The Solid State - Result Question 15
####15. A metal has an $f c c$ lattice. The edge length of the unit cell is $404 pm$. The density of the metal is $2.72 g cm^{-3}$. The molar mass of the metal is : $(N_A.$, Avogadro’s constant $.=6.02 \times 10^{23} mol^{-1})$
(a) $30 g mol^{-1}$
(b) $27 g mol^{-1}$
(c) $20 g mol^{-1}$
(d) $40 g mol^{-1}$
[NEET 2013]
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Solution:
- (b) Density is given by
$d=\frac{Z \times M}{N_A a^{3}}$; where $Z=$ number of formula units present in unit cell, which is 4 for $f c c$
$a=$ edge length of unit cell. $M=$ Molecular mass
$2.72=\frac{4 \times M}{6.02 \times 10^{23} \times(404 \times 10^{-10})^{3}}$
$(\because 1 pm=10^{-10} cm)$
$M=\frac{2.72 \times 6.02 \times(404)^{3}}{4 \times 10^{7}}=26.99=27 g mol^{-1}$
