Chemistry, asked by jagtapadesh588, 5 months ago

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​

Answers

Answered by pankajrastogi321
1

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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