Question

The mineral manganosite, manganese(ll) oxide, crystallizes in the rock salt structure the face-centered structure adopted by NaCl) with a density of 5.365 g/cm'. Find the unit cell edge length of manganosite.
A. 444.5 pm
B. 352.8 pm
C. 280.0 pm
D. 368.2 pm
E. 417.9 pm

Answer 1

Answer:

A. 444.5 pm

Explanation:

We know that:

$Density = \dfrac{mass \ of \ atoms \ in \ unit \ cell}{total \ volume \ of \ unit \ cell}$

i.e.

$\rho = \dfrac{n*M}{v_c * N_A}$

$\rho = \dfrac{n*M}{a^3 * N_A}$

in a face-centered cubic crystal, the number of atoms per unit cell is (n) = 4

The molar mass of manganese (II) oxide $[Mn(11)O] = 70.93 \ g/mol$

Density $\rho$ is given as 5.365 g/cm³

Avogadro constant $N_A$ = 6.023 × 10²³ atoms/mol

∴

$\rho = \dfrac{n*M}{a^3 * N_A}$

Making th edge length "a" the subject, we get:

$a^3 = \dfrac{n*M}{\rho* N_A}$

$a^3 = \dfrac{4*70.93 \ g/mol}{5.365 \ g/cm^3 *6.023 * 10^{23} \ atoms/mol }$

$a^3= 8.78 \times 10^{-23} \ cm^3$

$a= \sqrt[3]{8.78 \times 10^{-23} \ cm^3}$

a = 4.445 × 10⁻⁸ cm

a = 444.5 pm

Answer 2

The unit cell edge length of manganosite is equal to: A. 444.5 pm

Given the following data:

• Density of NaCl = 5.365 $g/cm^3$

We know that the molar mass of manganese (ll) oxide is equal to 70.93 g/mol.

Avogadro constant = $6.02 \times 10^{23}$

Since the rock salt is face-centered cubic crystal, the number of atoms per unit cell, n = 4

To find the unit cell edge length of manganosite:

For a crystal structure, density is given by the formula:

$Density = \frac{Mass\; of\;atoms\;in\;a\;unit\;cell}{Total\;volume\;of\;a\;unit\;cell}$

$\rho = \frac{nM}{N_Aa^3}$

Where:

• n is the number of atoms per unit cell.

• $N_A$ is Avogadro constant.

• a is the edge length.

• M is the mass.

Making "a" the subject of formula, we have:

$a=\sqrt[3]{\frac{nM}{\rho N_A} }$

Substituting the parameters into the formula, we have;

$a=\sqrt[3]{\frac{4 \times 70.93}{5.365 \times 6.02 \times 10^{23} }}\\\\a=\sqrt[3]{\frac{283.72}{3.23 \times 10^{24} }}\\\\a=\sqrt[3]{8.79 \times 10^{-23} }}\\\\a = 4.445 \times 10^{-8}\; meters$

Unit cell edge length, a = 444.5 pm

Read more: https://brainly.com/question/18320053