Question

Question. 1 An element A (atomic
mass = 60) has simple cubic lattice
of edge 100pm. Find the density of
crystal​

Answer 1

Answer:99.8g/cc

Explanation:

Atomic mass= 60 ;

a = 100pm

density= ( 60* 1)/((100)^3 * 10^-30 * 6.02*10^23)

= 99.61g/cc

Answer 2

The density of the crystal is 399 grams per centimeter cube.

Given,

The atomic mass of A is 60

The edge length of the simple cubic lattice is 100 pm

To Find,

The density of the crystal, d

Solution,

Let the no. of corners in each cube be g

However, each corner atom contributes (1/8)th to each cube because it is shared with eight additional cubes.

No. of faces in a cube = 6

Due to the fact that each atom on a face center is shared by two cubes, its contribution is (1/2).

Therefore, the total number of atoms in a cube is given by:

z = 8 × 1/8 + 6 × 1/2 = 4 atoms

Now, the density is found by the following formula:

$\implies d = \frac{z \times M}{N_A \times a^3}$

where 'd' is the density, 'z' is the number of atoms per unit cell, 'M' is the molecular or atomic mass in g/mol, 'NA' is the Avogadro’s number, and ‘a’ is the edge length.

We have been given the following:

z = 4,

M = 60g/mol,

NA = 6.022 × 10^(23) mol-1

a = 100 pm = 100 × 10^(-10) = 10^(-8)

Substitute these values to get the density of the crystal

$\implies d = \frac{4 \times 60}{6.022 \times 10^{23} \times (10^{-8})^3}\\\\\implies d = \frac{240}{6.022 \times 10^{23} \times 10^{-24}}\\\\\implies d = \frac{240}{6.022 \times 10^{-1}}\\\\\implies d = 39.854 \times 10^1 \hspace{0.1cm} gcm^{-3}\\\\ \implies d = 398.54 \hspace{0.1cm} gcm^{-3}\\\\\implies d \approx 399 \hspace{0.1cm} gcm^{-3}$

The density of the crystal is 399 grams per centimeter cube.

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