The edge length of unit cell of a metal having molecular weight 75 g / mol is 5 mathringA which crystallizes in cubic lattice. If the density is 2 g / cc, then find the radius of metal atom. (N A =6 × 1023). Give the answer in pm.

Question

The edge length of unit cell of a metal having molecular weight 75 g / mol is 5 mathringA which crystallizes in cubic lattice. If the density is 2 g / cc, then find the radius of metal atom. (N A =6 × 1023). Give the answer in pm. (A) 217 pm (B) 210 pm (C) 220 pm (D) 205 pm

Tardigrade Admin · Accepted Answer

ρ=(Z M/NA V) Z=(ρ NA V/M)=(2 × 6 × 1023 ×(5 × 10-8)3/75) Z=2, which represents b c c structure ∴ r=(√3/4) a=(√3/4) × 5=2.165 mathringA=216.5 pm ≈ 217 pm

Q. The edge length of unit cell of a metal having molecular weight $75 g / m o l$ is $5 \overset{˚}{A}$ which crystallizes in cubic lattice. If the density is $2 g / cc$ , then find the radius of metal atom. $(N_{A} = 6 \times 1 0^{23})$ . Give the answer in pm.

$217 p m$

$210 p m$

$220 p m$

$205 p m$

$ρ = \frac{ZM}{N _{A} V}$
$Z = \frac{ρ N _{A} V}{M} = \frac{2 \times 6 \times 1 0 ^{23} \times ( 5 \times 1 0 ^{- 8} ) ^{3}}{75}$
$Z = 2$ , which represents $b cc$ structure
$∴ r = \frac{3}{4} a = \frac{3}{4} \times 5 = 2.165 \overset{˚}{A} = 216.5 p m \approx 217 p m$

Q. The edge length of unit cell of a metal having molecular weight 75g/mol is 5A˚ which crystallizes in cubic lattice. If the density is 2g/cc, then find the radius of metal atom. (NA​=6×1023). Give the answer in pm.

217pm

210pm

220pm

205pm