How many lithium atoms are present in a unit cell with edge length 3.5 mathringA and density 0.53 g cm-3 ? (Atomic mass of Li = 6.94)

Question

How many lithium atoms are present in a unit cell with edge length 3.5 mathringA and density 0.53 g cm-3 ? (Atomic mass of Li = 6.94) (A) 2 (B) 1 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

d=(Z× M/NA× a3) Z=(a3× d× NA/M) Z=((3.5×10-8)3×0.53×6.023×1023/6.94) =1.97≈2

Q. How many lithium atoms are present in a unit cell with edge length $3.5$ $\overset{˚}{A}$ and density $0.53 g c m^{- 3}$ ?
(Atomic mass of $L i = 6.94$ )

$2$

$1$

$4$

$6$

$d = \frac{Z \times M}{N _{A} \times a ^{3}}$
$Z = \frac{a ^{3} \times d \times N _{A}}{M}$
$Z = \frac{( 3.5 \times 1 0 ^{- 8} ) ^{3} \times 0.53 \times 6.023 \times 1 0 ^{23}}{6.94}$
$= 1.97 \approx 2$

Q. How many lithium atoms are present in a unit cell with edge length 3.5 A˚ and density 0.53gcm−3 ? (Atomic mass of Li=6.94)

2

1

4

6

d=NA​×a3Z×M​ Z=Ma3×d×NA​​ Z=6.94(3.5×10−8)3×0.53×6.023×1023​ =1.97≈2

Q. How many lithium atoms are present in a unit cell with edge length $3.5$ $\overset{˚}{A}$ and density $0.53 g c m^{- 3}$ ?
(Atomic mass of $L i = 6.94$ )

$2$

$1$

$4$

$6$

$d = \frac{Z \times M}{N _{A} \times a ^{3}}$
$Z = \frac{a ^{3} \times d \times N _{A}}{M}$
$Z = \frac{( 3.5 \times 1 0 ^{- 8} ) ^{3} \times 0.53 \times 6.023 \times 1 0 ^{23}}{6.94}$
$= 1.97 \approx 2$