
How many unit cells are there in 1.00g cube shaped ideal crystal of AB(Mw=60) which has a NaCl type lattice
A. $$6.02 \times {10^{23}}$$
B. $$1.00 \times {20^{22}}$$
C. $$2.50 \times {10^{21}}$$
D. $$6.02 \times {10^{24}}$$
Answer
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Hint: A unit cell is the smallest repeating portion of the crystal lattice that on repetition forms a unit cell. The simplest cubic unit cell consists of eight atoms in it which are present at its corners. As the unit cell has three edges a,b, and c. so these three edges are not mutually perpendicular to each other.
Complete step by step answer:
The NaCl type lattice is a rock salt type structure. It is a face centered cubic unit cell in which the atoms are present at the corners and at the faces of the lattice. So this cell contains 4 atoms. That is 1 NaCl unit cell is made up of 4 atoms. As given in the question that
The molecular weight of AB crystal is = 60g/mol
So by mole concept
60g/mol of AB will contain = $$6.023 \times {10^{23}}$$atoms
Now 1g/mol AB will contain = $$\dfrac{{6.023 \times {{10}^{23}}}}{{60}} = 0.10038 \times {10^{23}}$$ atoms
4 atoms of the crystal lattice forms = 1 unit cell so,
$$0.10038 \times {10^{23}}$$ atoms will form = $$\dfrac{{0.10038 \times {{10}^{23}}}}{4} = 0.0250 \times {10^{23}} = 2.50 \times {10^{21}}$$ unit cells.
So the unit cells present in a 1.00g cube shaped ideal crystal of AB which has a NaCl type lattice is $$2.50 \times {10^{21}}$$ unit cells.
So the correct option for this answer is option C.
Note: The face centered cubic cell or fcc has 4 atoms in it. Its coordination number is twelve. In this the atoms are present in the corner and at the faces too. Copper, silver, gold they all crystallize in face centered cubic cells.
Complete step by step answer:
The NaCl type lattice is a rock salt type structure. It is a face centered cubic unit cell in which the atoms are present at the corners and at the faces of the lattice. So this cell contains 4 atoms. That is 1 NaCl unit cell is made up of 4 atoms. As given in the question that
The molecular weight of AB crystal is = 60g/mol
So by mole concept
60g/mol of AB will contain = $$6.023 \times {10^{23}}$$atoms
Now 1g/mol AB will contain = $$\dfrac{{6.023 \times {{10}^{23}}}}{{60}} = 0.10038 \times {10^{23}}$$ atoms
4 atoms of the crystal lattice forms = 1 unit cell so,
$$0.10038 \times {10^{23}}$$ atoms will form = $$\dfrac{{0.10038 \times {{10}^{23}}}}{4} = 0.0250 \times {10^{23}} = 2.50 \times {10^{21}}$$ unit cells.
So the unit cells present in a 1.00g cube shaped ideal crystal of AB which has a NaCl type lattice is $$2.50 \times {10^{21}}$$ unit cells.
So the correct option for this answer is option C.
Note: The face centered cubic cell or fcc has 4 atoms in it. Its coordination number is twelve. In this the atoms are present in the corner and at the faces too. Copper, silver, gold they all crystallize in face centered cubic cells.
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