Question

A compound ab has cubic structure and molecular mass 99. Its density is 3.4g/cm cube .what is the length of edge of unit cell

Answer 1

length of edge of unit cell is 3.64 A°

• density = (Z×M)/(V×N)

= (Z×M)/(a³×N)

• where Z = No. of atoms in unit cell = 1 ( for simple cubic)

• M = Atomic mass = 99 gram

• a = edge length

• N = No. of atoms in one mole

= 6.023× 10²³

• density = (Z×M)/(V×N)

= (Z×M)/(a³×N)

• 3.4 = (1×99)/( a³ ×6.023×10²³)

• a³ = 99/ (3.4 ×6.023×10²³)

• a³ = 4.8344×10^-²³

• a³ = 48.344×10^-²⁴

• a = 3.64 × 10^-8 cm

• a = 3.64 × 10^-10 m

• a = 3.64 A°

Answer 2

Length of the edge of the unit cell is $1.7 \times 10^{-8}$ cm.

Explanation:

For cubic structure $Z=1$

$M= 99 amu$

ρ $= 3.4g/cm$

$a=?$

Applying the formula of density of simple cubic structure,

ρ $=\frac{Z\times M}{N_{A}\times a^{3} }$

3.4    $=\frac{1\times 99}{6.022\times 10^{23} \times a^{3} }$

Following the Above equation,

$a^{3}$$=\frac{1\times 99}{6.022\times 10^{23} \times 3.4 }$

$a^{3}$$=4.836\times10^{-25} cm$

Therefore,

$a= 1.71\times10^{-8} cm$