A}}$ is the Avogadro’s number and it is equal to $6\text{ x 1}{{\text{0}}^{23}}$. The question says that the $Na$crystallizes B.C.C., which means that the sodium metal crystallizes at body-centered cubic crystal so there are two atoms in the unit cell ( one at the corner of the unit cell and one at the center of the unit cell). Therefore, Z will be 2.
We are given the edge length of the unit cell as $2\overset{\circ }{\mathop{\text{A}}}\,$and this value has to be converted into meters. So $\overset{\circ }{\mathop{\text{A}}}\,$ is the angstrom and it is equal to ${{10}^{-10}}$ meters. So the edge length will be $2\text{ x 1}{{\text{0}}^{-10}}m$. We know the molecular mass of the sodium metal is 23 g / mol so the value of M will be 23.
Now putting all the values in the formula of density, we get:
$d=\dfrac{\text{2 x 23}}{\text{6 x 1}{{\text{0}}^{23}}\text{ x (2}{{\text{)}}^{3}}\,\text{x 1}{{\text{0}}^{-30}}}$
$d=9.5\text{ x 1}{{\text{0}}^{6}}\text{ Kg / }{{\text{m}}^{3}}$
So the density of the sodium is $9.5\text{ x 1}{{\text{0}}^{6}}\text{ Kg / }{{\text{m}}^{3}}$.
Note:
If the number of atoms is not given the question then it calculated with the structure of the crystal, if the structure is simple then the number of atoms will be 1, if the structure is body-centered cubic then the number of atoms will be 2, if the structure is face-centered cubic then the number of atoms will be 4.
Complete answer:
We can calculate the density of the solid by using the formula$d=\dfrac{Z\text{ x M}}{{{N}_{A}}\text{ x }{{\text{a}}^{3}}}$, where d is the density, Z is the number of atoms, M is the molecular mass of the atom, a is the edge length, and ${{N}_{A}}$ is the Avogadro’s number.
So in this question we have ${{N}_{A}}$ is the Avogadro’s number and it is equal to $6\text{ x 1}{{\text{0}}^{23}}$. The question says that the $Na$crystallizes B.C.C., which means that the sodium metal crystallizes at body-centered cubic crystal so there are two atoms in the unit cell ( one at the corner of the unit cell and one at the center of the unit cell). Therefore, Z will be 2.
We are given the edge length of the unit cell as $2\overset{\circ }{\mathop{\text{A}}}\,$and this value has to be converted into meters. So $\overset{\circ }{\mathop{\text{A}}}\,$ is the angstrom and it is equal to ${{10}^{-10}}$ meters. So the edge length will be $2\text{ x 1}{{\text{0}}^{-10}}m$. We know the molecular mass of the sodium metal is 23 g / mol so the value of M will be 23.
Now putting all the values in the formula of density, we get:
$d=\dfrac{\text{2 x 23}}{\text{6 x 1}{{\text{0}}^{23}}\text{ x (2}{{\text{)}}^{3}}\,\text{x 1}{{\text{0}}^{-30}}}$
$d=9.5\text{ x 1}{{\text{0}}^{6}}\text{ Kg / }{{\text{m}}^{3}}$
So the density of the sodium is $9.5\text{ x 1}{{\text{0}}^{6}}\text{ Kg / }{{\text{m}}^{3}}$.
Note:
If the number of atoms is not given the question then it calculated with the structure of the crystal, if the structure is simple then the number of atoms will be 1, if the structure is body-centered cubic then the number of atoms will be 2, if the structure is face-centered cubic then the number of atoms will be 4.
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