
A solid having density of $9 \times {10^3}kg{m^{ - 3}}$ forms face-centered cubic crystals of edge length $200\sqrt 2 pm$ . What is the molar mass of the solid? [Avogadro constant $ \cong 6 \times {10^{23}}mo{l^{ - 1}},\pi \cong 3$]
A. $0.0216kgmo{l^{ - 1}}$
B. $0.0305kgmo{l^{ - 1}}$
C. $0.4320kgmo{l^{ - 1}}$
D. $0.0432kgmo{l^{ - 1}}$
Answer
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Hint: A face-centered cubic crystal is a crystal structure having atoms at each corner of the cubic, and an atom at each of the six face centers of the cube. Metals that possess face-centered structures include copper, aluminum, silver, gold, etc. The total number of the atoms in a face-centered cubic structure is four.
Complete answer: or Complete step by step answer:
The density of a unit cell is defined as the ratio of the mass and the volume of the unit cell. The mass of a unit cell is equal to the product of the number of atoms in a unit cell and the mass of each atom.
Let us assume, the number of atoms in a unit cell is z and the mass of each atom is m, therefore, the mass of the unit cell would be $z \times m$.
Let the mass of 1 mole of the element be M that means, the mass of ${N_A}$ atoms is M. now using the unitary method the mass of 1 atom will be $\dfrac{M}{{{N_A}}}$.
Now, if we consider that the side of the cell is equal to a, its volume will be, $V = {a^3}$.
Since density is the ratio of the mass of the unit cell to the volume of the unit cell, we have,
$Density = \dfrac{m}{V} = \dfrac{{z \times m}}{{{a^3}}} = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}$
Putting all the values provided is the question, we have,
$9 \times {10^3} = \dfrac{{4 \times M}}{{{{(200\sqrt 2 \times {{10}^{ - 12}})}^3} \times 6.022 \times {{10}^{23}}}}$
Which gives, M $ = 0.0305kgmo{l^{ - 1}}$.
So, the correct answer is Option B.
Note:
A unit cell is defined as the smallest group of atoms having the overall symmetry of a crystal and from which the entire lattice can be built up by the repetition of the unit cell in a three-dimensional space is termed as the unit cell. Apart from the face-centered cubic unit cell, we have many more types as well. For instance, body-centered cubic unit cells, etc.
Complete answer: or Complete step by step answer:
The density of a unit cell is defined as the ratio of the mass and the volume of the unit cell. The mass of a unit cell is equal to the product of the number of atoms in a unit cell and the mass of each atom.
Let us assume, the number of atoms in a unit cell is z and the mass of each atom is m, therefore, the mass of the unit cell would be $z \times m$.
Let the mass of 1 mole of the element be M that means, the mass of ${N_A}$ atoms is M. now using the unitary method the mass of 1 atom will be $\dfrac{M}{{{N_A}}}$.
Now, if we consider that the side of the cell is equal to a, its volume will be, $V = {a^3}$.
Since density is the ratio of the mass of the unit cell to the volume of the unit cell, we have,
$Density = \dfrac{m}{V} = \dfrac{{z \times m}}{{{a^3}}} = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}$
Putting all the values provided is the question, we have,
$9 \times {10^3} = \dfrac{{4 \times M}}{{{{(200\sqrt 2 \times {{10}^{ - 12}})}^3} \times 6.022 \times {{10}^{23}}}}$
Which gives, M $ = 0.0305kgmo{l^{ - 1}}$.
So, the correct answer is Option B.
Note:
A unit cell is defined as the smallest group of atoms having the overall symmetry of a crystal and from which the entire lattice can be built up by the repetition of the unit cell in a three-dimensional space is termed as the unit cell. Apart from the face-centered cubic unit cell, we have many more types as well. For instance, body-centered cubic unit cells, etc.
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