An atomic solid crystallizes in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by 60.3 pm. If the atomic weight of A is 48 , then density of the solid, is nearly:

Question

An atomic solid crystallizes in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by 60.3 pm. If the atomic weight of A is 48 , then density of the solid, is nearly: (A) 2.7 g / cc (B) 5.07 g / cc (C) 3.5 g / cc (D) 1.75 g / cc

Tardigrade Admin · Accepted Answer

Given a-2 r=60.3 and for bcc, 4 r=√3 ⇒ a -(√3/2) a =60.3 ⇒ a =450 pm Density (ρ)=(2 × 48/6.023 × 1023 ×(4.5)3 × 10-24) =1.75 g / cc

Q. An atomic solid crystallizes in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by $60.3 p m$ . If the atomic weight of $A$ is 48 , then density of the solid, is nearly:

2.7 g / cc

5.07 g / cc

3.5 g / cc

1.75 g / cc

Given $a - 2 r = 60.3$ and for bcc, $4 r = 3$

$\Rightarrow a - \frac{3}{2} a = 60.3$

$\Rightarrow a = 450 p m$

Density $(ρ) = \frac{2 \times 48}{6.023 \times 1 0 ^{23} \times ( 4.5 ) ^{3} \times 1 0 ^{- 24}}$

$= 1.75 g / cc$

Q. An atomic solid crystallizes in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by 60.3pm. If the atomic weight of A is 48 , then density of the solid, is nearly: