Question

8. An element (atomic mass = 100 g/mol) having bcc
structure has unit cell edge 400 pm. Then density of the
element is:
(a) 10.376 g/cm3
(b) 5.188 g/cm3
(c) 7.289 g/cm3
(d) 2.144 g/cm3
guys please help me​

Answer 1

Answer:

Given, M = 100 g/mol, a = 400pm

For BCC -

n=8×

8

1

+1=2

As we know,

Density=

a

3

×N

A

n×M

=

(400×10

−10

)

3

×6.02×10

23

2×100

=5.188 g cm

−3

(∵400pm=400×10

−10

cm)

Explanation:

pla mark me as brainlist

Answer 2

12th/Chemistry

The Solid State

Answer :

Atomic mass of element = 100g/mol

Type of unit cell = BCC

Edge length of unit cell = 400pm

We have to find density of the element$.$

★ Calculation involving unit cell dimensions,

• $\boxed{\bf{\blue{d=\dfrac{Z\times M}{a^3\times N_A\times 10^{-30}}}}}$

where,

Z = no. of atoms per unit cell

• For BCC structure : Z = 2

a = edge length in pm

d = density of solid

M = molar mass

N_A = Avogadro's number

$\leadsto\sf\:d=\dfrac{Z\times M}{a^3\times N_A\times 10^{-30}}$

$\leadsto\sf\:d = \dfrac{2\times 100}{400^3\times (6.022\times 10^{23})\times 10^{-30}}$

$\leadsto\sf\:d=\dfrac{200}{38.54}$

$\leadsto\:\boxed{\bf{\blue{d=5.18\:gcm^{-3}}}}$

∴ (B) is the correct answer.