Question

In a solid substance edge length of unit cell is 405 pm. density of solid 10.9 gram/cm^3 and molar mass of the 109 gram then find the radius of atom which form the unit cell in pm​

Answer 1

We know that,

Density-

d=

a

3

×NA

Z×M

Given :

Molar Mass(M)=108g/mol

Density (d)=10.5g/cm

3

Edge Length(a)=409pm

Z=

M

d×a

3

×N

A

Value of N

A

is N

A

=6.023×10

23

Z=

108

10.5×(409×10

−10

cm

3

)

3

×6.023×10

23

=4

Number of atoms =4

Hence the element is packed in FCC structure.

Answer 2

Answer:

$\boxed{\mathfrak{Radius \ of \ atom = 143.2 \ pm}}$

Given:

Length of unit cell (l) = 405 pm = $\sf 405 \times 10^{-10} \ cm$

Density ($\sf \rho$) = $\sf 10.9$ g/cm³

Molar mass (M) = 109 g

To Find:

Radius of atom (r)

Explanation:

To find the radius of atom first we need to know type of unit cell.

$\star$ Density of unit cell:

$\boxed{ \bold{\rho = \frac{Z \times M}{N_0 \times a^3}}}$

Where:

Z $\rightarrow$ Number of atoms present in one unit cell

M $\rightarrow$ Molar mass

$\sf N_0 \rightarrow$ Avogadro's number

a $\rightarrow$ Length of unit cell

By knowing the value of Z we can find the type of unit cell.

Substituting value of $\sf \rho$, M, $\sf N_0$ & a in the equation we get:

$\sf \implies 10.9 = \dfrac{Z \times 109}{6.02 \times {10}^{23} \times {(405 \times {10}^{ - 10} )}^{3} } \\ \\ \sf \implies 10.9 = \dfrac{Z \times 109}{6.02 \times {10}^{23} \times 66430125 \times {10}^{ - 30} } \\ \\ \sf \implies 10.9 = \dfrac{Z \times 109}{39,99,09,352.5 \times {10}^{ - 7} } \\ \\ \sf \implies 10.9 = \dfrac{Z \times 2.72 \times \cancel{ {10}^{ - 7} }}{ \cancel{{10}^{ - 7} } } \\ \\ \sf \implies Z = \dfrac{10.9}{2.72} \\ \\ \sf \implies Z = 4$

So, type of unit cell is ccp or fcc.

For a ccp or fcc unit cell relation between length of side of the unit cell with radius:

$\boxed{ \bold{r = \frac{ \sqrt{2} a}{4} }}$

Substituting value of a in the equation:

$\sf \implies r = \frac{ \sqrt{2} \times 405}{4} \\ \\ \sf \implies r = 143.2 \: pm$

$\therefore$

Radius of atom (r) = 143.2 pm