Question

silver crystalisses in fcc lattice if edge of the cell is 4.08×10^-8cm and density is 10.5g cm^-3 calculate atom mass of the solid​

Answer 1

Answer:

Explanation:

Given, edge length (a) =4.07×10−8

Density (d) = 10.5 g cm-3

For fcc number of atoms per unit cell (z) = 4

We know that Atomic Mass (M) =da3NAz

Thus, M = 10.5gcm3×4.07×10−8×6.033×1023mol−14

Or, M = 10.5g×67.419×10−24×6.022×1023mol−14

Or, M = 10.5g×67.419×10−1×6.022mol−14

Or, M =426.29707894gmol−1

Or, M=106.574gmol−1=107gmol−1

Question – 1.12 - A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Answer: Given, Atoms of Q are at the corners of the cube and P at the body-centre.

So, number of atoms Q in one unit cell =8×18=1

Number of atoms of P in one unit cell = 1

So, ratio of P and Q atoms = P : Q = 1 : 1

So, formula of given compound = PQ

Since it is bcc

Hence, coordination number of P and Q = 9

Question – 1.13 - Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.

Answer: Given , density (d) = 8.55 g cm-3

Atomic Mass (M) = 93 u = 93 g mol-1

Atomic radius (r) = ?

We know, Avogadro Number NA = 6.022 ×

1023mol-1

Since, given lattice is bcc

Hence, number of atoms per unit cell (z) = 2

We know that, d=zMa3NA

Or, 8.55 g cm-3 = 2×93gmol−1a3×6.022×1023mol−1

Or, a3=2×93g8.55gcm−3×6.022×1023

Or, a3=18651.4881×1023cm3

Or, a3=3.6124×10−23cm3

Or, a3=36.124×10−24cm3

Or, a=3.3057×10−8cm

Thus atomic radius of niobium = 14.31 nm