Question

Using X-ray diffraction, it is found that nickel (mass $=59$ g $mo{l}^{-1}$) crystallizes as ccp. The edge length of the unit cell is $3.5\stackrel{o}{A}$. If density of Ni crystal is $9.0$ g$/c{m}^3$, then value of Avogadro's number from the data is:

• A. $6.05\times {10}^{23}$
• B. $6.11\times {10}^{23}$
• C. $6.02\times {10}^{23}$
• D. $6.023\times {10}^{23}$

Answer 1

Answer: (B)

$6.11\times {10}^{23}$

For 1 mol of atoms in a lattice, density is given as:

$\rho =\frac{Z\times M}{a^3\times N_A}$

where Z=number of atoms in one unit cell, M=molar mass of atom, a=edge of unit cell, $N_A$= avogadro's number

given that $\rho =9g/cc,a=3.5A^{\circ }=3.5\times {10}^{-8}cm,M=59g/mol$

for ccp $Z=\frac{1}{2}\times 6+\frac{1}{8}\times 8=3+1=4$

upon substitution we get:

$9=\frac{4\times 59}{(3.5\times {10}^{-8}{)}^3\times N_A}$

or $N_A=\frac{4\times 59}{(3.5\times {10}^{-8}{)}^3\times 9}$

$N_A=6.11\times {10}^{23}mo{l}^{-1}$

the value of Avogadro's number from the data is $6.11\times {10}^{23}$