A face-centred cubic element (atomic mass 60) has a cell edge of 400 pm. What'is its density?

Question

A face-centred cubic element (atomic mass 60) has a cell edge of 400 pm. What'is its density? (A) 6.2 g cm-3 (B) 24.8 g cm-3 (C) 12.4 g cm-3 (D) 3.1 g cm-3

Tardigrade Admin · Accepted Answer

ρ = (Z × M/a3 × N0) = (40 ×60 textg textmol-1/(400 ×10-10 textcm )3 ×6.02 ×1023 textmol-1) = 6.2 textg textcm-3

Q. A face-centred cubic element (atomic mass 60) has a cell edge of 400 pm. What'is its density?

$6.2 g c m^{- 3}$

$24.8 g c m^{- 3}$

$12.4 g c m^{- 3}$

$3.1 g c m^{- 3}$

$ρ = \frac{Z \times M}{a ^{3} \times N _{0}}$
$= \frac{40 \times 60 g mol ^{- 1}}{( 400 \times 1 0 ^{- 10} cm ) ^{3} \times 6.02 \times 1 0 ^{23} mol ^{- 1}}$
$= 6.2 g cm^{- 3}$

Q. A face-centred cubic element (atomic mass 60) has a cell edge of 400 pm. What'is its density?

6.2gcm−3

24.8gcm−3

12.4gcm−3

3.1gcm−3

ρ=a3×N0​Z×M​ =(400×10−10cm)3×6.02×1023mol−140×60gmol−1​ =6.2gcm−3

$6.2 g c m^{- 3}$

$24.8 g c m^{- 3}$

$12.4 g c m^{- 3}$

$3.1 g c m^{- 3}$

$ρ = \frac{Z \times M}{a ^{3} \times N _{0}}$
$= \frac{40 \times 60 g mol ^{- 1}}{( 400 \times 1 0 ^{- 10} cm ) ^{3} \times 6.02 \times 1 0 ^{23} mol ^{- 1}}$
$= 6.2 g cm^{- 3}$