Question

The mass of a unit cell of a body- centred cubic crystal of metal is $72.2 \times {10^{ - 23}}{\text{g}}$. The atomic mass of the metal is:
A) $128.6{\text{g mo}}{{\text{l}}^{ - 1}}$
B) $108.7{\text{g mo}}{{\text{l}}^{ - 1}}$
C) $217.3{\text{g mo}}{{\text{l}}^{ - 1}}$
D) $57.86{\text{g mo}}{{\text{l}}^{ - 1}}$

Answer 1

Hint:
To answer this question, you must recall the concept of the atomic mass unit. amu represents an atomic mass unit which is an international standard unit to weigh an element. We must also know the formula to find the number of atoms in a unit cell.

Formula:
${\text{mass of one atom}} = \dfrac{{{\text{mass of unit cell}}}}{{{\text{number of atoms in one cell}}}}$

Complete step by step solution:
Atomic mass unit represents the mass of a single atom. It is defined as one twelfth of the mass of one atom of carbon-12 isotope. We can also say that one amu is equal to the mass of one proton (mass of proton and neutron are almost similar). The atomic mass unit represents the mass of an atom.
We know that, one atomic mass unit is given by, ${\text{1a}}{\text{.m}}{\text{.u}}{\text{. = 1}}{.66 \times 1}{{\text{0}}^{{\text{ - 24}}}}{\text{g}}$.
The mass of unit cell is given in the question as $72.2 \times {10^{ - 23}}{\text{g}} = \dfrac{{72.2 \times {{10}^{ - 23}}}}{{1.66 \times {{10}^{ - 24}}}}{\text{amu}} = 434.9{\text{amu}}$
We know that in a body centred cubic lattice, the effective number of atoms is 2.
So the mass of one atom of the metal can be given by, $\dfrac{{434.9}}{2} = 217.4{\text{amu}}$
We know that the mass of a mole of atoms in grams is equal to the mass of an atom in amu. So the atomic mass of this metal can be written as $217.4{\text{g}}$

Thus, the correct answer is C.

Note:
It should be known that in a face centred cubic unit cell, atoms are present at the corners and the face centres of the cube. So, we can write that,
${n_c} = $ number of atoms present at the corners of the unit cell $ = 8$
${n_f} = $number of atoms present at the six faces of the unit cell $ = 0$
${n_i} = $ number of atoms present completely inside the unit cell $ = 1$
${n_e} = $number of atoms present at the edge centres of the unit cell $ = 0$

Thus, the total number of atoms in a body centred unit cell is
$n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4}$
Substituting the values, we get
$n = \dfrac{8}{8} + \dfrac{0}{2} + \dfrac{1}{1} + \dfrac{0}{4}{\text{ }}$
$n = 1 + 0 + 1 + 0{\text{ }}$
Thus, $n = 2$