Question

an element crystallizes into fcc unit cell type structure. the edge length of unit cell is 150 picometre. if 150 gram of this element have 12 into 10 to the power 23 atoms then calculate the density of the element.



​

Answer 1

$\huge\bf\mathscr\pink{Your\: Answer}$

d = 148.1 g${cm}^{-3}$

step-by-step explanation:

Given,

crystal has fcc structure

therefore,

Z(number of lattice points in one unit cell) = 4

Now,

it is given that,

a = 150 pm = 150× ${10}^{-10}$ cm

w = mass of particles = 150 g

(in the given sample)

N = number of particles = 12 × ${10}^{23}$

(in the given sample)

d = density of element

Now,

we know that,

w/M = N/NA

where,

M = molar mass of the element

NA = Avagadro's Number

.°. w/N = M/NA

as,

d = Z×M / ${a}^{3}$ × NA

=> d = Z × w/ ${a}^{3}$× N

=> d = 4×150/ ${(150)}^{3}$× ${10}^{-30}$×12× ${10}^{23}$

=> d = 4/ ${(150)}^{2}$× ${10}^{-7}$×12

=> d = 4× ${10}^{7}$/ ${(150)}^{2}$×12

=> d = 148.1 g ${cm}^{-3}$

Answer 2

In fcc

z = 4

According to question we have given;

NA (Avogadro's Number) = 23 × ${10}^{23}$ atoms

a = 150 pm = 150 × ${10}^{-10}$ cm

d (density) = ?

d = $\dfrac{z \: \times \: M}{ {a}^{3} \: \times \: N}$ ...(1)

Now...

$\dfrac{w}{M}$ = $\dfrac{N}{NA}$

M = $\dfrac{w}{A}$

Here;

M = Molar Mass

N = Number of Particles

w = Mass of particle = 150 g

On putting value of M in (1) we get

d = $\dfrac{z \: \times \: w}{ {a}^{3} \: \times \: NA}$

d = $\dfrac{4 \: \times \:150}{ {(150)}^{3} \: \times \: {10}^{ - 30} \: \times \:12\:\times {10}^{23} }$

d = $\dfrac{600\: \times \: {10}^{7} }{ 3375000 \: \times \: 12 }$

d = 148.1 $g \: {cm}^{ - 3}$