Question

By X-ray diffraction it is found that nickel (at mass $=59gmo{l}^{-1}$), crystallized with $ccp$. The edge length of the unit cell is $3.52\mathring{A}$. If density of $Ni$ crystal is $8.94g/c{m}^3$. Then value of Avogadro's number from the date is:

• A. $5.05\times {10}^{23}$
• B. $4.11\times {10}^{23}$
• C. $3.02\times {10}^{23}$
• D. $6.023\times {10}^{23}$

Answer 1

The correct option is D $6.023\times {10}^{23}$

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

$M_{Ni}=59g/mol$

$Z_{ccp}=6\times \frac{1}{2}+1=4$

$a=3.52\mathring{A}$

$\rho =8.94$ $g/c{m}^3$

$N_A=?$

$8.94=\frac{4\times 59}{N_A\times (3.52{)}^3\times {10}^{-24}}$

$N_A=6.0\times {10}^{23}$