A metal has an fcc lattice. The edge length of the unit cell is 400 pm. The density of the metal is 2.8 g cm -3. The molar mass of the metal is g mol -1. (Avogadro's constant =6.00 × 1023 mol -1 )

Question

A metal has an fcc lattice. The edge length of the unit cell is 400 pm. The density of the metal is 2.8 g cm -3. The molar mass of the metal is g mol -1. (Avogadro's constant =6.00 × 1023 mol -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Metal has fcc lattice, ∴ Z=4 d =(Z × M / a 3 × N A ) ∴ 2.8=(4 × M /(4.00 × 10-8)3 × 6.00 × 1023) ∴ M =(2.8 ×(4.00)3 × 6.00 × 10-1/4) =26.88 g mol -1

Q. A metal has an fcc lattice. The edge length of the unit cell is $400 p m$ . The density of the metal is $2.8 g c m^{- 3}$ . The molar mass of the metal is ____ $g m o l^{- 1}$ .
(Avogadro's constant $= 6.00 \times 1 0^{23} m o l^{- 1}$ )

Metal has $f cc$ lattice,
$∴ Z = 4$
$d = \frac{Z \times M}{a ^{3} \times N _{A}}$
$∴ 2.8 = \frac{4 \times M}{( 4.00 \times 1 0 ^{- 8} ) ^{3} \times 6.00 \times 1 0 ^{23}}$
$∴ M = \frac{2.8 \times ( 4.00 ) ^{3} \times 6.00 \times 1 0 ^{- 1}}{4}$
$= 26.88 g m o l^{- 1}$

Q. A metal has an fcc lattice. The edge length of the unit cell is 400pm. The density of the metal is 2.8gcm−3. The molar mass of the metal is ____gmol−1. (Avogadro's constant =6.00×1023mol−1 )

Metal has fcc lattice, ∴Z=4 d=a3×NA​Z×M​ ∴2.8=(4.00×10−8)3×6.00×10234×M​ ∴M=42.8×(4.00)3×6.00×10−1​ =26.88gmol−1

Q. A metal has an fcc lattice. The edge length of the unit cell is $400 p m$ . The density of the metal is $2.8 g c m^{- 3}$ . The molar mass of the metal is ____ $g m o l^{- 1}$ .
(Avogadro's constant $= 6.00 \times 1 0^{23} m o l^{- 1}$ )

Metal has $f cc$ lattice,
$∴ Z = 4$
$d = \frac{Z \times M}{a ^{3} \times N _{A}}$
$∴ 2.8 = \frac{4 \times M}{( 4.00 \times 1 0 ^{- 8} ) ^{3} \times 6.00 \times 1 0 ^{23}}$
$∴ M = \frac{2.8 \times ( 4.00 ) ^{3} \times 6.00 \times 1 0 ^{- 1}}{4}$
$= 26.88 g m o l^{- 1}$