
A metal has fcc lattice. The edge length of the unit cell is 404pm. The density of metal is $2.72g/c{m^3}$ . The molar mass of the metal is (N=6.02\times 10 mol).
A) 30 g/mol
B) 27 g/mol
C) 20 g/mol
D) 40 g/mol
Answer
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Hint:: In order to answer this question, you must recall the Solid state chapter’s that portion in which you have calculated the different parameters for different lattices. The face-centered cubic system has lattice points on the faces of the cube, that each gives exactly one half contribution, in addition to the corner lattice points, giving a total of 4 lattice points per unit cell. You must revise all the formulae and use them correctly with correct units and by making all the mathematical equations precisely.
Complete step by step answer:
Step 1: In this step lets enlist all the given quantities:
Density, d = $2.72g/c{m^3}$
z = number of effective constituent particles in one unit cell
M = molecular weight
a = 404pm edge length of unit cell
${N_A}$ = $6.023 \times {10^{23}}$
Step 2: In this step we will use the formula for density to find M (molecular weight):
The formula for density is given by: $d\, = \,\frac{{zM}}{{{a^3}{N_A}^{}}}$
$ \Rightarrow $ $M\, = \,\frac{{d{a^3}{N_A}}}{z}$
We know that for an fcc structure, z = 4
$ \Rightarrow $ $M\, = \,\frac{{2.72 \times {{(404 \times {{10}^{ - 10}})}^3} \times 6.023 \times {{10}^{23}}}}{4}$
$ \Rightarrow $ $M\, = \,27g/mol$
So here we got the required molecular mass and with the correct units.
Hence the correct answer is option B.
Note: Face-centered cubic lattice (fcc or cubic-F), like all lattices, has lattice points at the eight corners of the unit cell plus additional points at the centers of each face of the unit cell. It has unit cell vectors a =b =c and interaxial angles α=β=γ=90°. The simplest crystal structures are those in which there is only a single atom at each lattice point. In the fcc structures the spheres fill 74 % of the volume. The number of atoms in a unit cell is four ( 8×1/8 + 6×1/2 = 4 ). There are 26 metals that have the fcc lattice.
Complete step by step answer:
Step 1: In this step lets enlist all the given quantities:
Density, d = $2.72g/c{m^3}$
z = number of effective constituent particles in one unit cell
M = molecular weight
a = 404pm edge length of unit cell
${N_A}$ = $6.023 \times {10^{23}}$
Step 2: In this step we will use the formula for density to find M (molecular weight):
The formula for density is given by: $d\, = \,\frac{{zM}}{{{a^3}{N_A}^{}}}$
$ \Rightarrow $ $M\, = \,\frac{{d{a^3}{N_A}}}{z}$
We know that for an fcc structure, z = 4
$ \Rightarrow $ $M\, = \,\frac{{2.72 \times {{(404 \times {{10}^{ - 10}})}^3} \times 6.023 \times {{10}^{23}}}}{4}$
$ \Rightarrow $ $M\, = \,27g/mol$
So here we got the required molecular mass and with the correct units.
Hence the correct answer is option B.
Note: Face-centered cubic lattice (fcc or cubic-F), like all lattices, has lattice points at the eight corners of the unit cell plus additional points at the centers of each face of the unit cell. It has unit cell vectors a =b =c and interaxial angles α=β=γ=90°. The simplest crystal structures are those in which there is only a single atom at each lattice point. In the fcc structures the spheres fill 74 % of the volume. The number of atoms in a unit cell is four ( 8×1/8 + 6×1/2 = 4 ). There are 26 metals that have the fcc lattice.
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