Question

A body centred cubic element of density 10.3g/cm*3 has a cell edge of 314Pm . Calculate the atomic mass of the element

Answer 1

Answer:

96.03 gram

Explanation:

Here, D = 10.3

a= 314

N = 6.023×10²³

d= Z m/a³ × na

m = d.a³ ×na/Z

m= 10.3×(3.14×10-¹⁰)³ × 6.023×10²³

m = 96.03 gram

Answer 2

Answer:

The body-centered cubic element's atomic mass, M, measured is $96u$.

Explanation:

Given,

The body-centered cubic ( bcc ) element's density, d = $10.3g/cm^{3}$

The edge length of the element, a = $314pm$ = $3.14\times 10^{-8}cm$

The atomic mass of the element, M =?

As we know,

• The atomic mass of an element can be calculated by the density formula given below:
• d = $\frac{Z\times M}{N_{A}\times a^{3} }$
• M = $\frac{d \times N_{A} \times a^{3} }{Z}$     -------equation (1)

Here,

• $N_{A}$ = Avogadro number = $6.022\times 10^{23}$
• Z = The number of atoms per unit cell

And, for bcc, Z = $2$

After putting the values in the equation (1), we get:

• M = $\frac{10.3 \times 6.022\times 10^{23} \times (3.14\times 10^{-8}) ^{3} }{2}$
• M = $\frac{10.3 \times 6.022\times 10^{23} \times 30.95\times 10^{-24} }{2}$
• M = $\frac{1919.72\times 10^{-1}}{2}$

• M = $95.98u$

Hence, the atomic mass of the element, M = $95.98u$ ≈ $96u$.