The density of iridium is 22.4 g/cm^3. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol
Answers
The density of iridium is 22.4 g/cm^3. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol
Explanation:
Density of iridium,
ρ = 22 ⋅ 4 g / c m 3
Atomic weight of iridium,
A I r = 192 ⋅ 2 g / m o l
For FCC crystal number of atom is,
n = 4
The Volume of unit FCC unit cell is given as,
V = 16 R 3 √ 2
The expression for density will be given as
ρ = m V
m is the mass of iridium which is given as
m = n A I r
The expression is as below:
ρ = n A I r 16 R 3 √ 2 N A
Here N ₐ
is the no. of mole of iridium
When we substitute the value in the given expression, we get:
22 ⋅ 4 g / c m 3 = 4 × 192 ⋅ 2 g / m o l /16 R 3 √ 2 × 6 ⋅ 023 × 10 2²³ m o l R = 1 ⋅ 36 × 10 − ⁸c m
the required radius will be:
1 ⋅ 36 × 10 −⁸c m
.
Answer:
136 pm
Explanation:
given:- density= 22.4 g/cm³
r=?
n(fcc)=4
m=192.2 g/mol
we know that
density= nm/a³nA
putting vallues
22.4=4×192.2÷a³×6.022×10²³
a³=4×192.2÷22.4×6.022×10²³
768.8÷134.89×10²³
a³=5.699×10⁻²³
taking cube root
a=∛5.699×10⁻²³
(5.699×10⁻²³)1/3
(56.99×10⁻¹×10⁻²³)
(56.99)¹/3×(10⁻²⁴)1/3
a=(56.99)1/3×10⁻⁸
(56.99)1/3
1/3㏒(56.99) =1/3 (1.7558)
=0.5862
antilog= 3.857
a=3.857×10⁻⁸
now
in fcc
r=a/2√2
r=3.857×10⁻⁸/2×1.41
3.857×10⁻⁸/2.82
r=1.3677×10⁻⁸cm₃
r=1.3677×10⁻⁸×10⁻²m
136.77×10⁻¹⁰m
136.77 ₓ 10⁻² ₓ 10⁻¹⁰m
136.77 ₓ 10⁻¹²m
136.77 pm