Density of lithium atom is 0-53 g- cm3- The edge length of Li is 3-5-A- What is the number of lithium atoms in a unit cell- Atomic mass of lithium is 6-94

a=3.5 A , ρ= 0.53g/cm^3 , M_A=6.94 , Z=? ρ=Z× M_AN_A× a^3 0.53= Z× 6.946.023× 10^23× 42.875× 10^ 24 Z=2 Therefore, 2 at ...

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Standard XII

Chemistry

Density

Density of li...

Question

Density of lithium atom is $0.53 g / c m^{3}$ . The edge length of $L i$ is $3.5 \circ A$ . What is the number of lithium atoms in a unit cell? (Atomic mass of lithium is $6.94$ )

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Solution

$a = 3.5 ˚ A$ , $ρ = 0.53 g / c m^{3}$ , $M_{A} = 6.94$ , $Z = ?$
$ρ = \frac{Z \times M_{A}}{N_{A} \times a^{3}}$
$0.53 = \frac{Z \times 6.94}{6.023 \times 10^{23} \times 42.875 \times 10^{- 24}}$
$Z = 2$

Therefore, $2$ atoms of lithium are present in an unit cell.

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Lithium has a BCC structure. Its density is 530 kg $m^{- 3}$ and its atomic mass is 6.94 g $m o l^{- 1}$ . Calculate the edge length of a unit cell of Lithium metal $(N_{A} = 6.02 \times 10^{23} m o l^{- 1})$

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Density of lithium atom is 0.53 g/cm3. The edge length of Li is 3.5∘A. What is the number of lithium atoms in a unit cell? (Atomic mass of lithium is 6.94)

a=3.5˚A, ρ=0.53g/cm3, MA=6.94, Z=?ρ=Z×MANA×a30.53=Z×6.946.023×1023×42.875×10−24Z=2Therefore, 2 atoms of lithium are present in an unit cell.

Density of lithium atom is $0.53 g / c m^{3}$ . The edge length of $L i$ is $3.5 \circ A$ . What is the number of lithium atoms in a unit cell? (Atomic mass of lithium is $6.94$ )

$a = 3.5 ˚ A$ , $ρ = 0.53 g / c m^{3}$ , $M_{A} = 6.94$ , $Z = ?$
$ρ = \frac{Z \times M_{A}}{N_{A} \times a^{3}}$
$0.53 = \frac{Z \times 6.94}{6.023 \times 10^{23} \times 42.875 \times 10^{- 24}}$
$Z = 2$

Therefore, $2$ atoms of lithium are present in an unit cell.