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

Question

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

Tardigrade Admin · Accepted Answer

d =( Z × M / N A × a 3) ⇒ Z =( a 3 × d × N A / M ) Z=((3.5 × 10-8)3 × 0.53 × 6.023 × 1023/6.94) = 1.97002 ≈ 2

Q. How many lithium atoms are present in a cubic 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

$d = \frac{Z \times M}{N _{A} \times a ^{3}}$

$\Rightarrow Z = \frac{a ^{3} \times d \times N _{A}}{M}$

$Z = \frac{( 3.5 \times 1 0 ^{- 8} ) ^{3} \times 0.53 \times 6.023 \times 1 0 ^{23}}{6.94}$

$= 1.97002 \approx 2$

Q. How many lithium atoms are present in a cubic unit cell with edge length 3.5A˚ 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​ =1.97002≈2

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

$d = \frac{Z \times M}{N _{A} \times a ^{3}}$

$\Rightarrow Z = \frac{a ^{3} \times d \times N _{A}}{M}$

$Z = \frac{( 3.5 \times 1 0 ^{- 8} ) ^{3} \times 0.53 \times 6.023 \times 1 0 ^{23}}{6.94}$

$= 1.97002 \approx 2$