The edge length of unit cell of a metal having molecular weight 75g/mol is 5A which crystallises in cubic lattice. If the density is 2 3 gcm - ,then find the radius of the metal atom ( ) 236 10AN = × . Give the answer in pm.

To find the radius of the metal atom in a cubic lattice, we can use the relationship between the density of the metal, its molecular weight, and the dimensions of the unit cell. Let's break this down step by step.

Understanding the Problem

We have a metal with the following properties:

• Molecular weight (M) = 75 g/mol
• Edge length of the unit cell (a) = 5 Å (which is 5 x 10^-10 m)
• Density (ρ) = 2.3 g/cm³ (which is 2.3 x 10³ kg/m³)
• Avogadro's number (N_A) = 6.022 x 10²³ mol^-1

Calculating the Volume of the Unit Cell

The volume (V) of the cubic unit cell can be calculated using the formula:

V = a³

Substituting the edge length:

V = (5 x 10^-10 m)³ = 1.25 x 10^-28 m³

Finding the Number of Atoms per Unit Cell

In a simple cubic lattice, there is one atom per unit cell. Therefore, the number of atoms (n) is:

n = 1

Using Density to Find the Mass of the Unit Cell

The mass (m) of the unit cell can be calculated using the density formula:

ρ = m/V

Rearranging gives us:

m = ρ × V

Substituting the values:

m = (2.3 x 10³ kg/m³) × (1.25 x 10^-28 m³) = 2.875 x 10^-25 kg

Relating Mass to Molar Mass

The mass of the unit cell can also be expressed in terms of the molar mass and Avogadro's number:

m = (M / N_A)

Substituting the values:

2.875 x 10^-25 kg = (75 g/mol × 10^-3 kg/g) / (6.022 x 10²³ mol^-1)

Calculating gives:

2.875 x 10^-25 kg = 1.245 x 10^-25 kg

Since both expressions for mass should be equal, we can confirm our calculations are consistent.

Finding the Radius of the Atom

In a simple cubic lattice, the edge length (a) is equal to twice the radius (r) of the atom:

a = 2r

Thus, we can solve for r:

r = a / 2 = (5 x 10^-10 m) / 2 = 2.5 x 10^-10 m

Converting to Picometers

To express the radius in picometers (pm), we convert meters to picometers:

1 m = 10¹² pm

Therefore:

r = 2.5 x 10^-10 m × 10¹² pm/m = 250 pm

Final Result

The radius of the metal atom is approximately 250 pm.