How many unit cells are present in the cube shaped ideal crystal of $\text{NaCl}$ mass $1.00$ gm?
Answer
602.7k+ views
Hint: The unit cell of a crystal structure can be defined as the building block of a crystal structure which on repetition in the three dimensions would result in the formation of the crystal lattice.
Complete step by step answer:
The crystal lattice exists in different unit cells such as, the primitive unit cell, the body-centred unit cell, the face-centred unit cell, and the end-centred unit cells.
So for the face centred one, as there are 6 atoms on each face of the unit cell that is shared by two other unit cells, we get 3 atoms from there and 1 atom from the corners which makes a total of 4 atoms in the face-centred cubic cell.
Now, the molecular mass of $\text{NaCl}$ = $23+35.5$= $58.5$ grams = $6.023\times {{10}^{23}}$ molecules of sodium chloride.
Therefore, 1 gram of $\text{NaCl}$ = $\dfrac{6.023}{58.5}\times {{10}^{23}}$ molecules = $1.02\times {{10}^{22}}$ molecules of sodium chloride.
Now, as each unit cell of a face-centred cubic unit cell contains 4 molecules of sodium chloride, therefore the number of unit cells present in $1.00$ gm of sodium chloride
= $\dfrac{1.02\times {{10}^{22}}}{4}$ = $2.57\times {{10}^{21}}$ unit cells of sodium chloride.
Note:
Each unit cell contains a certain number of constituent particles. For example, in the primitive unit the corners of each cell is shared by eight different atoms and hence each corner gets $\dfrac{1}{8}$ of the atom. As there are 8 corners so the total unit cell gets 1 atom in total.
Complete step by step answer:
The crystal lattice exists in different unit cells such as, the primitive unit cell, the body-centred unit cell, the face-centred unit cell, and the end-centred unit cells.
So for the face centred one, as there are 6 atoms on each face of the unit cell that is shared by two other unit cells, we get 3 atoms from there and 1 atom from the corners which makes a total of 4 atoms in the face-centred cubic cell.
Now, the molecular mass of $\text{NaCl}$ = $23+35.5$= $58.5$ grams = $6.023\times {{10}^{23}}$ molecules of sodium chloride.
Therefore, 1 gram of $\text{NaCl}$ = $\dfrac{6.023}{58.5}\times {{10}^{23}}$ molecules = $1.02\times {{10}^{22}}$ molecules of sodium chloride.
Now, as each unit cell of a face-centred cubic unit cell contains 4 molecules of sodium chloride, therefore the number of unit cells present in $1.00$ gm of sodium chloride
= $\dfrac{1.02\times {{10}^{22}}}{4}$ = $2.57\times {{10}^{21}}$ unit cells of sodium chloride.
Note:
Each unit cell contains a certain number of constituent particles. For example, in the primitive unit the corners of each cell is shared by eight different atoms and hence each corner gets $\dfrac{1}{8}$ of the atom. As there are 8 corners so the total unit cell gets 1 atom in total.
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