when iron crystallises it form BCC cell the unit cell edge length in 288 pm calculate the density of iron molar mass of iron is 55.88 g/mol
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Answer:
a = 286.65 pm
=286.65×10
−10
cm
Density (ϵ)=7.874 g cm
−3
At mass of Fe= 56.0u
Z=2(for body centered cubic unit cell)
Avogadro number (N
0
)=?
p=
a
3
×N
0
Z×M
7.874 gcm
−3
=
(286.65×10
−10
cm)
3
×N
0
2×(56.0gmol
−1
)
N
0
=
(286.65×10
−10
cm
3
)×(7.874gcm
−3
)
2×(56 gmol
−1
)
=6.022×10
23
mol
−1
Explanation:
a = 286.65 pm
=286.65×10
−10
cm
Density (ϵ)=7.874 g cm
−3
At mass of Fe= 56.0u
Z=2(for body centered cubic unit cell)
Avogadro number (N
0
)=?
p=
a
3
×N
0
Z×M
7.874 gcm
−3
=
(286.65×10
−10
cm)
3
×N
0
2×(56.0gmol
−1
)
N
0
=
(286.65×10
−10
cm
3
)×(7.874gcm
−3
)
2×(56 gmol
−1
)
=6.022×10
23
mol
−1
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